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1lgsujylg | maths | vector-algebra | scalar-and-vector-triple-product | <p>If four distinct points with position vectors $$\vec{a}, \vec{b}, \vec{c}$$ and $$\vec{d}$$ are coplanar, then $$[\vec{a} \,\,\vec{b} \,\,\vec{c}]$$ is equal to :</p> | [{"identifier": "A", "content": "$$[\\vec{d} \\,\\,\\,\\,\\,\\vec{b} \\,\\,\\,\\,\\,\\vec{a}]+[\\vec{a} \\,\\,\\,\\,\\,\\vec{c} \\,\\,\\,\\,\\,\\vec{d}]+[\\vec{d} \\,\\,\\,\\,\\,\\vec{b} \\,\\,\\,\\,\\,\\vec{c}]$$"}, {"identifier": "B", "content": "$$[\\vec{b} \\,\\,\\,\\,\\,\\vec{c} \\,\\,\\,\\,\\,\\vec{d}]+[\\vec{d} ... | ["D"] | null | $$
\begin{aligned}
& {[\vec{b}-\vec{a} \,\,\,\,\,\vec{c}-\vec{a} \,\,\,\,\,\vec{d}-\vec{a}]=0} \\\\
& (\vec{b}-\vec{a}) \cdot[(\vec{c}-\vec{a}) \times(\vec{d}-\vec{a})]=0 \\\\
& (\vec{b}-\vec{a}) \cdot(\vec{c} \times \vec{d}-\vec{c} \times \vec{a}-\vec{a} \times \vec{d})=0 \\\\
& {[\vec{b}\,\,\,\,\, \vec{c} \,\,\,\,\,\... | mcq | jee-main-2023-online-11th-april-evening-shift | 8,675 |
1lgyl6v9x | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let the vectors $$\vec{u}_{1}=\hat{i}+\hat{j}+a \hat{k}, \vec{u}_{2}=\hat{i}+b \hat{j}+\hat{k}$$ and $$\vec{u}_{3}=c \hat{i}+\hat{j}+\hat{k}$$ be coplanar. If the vectors $$\vec{v}_{1}=(a+b) \hat{i}+c \hat{j}+c \hat{k}, \vec{v}_{2}=a \hat{i}+(b+c) \hat{j}+a \hat{k}$$ and $$\vec{v}_{3}=b \hat{i}+b \hat{j}+(c+a) \hat{... | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "4"}] | ["A"] | null | Since, $\vec{u}_1, \vec{u}_2, \vec{u}_3$ are coplanar.
<br/><br/>So, $\left[\begin{array}{lll}\vec{u}_1 & \vec{u}_2 & \vec{u}_3\end{array}\right]=0$
<br/><br/>$$
\begin{aligned}
& \Rightarrow\left|\begin{array}{lll}
1 & 1 & a \\
1 & b & 1 \\
c & 1 & 1
\end{array}\right|=0 \\\\
& \Rightarrow 1(b-1)-1(1-c)+a(1-b c)=0 \\\... | mcq | jee-main-2023-online-8th-april-evening-shift | 8,676 |
1lh21ilhv | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let the position vectors of the points A, B, C and D be
$$5 \hat{i}+5 \hat{j}+2 \lambda \hat{k}, \hat{i}+2 \hat{j}+3 \hat{k},-2 \hat{i}+\lambda \hat{j}+4 \hat{k}$$ and $$-\hat{i}+5 \hat{j}+6 \hat{k}$$. Let the set $$S=\{\lambda \in \mathbb{R}$$ :
the points A, B, C and D are coplanar $$\}$$.
<br/><br/>Then $$\sum_\... | [{"identifier": "A", "content": "$$\\frac{37}{2}$$"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "41"}] | ["D"] | null | Given, position vectors of the points $A, B, C$ and $D$ be
<br/><br/>$5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}}, \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}},-2 \hat{\mathbf{i}}+\lambda \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$
<... | mcq | jee-main-2023-online-6th-april-morning-shift | 8,677 |
1lh2y5tet | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let the vectors $$\vec{a}, \vec{b}, \vec{c}$$ represent three coterminous edges of a parallelopiped of volume V. Then the volume of the parallelopiped, whose coterminous edges are represented by $$\vec{a}, \vec{b}+\vec{c}$$ and $$\vec{a}+2 \vec{b}+3 \vec{c}$$ is equal to :</p> | [{"identifier": "A", "content": "3 V"}, {"identifier": "B", "content": "2 V"}, {"identifier": "C", "content": "6 V"}, {"identifier": "D", "content": "V"}] | ["D"] | null | Given that the volume $V$ of the parallelepiped formed by the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ is represented by the scalar triple product $[\vec{a},\vec{b},\vec{c}]$, which is the determinant of the 3 x 3 matrix with vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ as its rows (or columns).
<br/><br/>When the v... | mcq | jee-main-2023-online-6th-april-evening-shift | 8,678 |
1lh2yfbhy | maths | vector-algebra | scalar-and-vector-triple-product | <p>The sum of all values of $$\alpha$$, for which the points whose position vectors are $$\hat{i}-2 \hat{j}+3 \hat{k}, 2 \hat{i}-3 \hat{j}+4 \hat{k},(\alpha+1) \hat{i}+2 \hat{k}$$ and $$9 \hat{i}+(\alpha-8) \hat{j}+6 \hat{k}$$ are coplanar, is equal to :</p> | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$$-$$2"}, {"identifier": "D", "content": "2"}] | ["D"] | null | Let $\overrightarrow{O A}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
<br/><br/>$$
\begin{aligned}
& \overrightarrow{O B}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\\\
& \overrightarrow{O C}=(a+1) \hat{\mathbf{i}}+2 \hat{\mathbf{k}}
\end{aligned}
$$
<br/><br/>and $ \overrightarrow{O D}=9 \ha... | mcq | jee-main-2023-online-6th-april-evening-shift | 8,679 |
lsbl7k96 | maths | vector-algebra | scalar-and-vector-triple-product | Let $\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+\hat{k}, $
<br/>$\overrightarrow{\mathrm{b}}=3(\hat{i}-\hat{j}+\hat{k})$.
<br/>Let $\overrightarrow{\mathrm{c}}$ be the vector such that $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}$ and $\vec{a} \cdot \vec{c}=3$.
<br/> T... | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "36"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "20"}] | ["C"] | null | <p>$$\begin{aligned}
& \vec{a} \cdot[(\vec{c} \times \vec{b})-\vec{b}-\vec{c}] \\
& \vec{a} \cdot(\vec{c} \times \vec{b})-\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c} \quad \text{..... (i)}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { given } \vec{a} \times \vec{c}=\vec{b} \\
& \Rightarrow(\vec{a} \times \vec{c}) ... | mcq | jee-main-2024-online-27th-january-morning-shift | 8,680 |
ntSo8ltNI3kPao4H | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If $$\left| {\overrightarrow a } \right| = 5,\left| {\overrightarrow b } \right| = 4,\left| {\overrightarrow c } \right| = 3$$ thus what will be the value of $$\left| {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right|,$$ given that $$\overrigh... | [{"identifier": "A", "content": "$$25$$"}, {"identifier": "B", "content": "$$50$$ "}, {"identifier": "C", "content": "$$-25$$"}, {"identifier": "D", "content": "$$-50$$"}] | ["A"] | null | We have, $$\overrightarrow a + \overrightarrow b + \overrightarrow c = \overrightarrow 0 $$
<br><br>$$ \Rightarrow {\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)^2} = 0$$
<br><br>$$ \Rightarrow {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left|... | mcq | aieee-2002 | 8,682 |
LrCwfitL3Reaie1v | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | $$\overrightarrow a \,,\overrightarrow b \,,\overrightarrow c $$ are $$3$$ vectors, such that <br/><br/>$$\overrightarrow a + \overrightarrow b + \overrightarrow c = 0$$ , $$\left| {\overrightarrow a } \right| = 1\,\,\,\left| {\overrightarrow b } \right| = 2,\,\,\,\left| {\overrightarrow c } \right| = 3,$$,
<br/><b... | [{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$$0$$"}, {"identifier": "C", "content": "$$-7$$ "}, {"identifier": "D", "content": "$$7$$"}] | ["C"] | null | $$\overrightarrow a + \overrightarrow b + \overrightarrow c = 0$$
<br><br>$$ \Rightarrow \left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right).\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) = 0$$
<br><br>$${\left| {\overrightarrow a } \right|^2} + {\left| {\ove... | mcq | aieee-2003 | 8,683 |
WBKKBa1dT8vFmppP | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | A particle acted on by constant forces $$4\widehat i + \widehat j - 3\widehat k$$ and $$3\widehat i + \widehat j - \widehat k$$ is displaced from the point $$\widehat i + 2\widehat j + 3\widehat k$$ to the point $$\,5\widehat i + 4\widehat j + \widehat k.$$ The total work done by the forces is : | [{"identifier": "A", "content": "$$50$$ units "}, {"identifier": "B", "content": "$$20$$ units "}, {"identifier": "C", "content": "$$30$$ units "}, {"identifier": "D", "content": "$$40$$ units "}] | ["D"] | null | The work done by a force on a particle is given by the dot product of the force and the displacement vector of the particle. The displacement vector can be found by subtracting the initial position from the final position:
<br/><br/>
$$\mathbf{displacement} = \mathbf{final\ position} - \mathbf{initial\ position} = (5\w... | mcq | aieee-2004 | 8,684 |
R1ZCf7lzVetrtUe4 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | The values of a, for which the points $$A, B, C$$ with position vectors $$2\widehat i - \widehat j + \widehat k,\,\,\widehat i - 3\widehat j - 5\widehat k$$ and $$a\widehat i - 3\widehat j + \widehat k$$ respectively are the vertices of a right angled triangle with $$C = {\pi \over 2}$$ are : | [{"identifier": "A", "content": "$$2$$ and $$1$$ "}, {"identifier": "B", "content": "$$-2$$ and $$-1$$ "}, {"identifier": "C", "content": "$$-2$$ and $$1$$ "}, {"identifier": "D", "content": "$$2$$ and $$-1$$ "}] | ["A"] | null | $$\overrightarrow {CA} = \left( {2 - a} \right)\widehat i + 2\widehat j;$$
<br><br>$$\overrightarrow {CB} = \left( {1 - a} \right)\widehat i - 6\widehat k$$
<br><br>$$\overrightarrow {CA} .\overrightarrow {CB} = 0$$
<br><br>$$\,\,\,\,\,\,\,\, \Rightarrow \left( {2 - a} \right)\left( {1 - a} \right) = 0$$
<br><br>$$ ... | mcq | aieee-2006 | 8,686 |
DRDbNWOhlsTgTTBc | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If the vectors $$\overrightarrow a = \widehat i - \widehat j + 2\widehat k,\,\,\,\,\,\overrightarrow b = 2\widehat i + 4\widehat j + \widehat k\,\,\,$$ and $$\,\overrightarrow c = \lambda \widehat i + \widehat j + \mu \widehat k$$ are mutually orthogonal, then $$\,\left( {\lambda ,\mu } \right)$$ is equal to : | [{"identifier": "A", "content": "$$(2, -3)$$"}, {"identifier": "B", "content": "$$(-2, 3)$$"}, {"identifier": "C", "content": "$$(3, -2)$$"}, {"identifier": "D", "content": "$$(-3, 2)$$"}] | ["D"] | null | Since, $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ are mutually orthogonal
<br><br> $$\overrightarrow a .\overrightarrow b = 0,\,\,\overrightarrow b .\overrightarrow c = 0,\,\,\overrightarrow c .\overrightarrow a = 0$$
<br><br>$$ \Rightarrow 2\lambda + 4 + \mu = 0\,\,\,\,\,\,\,\,\,\,\,...... | mcq | aieee-2010 | 8,687 |
LYE5VbPrgMavFhGX | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$ and $$\overrightarrow b $$ be two unit vectors. If the vectors $$\,\overrightarrow c = \widehat a + 2\widehat b$$ and $$\overrightarrow d = 5\widehat a - 4\widehat b$$ are perpendicular to each other, then the angle between $$\overrightarrow a $$ and $$\overrightarrow b $$ is : | [{"identifier": "A", "content": "$${\\pi \\over 6}$$ "}, {"identifier": "B", "content": "$${\\pi \\over 2}$$"}, {"identifier": "C", "content": "$${\\pi \\over 3}$$"}, {"identifier": "D", "content": "$${\\pi \\over 4}$$"}] | ["C"] | null | Let $$\overrightarrow c = \widehat a + 2\widehat b$$ and $$\overrightarrow d = 5\widehat a - 4\widehat b$$
<br><br>Since $$\overrightarrow c $$ and $$\overrightarrow d $$ are perpendicular to each other
<br><br>$$\therefore$$ $$\overrightarrow c .\overrightarrow d = 0 \Rightarrow \left( {\widehat a + 2\widehat b} \... | mcq | aieee-2012 | 8,688 |
J6fh5bJMpbtUOm1J | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$ABCD$$ be a parallelogram such that $$\overrightarrow {AB} = \overrightarrow q ,\overrightarrow {AD} = \overrightarrow p $$ and $$\angle BAD$$ be an acute angle. If $$\overrightarrow r $$ is the vector that coincide with the altitude directed from the vertex $$B$$ to the side $$AD,$$ then $$\overrightarrow r $... | [{"identifier": "A", "content": "$$\\overrightarrow r = 3\\overrightarrow q - {{3\\left( {\\overrightarrow p .\\overrightarrow q } \\right)} \\over {\\left( {\\overrightarrow p .\\overrightarrow p } \\right)}}\\overrightarrow p $$ "}, {"identifier": "B", "content": "$$\\overrightarrow r = - \\overrightarrow q + {{... | ["B"] | null | Let $$ABCD$$ be a parallelogram such that
<br><br>$$\overrightarrow {AB} = \overrightarrow q ,\overrightarrow {AD} = \overrightarrow p $$ and $$\angle BAD$$ be an acute angle.
<br><br>We have
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263647/exam_images/vxui8byefbr... | mcq | aieee-2012 | 8,689 |
MyLIse0cRI3zW4o3T2Zji | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively 3$$\widehat i$$ + $$\widehat j$$ $$-$$ $$\widehat k$$, $$-$$$$\widehat i$$ + 3$$\widehat j$$ + p$$\widehat k$$ and 5$$\widehat i$$ + q$$\widehat j$$ $$-$$ 4$$\widehat k$$, then the point (p, q) lies
on a line : | [{"identifier": "A", "content": "parallel to x-axis. "}, {"identifier": "B", "content": "parallel to y-axis."}, {"identifier": "C", "content": "making an acute angle with the positive direction of x-axis."}, {"identifier": "D", "content": "making an obtuse angle with the positive direction of x-axis. "}] | ["C"] | null | Given,
<br><br>$$\overrightarrow A = 3\widehat i + \widehat j - \widehat k$$
<br><br>$$\overrightarrow B = - \widehat i + 3\widehat j - p\widehat k$$
<br><br>$$\overrightarrow C = 5\widehat i + 9\widehat j - 4\widehat k$$
<br><br>$$ \therefore $$ $$\overrightarrow {AB} = - 4\widehat i + 2\wideha... | mcq | jee-main-2016-online-9th-april-morning-slot | 8,690 |
CknOj9CXeWNGkcOX | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow u $$ be a vector coplanar with the vectors $$\overrightarrow a = 2\widehat i + 3\widehat j - \widehat k$$ and $$\overrightarrow b = \widehat j + \widehat k$$. If $$\overrightarrow u $$ is perpendicular to $$\overrightarrow a $$ and $$\overrightarrow u .\overrightarrow b = 24$$, then $${\left| {... | [{"identifier": "A", "content": "336"}, {"identifier": "B", "content": "315"}, {"identifier": "C", "content": "256"}, {"identifier": "D", "content": "84"}] | ["A"] | null | You should know that, when $$\overrightarrow u $$ is coplanar with $$\overrightarrow a $$ and $$\overrightarrow b $$ then we can write $$\overrightarrow u = x\overrightarrow a + y\overrightarrow b $$
<br><br>Here, $$\overrightarrow u $$ is perpendicular with $$\overrightarrow a $$ then,
<br><br>$$\overrightarrow u... | mcq | jee-main-2018-offline | 8,691 |
U1UOPS1llBEOw9OxMNW7E | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\sqrt 3 \widehat i + \widehat j,$$ $$\widehat i + \sqrt 3 \widehat j$$ and $$\beta \widehat i + \left( {1 - \beta } \right)\widehat j$$ respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is ... | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}] | ["B"] | null | Angle bisector is x $$-$$ y = 0
<br><br>$$ \Rightarrow $$ $${{\left| {\beta - \left( {1 - \beta } \right)} \right|} \over {\sqrt 2 }} = {3 \over {\sqrt 2 }}$$
<br><br>$$ \Rightarrow $$ $$\left| {2\beta - 1} \right| = 3$$
<br><br>$$ \Rightarrow $$ $$\beta $$ = 2 or $$-$$ 1 | mcq | jee-main-2019-online-11th-january-evening-slot | 8,692 |
wXePTqL6zttwwsSMnl3rsa0w2w9jwy0f6ft | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let A (3, 0, –1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G
divides BM in the ratio, 2 : 1, then cos ($$\angle $$GOA) (O being the origin) is equal to : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt {15} }}$$"}, {"identifier": "B", "content": "$${1 \\over {6\\sqrt {10} }}$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt {30} }}$$"}, {"identifier": "D", "content": "$${1 \\over {2\\sqrt {15} }}$$"}] | ["A"] | null | G is the centroid of $$\Delta $$ABC<br><br>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266720/exam_images/loddnxhki4jx9azlbmpr.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265343/exam_images/jmidgr... | mcq | jee-main-2019-online-10th-april-morning-slot | 8,693 |
5UABRSZyRW07gN53tq18hoxe66ijvww95a3 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If a unit vector $$\overrightarrow a $$ makes angles $$\pi $$/3 with $$\widehat i$$ , $$\pi $$/ 4
with $$\widehat j$$ and $$\theta $$$$ \in $$(0, $$\pi $$) with $$\widehat k$$, then a value of $$\theta $$
is :- | [{"identifier": "A", "content": "$${{5\\pi } \\over {6}}$$"}, {"identifier": "B", "content": "$${{5\\pi } \\over {12}}$$"}, {"identifier": "C", "content": "$${{2\\pi } \\over {3}}$$"}, {"identifier": "D", "content": "$${{\\pi } \\over {4}}$$"}] | ["C"] | null | A unit vector $$\overrightarrow a $$ makes angles $$\pi $$/3 with $$\widehat i$$
<br><br>$$ \therefore $$ $$\alpha $$ = $$\pi $$/3
<br><br> and $$\pi $$/ 4
with $$\widehat j$$
<br><br>$$ \therefore $$ $$\beta $$ = $$\pi $$/ 4
<br><br>and $$\theta $$$$ \in $$(0, $$\pi $$) with $$\widehat k$$
<br><br>$$ \therefore $$ $$... | mcq | jee-main-2019-online-9th-april-evening-slot | 8,694 |
6C4xzfmrghEfZNGqxuFTE | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = 2\widehat i + {\lambda _1}\widehat j + 3\widehat k,\,\,$$ $$\overrightarrow b = 4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k,$$ and $$\overrightarrow c = 3\widehat i + 6\widehat j + \left( {{\lambda _3} - 1} \right)\widehat k$$ be three vectors such that $$\o... | [{"identifier": "A", "content": "(1, 5, 1)"}, {"identifier": "B", "content": "(1, 3, 1)"}, {"identifier": "C", "content": "$$\\left( { - {1 \\over 2},4,0} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 2},4, - 2} \\right)$$"}] | ["C"] | null | Given $$\overrightarrow b = 2\overrightarrow a $$
<br><br>$$ \therefore $$ $$4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k = 4\widehat i + 2{\lambda _1}\widehat j + 6\widehat k$$
<br><br>$$ \Rightarrow 3 - {\lambda _2} = 2{\lambda _1} \Rightarrow 2{\lambda _1} + {\lambda _2} = 3\,\,...(1)$$
<... | mcq | jee-main-2019-online-10th-january-morning-slot | 8,695 |
fzUXOCvAXq7qZfRqzTtn7 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i + \widehat j + \sqrt 2 \widehat k,$$ $$\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + \sqrt 2 \widehat k$$, $$\overrightarrow c = 5\widehat i + \widehat j + \sqrt 2 \widehat k$$ be three vectors such that the projection vector of $$\overrightarrow b $$ on $$\o... | [{"identifier": "A", "content": "$$\\sqrt {32} $$"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "$$\\sqrt {22} $$"}, {"identifier": "D", "content": "4"}] | ["B"] | null | Projection of $$\overrightarrow b $$ on $$\overrightarrow a $$ is $$\overrightarrow a $$
<br><br>$$ \therefore $$ $${{\overrightarrow b \cdot \overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = \left| {\overrightarrow a } \right|$$
<br><br>$$ \Rightarrow $$ $${{{b_1} + {b_2} ... | mcq | jee-main-2019-online-9th-january-evening-slot | 8,696 |
gBxihkaJRK9XmHPQAQ7k9k2k5e2n780 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | A vector $$\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k\left( {\alpha ,\beta \in R} \right)$$ lies in the plane of the vectors, $$\overrightarrow b = \widehat i + \widehat j$$ and $$\overrightarrow c = \widehat i - \widehat j + 4\widehat k$$. If $$\overrightarrow a $$ bisects the angle be... | [{"identifier": "A", "content": "$$\\overrightarrow a .\\widehat i + 3 = 0$$"}, {"identifier": "B", "content": "$$\\overrightarrow a .\\widehat k - 4 = 0$$"}, {"identifier": "C", "content": "$$\\overrightarrow a .\\widehat i + 1 = 0$$"}, {"identifier": "D", "content": "$$\\overrightarrow a .\\widehat k + 2 = 0$$"}] | ["B"] | null | Angle bisector $$\overrightarrow a = \lambda \left( {\widehat b + \widehat c} \right)$$
<br><br>= $$\lambda \left( {{{\widehat i + \widehat j} \over {\sqrt 2 }} + {{\widehat i - \widehat j + 4\widehat k} \over {3\sqrt 2 }}} \right)$$
<br><br>$$ \Rightarrow $$ $$\overrightarrow a = {\lambda \over {3\sqrt 2 }}\left( {... | mcq | jee-main-2020-online-7th-january-morning-slot | 8,697 |
wxnynWl5nmUW9g7btjjgy2xukewt1roa | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be three unit vectors such that
<br/>$${\left| {\overrightarrow a - \overrightarrow b } \right|^2}$$ + $${\left| {\overrightarrow a - \overrightarrow c } \right|^2}$$ = 8.
<br/><br/>Then $${\left| {\overrightarrow a + 2\overrightarrow b } ... | [] | null | 2 | Given, $$\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| = 1$$
<br><br>$${\left| {\overrightarrow a - \overrightarrow b } \right|^2}$$ + $${\left| {\overrightarrow a - \overrightarrow c } \right|^2}$$ = 8
<br><br>$$ \Rightarrow $$ $${\left| {\overrightar... | integer | jee-main-2020-online-2nd-september-morning-slot | 8,698 |
qU27zRsydVCHN6lGyOjgy2xukf3zyzqz | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let a, b c $$ \in $$ R be such that a<sup>2</sup>
+ b<sup>2</sup>
+ c<sup>2</sup>
= 1. If <br/>$$a\cos \theta = b\cos \left( {\theta + {{2\pi } \over 3}} \right) = c\cos \left( {\theta + {{4\pi } \over 3}} \right)$$,
<br/>where
$${\theta = {\pi \over 9}}$$, then the angle between the vectors
$$a\widehat i + b\w... | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$${{\\pi \\over 9}}$$"}, {"identifier": "C", "content": "$${{{2\\pi } \\over 3}}$$"}, {"identifier": "D", "content": "$${{\\pi \\over 2}}$$"}] | ["D"] | null | Let, $$\overrightarrow {{a_1}} = a\widehat i + b\widehat j + c\widehat k$$<br><br>and $$\overrightarrow {{a_2}} = b\widehat i + c\widehat j + a\widehat k$$<br><br>We know, Angle between two vectors<br><br>$$\cos \alpha = {{\overrightarrow {{a_1}} \,.\,\overrightarrow {{a_2}} } \over {|\overrightarrow {{a_1}} \,|.|\,... | mcq | jee-main-2020-online-3rd-september-evening-slot | 8,699 |
oBUoyZ4YjQ6r5Z8sqVjgy2xukg4n5m60 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If $$\overrightarrow x $$ and $$\overrightarrow y $$ be two non-zero vectors such that
$$\left| {\overrightarrow x + \overrightarrow y } \right| = \left| {\overrightarrow x } \right|$$ and $${2\overrightarrow x + \lambda \overrightarrow y }$$ is perpendicular to $${\overrightarrow y }$$,
then the value of $$\lambda $... | [] | null | 1 | $$\left| {\overrightarrow x + \overrightarrow y } \right| = \left| {\overrightarrow x } \right|$$
<br>Squaring both sides we get
<br><br>$${\left| {\overrightarrow x } \right|^2} + 2\overrightarrow x .\overrightarrow y + {\left| {\overrightarrow y } \right|^2} = {\left| {\overrightarrow x } \right|^2}$$
<br><br>$$ \R... | integer | jee-main-2020-online-6th-september-evening-slot | 8,701 |
hnJwMJTUDbFeYVoTKR1kmknwag0 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow x $$ be a vector in the plane containing vectors $$\overrightarrow a = 2\widehat i - \widehat j + \widehat k$$ and $$\overrightarrow b = \widehat i + 2\widehat j - \widehat k$$. If the vector $$\overrightarrow x $$ is perpendicular to $$\left( {3\widehat i + 2\widehat j - \widehat k} \right)$$ a... | [] | null | 486 | Let, $$\overrightarrow x = k(\overrightarrow a + \lambda \overrightarrow b )$$<br><br>$$\overrightarrow x$$ is perpendicular to $$3\widehat i + 2\widehat j - \widehat k$$<br><br><b>I.</b> k{(2 + $$\lambda$$)3 + (2$$\lambda$$ $$-$$ 1)2 + (1 $$-$$ $$\lambda$$)($$-$$1) = 0<br><br>$$ \Rightarrow $$ 8$$\lambda$$ + 3 = 0<... | integer | jee-main-2021-online-17th-march-evening-shift | 8,702 |
uSK38CUUVnLZXt4bdl1kmm3d8s3 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | In a triangle ABC, if $$|\overrightarrow {BC} | = 8,|\overrightarrow {CA} | = 7,|\overrightarrow {AB} | = 10$$, then the projection of the vector $$\overrightarrow {AB} $$ on $$\overrightarrow {AC} $$ is equal to : | [{"identifier": "A", "content": "$${{25} \\over 4}$$"}, {"identifier": "B", "content": "$${{127} \\over 20}$$"}, {"identifier": "C", "content": "$${{85} \\over 14}$$"}, {"identifier": "D", "content": "$${{115} \\over 16}$$"}] | ["C"] | null | <picture><source media="(max-width: 1728px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266979/exam_images/tvnba426ygdio3l2ckwo.webp"><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264573/exam_images/vlfqrjsuvhpnf9m6eq5m.webp"><source media="(max-wi... | mcq | jee-main-2021-online-18th-march-evening-shift | 8,703 |
1krrv1lee | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | In a triangle ABC, if $$\left| {\overrightarrow {BC} } \right| = 3$$, $$\left| {\overrightarrow {CA} } \right| = 5$$ and $$\left| {\overrightarrow {BA} } \right| = 7$$, then the projection of the vector $$\overrightarrow {BA} $$ on $$\overrightarrow {BC} $$ is equal to : | [{"identifier": "A", "content": "$${{19} \\over 2}$$"}, {"identifier": "B", "content": "$${{13} \\over 2}$$"}, {"identifier": "C", "content": "$${{11} \\over 2}$$"}, {"identifier": "D", "content": "$${{15} \\over 2}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264070/exam_images/ep6a24u2iigywhjyduzo.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 20th July Evening Shift Mathematics - Vector Algebra Question 131 English Explanation"> <br><br>Pr... | mcq | jee-main-2021-online-20th-july-evening-shift | 8,705 |
1krrw91t0 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | For p > 0, a vector $${\overrightarrow v _2} = 2\widehat i + (p + 1)\widehat j$$ is obtained by rotating the vector $${\overrightarrow v _1} = \sqrt 3 p\widehat i + \widehat j$$ by an angle $$\theta$$ about origin in counter clockwise direction. If $$\tan \theta = {{\left( {\alpha \sqrt 3 - 2} \right)} \over {\lef... | [] | null | 6 | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265929/exam_images/pfsd2oapdy3p6fqplgh1.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264150/exam_images/yosxjcnh7ins9tydvyvk.webp"><img src="https://res.c... | integer | jee-main-2021-online-20th-july-evening-shift | 8,706 |
1krzrafms | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If $$\left( {\overrightarrow a + 3\overrightarrow b } \right)$$ is perpendicular to $$\left( {7\overrightarrow a - 5\overrightarrow b } \right)$$ and $$\left( {\overrightarrow a - 4\overrightarrow b } \right)$$ is perpendicular to $$\left( {7\overrightarrow a - 2\overrightarrow b } \right)$$, then the angle between... | [] | null | 60 | $$\left( {\overrightarrow a + 3\overrightarrow b } \right) \bot \left( {7\overrightarrow a - 5\overrightarrow b } \right)$$<br><br>$$ \therefore $$ $$\left( {\overrightarrow a + 3\overrightarrow b } \right)\,.\,\left( {7\overrightarrow a - 5\overrightarrow b } \right) = 0$$<br><br>$$ \Rightarrow $$ $$7{\left| {\ove... | integer | jee-main-2021-online-25th-july-evening-shift | 8,707 |
1ktd1se61 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | A hall has a square floor of dimension 10 m $$\times$$ 10 m (see the figure) and vertical walls. If the angle GPH between the diagonals AG and BH is $${\cos ^{ - 1}}{1 \over 5}$$, then the height of the hall (in meters) is :<br/><br/><img src="data:image/png;base64,UklGRkQOAABXRUJQVlA4IDgOAADwVwCdASocARIBPm00l0ekIyKhJX... | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "2$$\\sqrt {10} $$"}, {"identifier": "C", "content": "5$$\\sqrt {3} $$"}, {"identifier": "D", "content": "5$$\\sqrt {2} $$"}] | ["D"] | null | $$A(\widehat j)\,.\,B(10\widehat i)$$<br><br>$$H(h\widehat j + 10\widehat k)$$<br><br>$$G(10\widehat i + h\widehat j + 10\widehat k)$$<br><br>$$\overrightarrow {AG} = 10\widehat i + h\widehat j + 10\widehat k$$<br><br>$$\overrightarrow {BH} = - 10\widehat i + h\widehat j + 10\widehat k$$<br><br>$$\cos \theta = {{\o... | mcq | jee-main-2021-online-26th-august-evening-shift | 8,708 |
1ktd3i9x4 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If the projection of the vector $$\widehat i + 2\widehat j + \widehat k$$ on the sum of the two vectors $$2\widehat i + 4\widehat j - 5\widehat k$$ and $$ - \lambda \widehat i + 2\widehat j + 3\widehat k$$ is 1, then $$\lambda$$ is equal to __________. | [] | null | 5 | $$\overrightarrow a = \widehat i + 2\widehat j + \widehat k$$<br><br>$$\overrightarrow b = (2 - \lambda )\widehat i + 6\widehat j - 2\widehat k$$<br><br>$${{\overrightarrow a \,.\,\overrightarrow b } \over {|\overrightarrow b |}} = 1,\overrightarrow a \,.\,\overrightarrow b = 12 - \lambda $$<br><br>$$\left( {\overri... | integer | jee-main-2021-online-26th-august-evening-shift | 8,709 |
1ktip5iva | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$ and $$\overrightarrow b $$ be two vectors <br/>such that $$\left| {2\overrightarrow a + 3\overrightarrow b } \right| = \left| {3\overrightarrow a + \overrightarrow b } \right|$$ and the angle between $$\overrightarrow a $$ and $$\overrightarrow b $$ is 60$$^\circ$$. If $${1 \over 8}\overrig... | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "8"}] | ["C"] | null | $${\left| {3\overrightarrow a + \overrightarrow b } \right|^2} = {\left| {2\overrightarrow a + 3\overrightarrow b } \right|^2}$$<br><br>$$\left( {3\overrightarrow a + \overrightarrow b } \right).\left( {3\overrightarrow a + \overrightarrow b } \right) = \left( {2\overrightarrow a + 3\overrightarrow b } \right).\le... | mcq | jee-main-2021-online-31st-august-morning-shift | 8,710 |
1l6nnts5x | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>Let S be the set of all a $$\in R$$ for which the angle between the vectors $$
\vec{u}=a\left(\log _{e} b\right) \hat{i}-6 \hat{j}+3 \hat{k}$$ and $$\vec{v}=\left(\log _{e} b\right) \hat{i}+2 \hat{j}+2 a\left(\log _{e} b\right) \hat{k}$$, $$(b>1)$$ is acute. Then S is equal to :</p> | [{"identifier": "A", "content": "$$\\left(-\\infty,-\\frac{4}{3}\\right)$$"}, {"identifier": "B", "content": "$$\\Phi $$"}, {"identifier": "C", "content": "$$\\left(-\\frac{4}{3}, 0\\right)$$"}, {"identifier": "D", "content": "$$\\left(\\frac{12}{7}, \\infty\\right)$$"}] | ["B"] | null | <p>$$\overrightarrow u = a({\log _e}b)\widehat i - 6\widehat j + 3\widehat k$$</p>
<p>$$\overrightarrow v = ({\log _e}b)\widehat i + 2\widehat j + 2a({\log _e}b)\widehat k$$</p>
<p>For acute angle $$\overrightarrow u \,.\,\overrightarrow v > 0$$</p>
<p>$$ \Rightarrow a{({\log _e}b)^2} - 12 + 6a({\log _e}b) > 0$$</p>... | mcq | jee-main-2022-online-28th-july-evening-shift | 8,712 |
1ldo522uv | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}$$ and $$\vec{b}=\hat{i}+3 \hat{j}+5 \hat{k}$$ be two vectors. Then which one of the following statements is TRUE ?</p> | [{"identifier": "A", "content": "Projection of $$\\vec{a}$$ on $$\\vec{b}$$ is $$\\frac{-13}{\\sqrt{35}}$$ and the direction of the projection vector is opposite to the direction \nof $$\\vec{b}$$."}, {"identifier": "B", "content": "Projection of $$\\vec{a}$$ on $$\\vec{b}$$ is $$\\frac{13}{\\sqrt{35}}$$ and the direct... | ["A"] | null | $\begin{aligned} & \text { Projection of }\vec{a} \text { on } \vec{b} =\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \\\\ & = \frac{(5 \hat{i}-\hat{j}-3 \hat{k}) \cdot(\hat{i}+3 \hat{j}+5 \hat{k})}{\sqrt{1^2+3^2+5^2}}=\frac{5-3-15}{\sqrt{35}} \\\\ & = \frac{-13}{\sqrt{35}}\end{aligned}$
<br/><br/>Negative sign indicates tha... | mcq | jee-main-2023-online-1st-february-evening-shift | 8,713 |
1ldo6n5a3 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=2 \hat{i}-7 \hat{j}+5 \hat{k}, \vec{b}=\hat{i}+\hat{k}$$ and $$\vec{c}=\hat{i}+2 \hat{j}-3 \hat{k}$$ be three given vectors. If $$\overrightarrow{\mathrm{r}}$$ is a vector such that $$\vec{r} \times \vec{a}=\vec{c} \times \vec{a}$$ and $$\vec{r} \cdot \vec{b}=0$$, then $$|\vec{r}|$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{11}{7}$$"}, {"identifier": "B", "content": "$$\\frac{11}{5} \\sqrt{2}$$"}, {"identifier": "C", "content": "$$\\frac{\\sqrt{914}}{7}$$"}, {"identifier": "D", "content": "$$\\frac{11}{7} \\sqrt{2}$$"}] | ["D"] | null | $\begin{aligned} & \vec{r} \times \vec{a}=\vec{c} \times \vec{a} \\\\ & \Rightarrow(\vec{r}-\vec{c}) \times \vec{a}=0 \Rightarrow \vec{r}-\vec{c}=\lambda \vec{a}((\vec{r}-\vec{c} ) \text{and} \overrightarrow{a} \text { are parallel }) \\\\ & \Rightarrow \vec{r}=\vec{c}+\lambda \vec{a} \\\\ & \Rightarrow \vec{r} \cdot ... | mcq | jee-main-2023-online-1st-february-evening-shift | 8,714 |
1ldpsrr3l | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=2 \hat{i}+\hat{j}+\hat{k}$$, and $$\vec{b}$$ and $$\vec{c}$$ be two nonzero vectors such that $$|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$$ and $$\vec{b} \cdot \vec{c}=0$$. Consider the following two statements:</p>
<p>(A) $$|\vec{a}+\lambda \vec{c}| \geq|\vec{a}|$$ for all $$\lambda \in \mat... | [{"identifier": "A", "content": "only (B) is correct"}, {"identifier": "B", "content": "both (A) and (B) are correct"}, {"identifier": "C", "content": "only (A) is correct"}, {"identifier": "D", "content": "neither (A) nor (B) is correct"}] | ["C"] | null | $|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$
<br/><br/>$$ \Rightarrow $$ $|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}+\vec{b}-\vec{c}|^{2}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow |\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \vec{a}) \\\\
& =|\vec{a}|^{2}+|\vec{... | mcq | jee-main-2023-online-31st-january-morning-shift | 8,715 |
1ldv14lsr | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>The vector $$\overrightarrow a = - \widehat i + 2\widehat j + \widehat k$$ is rotated through a right angle, passing through the y-axis in its way and the resulting vector is $$\overrightarrow b $$. Then the projection of $$3\overrightarrow a + \sqrt 2 \overrightarrow b $$ on $$\overrightarrow c = 5\widehat i + ... | [{"identifier": "A", "content": "$$\\sqrt6$$"}, {"identifier": "B", "content": "2$$\\sqrt3$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "3$$\\sqrt2$$"}] | ["D"] | null | <p>First, we write $\overrightarrow{b}$ as a linear combination of $\overrightarrow{a}$ and $\overrightarrow{j}$ since $\overrightarrow{b}$ is a rotation of $\overrightarrow{a}$ about the y-axis.</p>
<p>$\vec{b}=\lambda \vec{a}+\mu \hat{j}=\lambda(-\hat{i}+2 \hat{j}+\hat{k})+\mu \hat{j}=-\lambda \hat{i}+(2 \lambda+\mu ... | mcq | jee-main-2023-online-25th-january-morning-shift | 8,717 |
lsblj5uf | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | The least positive integral value of $\alpha$, for which the angle between the vectors $\alpha \hat{i}-2 \hat{j}+2 \hat{k}$ and $\alpha \hat{i}+2 \alpha \hat{j}-2 \hat{k}$ is acute, is ___________. | [] | null | 5 | <p>$$\begin{aligned}
& \cos \theta=\frac{(\alpha \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(\alpha \hat{\mathrm{i}}+2 \alpha \hat{\mathrm{j}}-2 \hat{\mathrm{k}})}{\sqrt{\alpha^2+4+4} \sqrt{\alpha^2+4 \alpha^2+4}} \\
& \cos \theta=\frac{\alpha^2-4 \alpha-4}{\sqrt{\alpha^2+8} \sqrt{5 \alpha^2+4}} \\
& ... | integer | jee-main-2024-online-27th-january-morning-shift | 8,719 |
jaoe38c1lsfkf8y5 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>Let a unit vector $$\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}$$ make angles $$\frac{\pi}{2}, \frac{\pi}{3}$$ and $$\frac{2 \pi}{3}$$ with the vectors $$\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}, \frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$$ and $$\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j... | [{"identifier": "A", "content": "$$\\frac{11}{2}$$\n"}, {"identifier": "B", "content": "$$\\frac{5}{2}$$"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "9"}] | ["B"] | null | <p>Unit vector $$\hat{\mathrm{u}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$$</p>
<p>$$\begin{aligned}
& \overrightarrow{\mathrm{p}}_1=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}, \overrightarrow{\mathrm{p}}_2=\frac{1}{\sqrt{2}} \hat{\mathrm{j}}+\fra... | mcq | jee-main-2024-online-29th-january-evening-shift | 8,720 |
lv0vxcgi | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>Let a unit vector which makes an angle of $$60^{\circ}$$ with $$2 \hat{i}+2 \hat{j}-\hat{k}$$ and an angle of $$45^{\circ}$$ with $$\hat{i}-\hat{k}$$ be $$\vec{C}$$. Then $$\vec{C}+\left(-\frac{1}{2} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{\sqrt{2}}{3} \hat{k}\right)$$ is:</p> | [{"identifier": "A", "content": "$$-\\frac{\\sqrt{2}}{3} \\hat{i}+\\frac{\\sqrt{2}}{3} \\hat{j}+\\left(\\frac{1}{2}+\\frac{2 \\sqrt{2}}{3}\\right) \\hat{k}$$\n"}, {"identifier": "B", "content": "$$\\left(\\frac{1}{\\sqrt{3}}+\\frac{1}{2}\\right) \\hat{i}+\\left(\\frac{1}{\\sqrt{3}}-\\frac{1}{3 \\sqrt{2}}\\right) \\hat{... | ["C"] | null | <p>$$\begin{aligned}
& \text { Let } \vec{C}=a \hat{i}+b \hat{j}+c \hat{k} \\
& (a \hat{i}+b \hat{j}+c \hat{k}) \cdot(2 \hat{i}+2 \hat{j}-\hat{k})=1 \times 3 \times \frac{1}{2} \\
& 2 a+2 b-c=\frac{3}{2} \qquad \text{... (1)}\\
& (a \hat{i}+b \hat{j}+c \hat{k}) \cdot(\hat{i}-\hat{k})=1 \times \sqrt{2} \times \frac{1}{\... | mcq | jee-main-2024-online-4th-april-morning-shift | 8,721 |
lv2erzmn | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>For $$\lambda>0$$, let $$\theta$$ be the angle between the vectors $$\vec{a}=\hat{i}+\lambda \hat{j}-3 \hat{k}$$ and $$\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$$. If the vectors $$\vec{a}+\vec{b}$$ and $$\vec{a}-\vec{b}$$ are mutually perpendicular, then the value of (14 cos $$\theta)^2$$ is equal to</p> | [{"identifier": "A", "content": "25"}, {"identifier": "B", "content": "50"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "40"}] | ["A"] | null | <p>$$\begin{aligned}
& \text { Given } \vec{a}=\hat{i}+\lambda \hat{j}-3 \hat{k} \\
& \vec{b}=3 \hat{i}-\hat{j}+2 \hat{k} \\
& \vec{a}+\vec{b}=4 \hat{i}+(\lambda-1) \hat{j}-\hat{k} \\
& \vec{a}-\vec{b}=-2 \hat{i}+(\lambda+1) \hat{j}-5 \hat{k} \\
& (\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=0 \\
& -8+\lambda^2-1+5=0
\end{... | mcq | jee-main-2024-online-4th-april-evening-shift | 8,722 |
lv3veez6 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}+3 \hat{j}-5 \hat{k}$$ and $$\overrightarrow{\mathrm{c}}=3 \hat{i}-\hat{j}+\lambda \hat{k}$$ be three vectors. Let $$\overrightarrow{\mathrm{r}}$$ be a unit vector along $$\vec{b}+\vec{c}$$. If $$\vec{r} \cdot \vec{a}... | [{"identifier": "A", "content": "21"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "27"}, {"identifier": "D", "content": "30"}] | ["B"] | null | <p>$$\begin{aligned}
& \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \\
& \vec{b}=2 \hat{i}+3 \hat{j}-5 \hat{k} \\
& \vec{c}=3 \hat{i}-\hat{j}+\lambda \hat{k} \\
& \vec{b}+\vec{c}=5 \hat{i}+2 \hat{j}+(\lambda-5) \hat{k}
\end{aligned}$$</p>
<p>$$\vec{r}$$ is a unit vector along $$\vec{b}+\vec{c}$$</p>
<p>$$\therefore \quad \vec{r... | mcq | jee-main-2024-online-8th-april-evening-shift | 8,723 |
lv5grw8k | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>The set of all $$\alpha$$, for which the vectors $$\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}$$ and $$\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}$$ are inclined at an obtuse angle for all $$t \in \mathbb{R}$$, is</p> | [{"identifier": "A", "content": "$$[0,1)$$\n"}, {"identifier": "B", "content": "$$\\left(-\\frac{4}{3}, 0\\right]$$\n"}, {"identifier": "C", "content": "$$(-2,0]$$\n"}, {"identifier": "D", "content": "$$\\left(-\\frac{4}{3}, 1\\right)$$"}] | ["B"] | null | <p>Given $$\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}$$</p>
<p>and $$\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}$$</p>
<p>angle between $$\vec{a}$$ and $$\vec{b}$$ is given by</p>
<p>$$\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$$</p>
<p>We have, $$\cos \theta < 0(\because$$ angle between $$\vec{a... | mcq | jee-main-2024-online-8th-april-morning-shift | 8,724 |
EmeP0gA9DJPiHDRu | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If $$\left| {\overrightarrow a } \right| = 4,\left| {\overrightarrow b } \right| = 2$$ and the angle between $${\overrightarrow a }$$ and $${\overrightarrow b }$$ is $$\pi /6$$ then $${\left( {\overrightarrow a \times \overrightarrow b } \right)^2}$$ is equal to : | [{"identifier": "A", "content": "$$48$$ "}, {"identifier": "B", "content": "$$16$$"}, {"identifier": "C", "content": "$$\\overrightarrow a $$ "}, {"identifier": "D", "content": "none of these "}] | ["B"] | null | $${\left( {\overrightarrow a \times \overrightarrow b } \right)^2} = {\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2}\,\,{\sin ^2}{\pi \over 6}$$
<br><br>$$ = 16 \times 4 \times {1 \over 4} = 16$$ | mcq | aieee-2002 | 8,726 |
SUoKDdZSEYkrRjwJ | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If the vectors $$\overrightarrow c ,\overrightarrow a = x\widehat i + y\widehat j + z\widehat k$$ and $$\widehat b = \widehat j$$ are such that $$\overrightarrow a ,\overrightarrow c $$ and $$\overrightarrow b $$ form a right handed system then $${\overrightarrow c }$$ is : | [{"identifier": "A", "content": "$$z\\widehat i - x\\widehat k$$ "}, {"identifier": "B", "content": "$$\\overrightarrow 0 $$ "}, {"identifier": "C", "content": "$$y\\widehat j$$ "}, {"identifier": "D", "content": "$$ - z\\widehat i + x\\widehat k$$ "}] | ["A"] | null | Since $$\overrightarrow a ,\overrightarrow c ,\overrightarrow b $$ form a right handed system,
<br><br>$$\therefore$$ $$\overrightarrow c = \overrightarrow b \times \overrightarrow a = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
0 & 1 & 0 \cr
x & y & z \cr
... | mcq | aieee-2002 | 8,727 |
FRP6dd3QwqsKilpn | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If $$\overrightarrow a \times \overrightarrow b = \overrightarrow b \times \overrightarrow c = \overrightarrow c \times \overrightarrow a $$ then $$\overrightarrow a + \overrightarrow b + \overrightarrow c = $$ | [{"identifier": "A", "content": "$$abc$$ "}, {"identifier": "B", "content": "$$-1$$"}, {"identifier": "C", "content": "$$0$$"}, {"identifier": "D", "content": "$$2$$"}] | ["C"] | null | Let $$\overrightarrow a + \overrightarrow b + \overrightarrow c = \overrightarrow r .$$ Then
<br><br>$$\overrightarrow a \times \left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) = \overrightarrow a \times \overrightarrow r $$
<br><br>$$ \Rightarrow 0 + \overrightarrow a \times \overrig... | mcq | aieee-2003 | 8,729 |
WpRf7gPaDyKnEwUr | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | A tetrahedron has vertices at $$O(0,0,0), A(1,2,1) B(2,1,3)$$ and $$C(-1,1,2).$$ Then the angle between the faces $$OAB$$ and $$ABC$$ will be : | [{"identifier": "A", "content": "$${90^ \\circ }$$ "}, {"identifier": "B", "content": "$${\\cos ^{ - 1}}\\left( {{{19} \\over {35}}} \\right)$$ "}, {"identifier": "C", "content": "$${\\cos ^{ - 1}}\\left( {{{17} \\over {31}}} \\right)$$"}, {"identifier": "D", "content": "$${30^ \\circ }$$"}] | ["B"] | null | Vector perpendicular to the face $$OAB$$
<br><br>$$ = \overrightarrow {OA} \times \overrightarrow {OB} = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 2 & 1 \cr
2 & 1 & 3 \cr
} } \right| = 5\widehat i - \widehat j - 3\widehat k$$
<br><br>Vector perpendicu... | mcq | aieee-2003 | 8,730 |
8fOb3BJc4CXuZI26 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ be non-zero vectors such that $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a \,\,.$$ If $$\theta $$ is ... | [{"identifier": "A", "content": "$${{2\\sqrt 2 } \\over 3}$$ "}, {"identifier": "B", "content": "$${{\\sqrt 2 } \\over 3}$$"}, {"identifier": "C", "content": "$${2 \\over 3}$$ "}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["A"] | null | Given $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
<br><br>Clearly $$\overrightarrow a $$ and $$\overrightarrow b $$ are noncollinear
<br><br>$$ \Rightarrow \left( {\overr... | mcq | aieee-2004 | 8,732 |
bnFroekS8xy90KW5 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | For any vector $${\overrightarrow a }$$ , the value of $${\left( {\overrightarrow a \times \widehat i} \right)^2} + {\left( {\overrightarrow a \times \widehat j} \right)^2} + {\left( {\overrightarrow a \times \widehat k} \right)^2}$$ is equal to : | [{"identifier": "A", "content": "$$3{\\overrightarrow a ^2}$$ "}, {"identifier": "B", "content": "$${\\overrightarrow a ^2}$$"}, {"identifier": "C", "content": "$$2{\\overrightarrow a ^2}$$"}, {"identifier": "D", "content": "$$4{\\overrightarrow a ^2}$$"}] | ["C"] | null | Let $$\overrightarrow a = x\overrightarrow i + y\overrightarrow j + z\overrightarrow k $$
<br><br>$$\overrightarrow a \times \overrightarrow i = z\overrightarrow j - y\overrightarrow k $$
<br><br>$$ \Rightarrow {\left( {\overrightarrow a \times \overrightarrow i } \right)^2} = {y^2} + {z^2}$$
<br><br>Similarly, ... | mcq | aieee-2005 | 8,733 |
KRG2QzjpRTUqQrKh | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If $$\widehat u$$ and $$\widehat v$$ are unit vectors and $$\theta $$ is the acute angle between them, then $$2\widehat u \times 3\widehat v$$ is a unit vector for : | [{"identifier": "A", "content": "no value of $$\\theta $$ "}, {"identifier": "B", "content": "exactly one value of $$\\theta $$ "}, {"identifier": "C", "content": "exactly two values of $$\\theta $$ "}, {"identifier": "D", "content": "more than two values of $$\\theta $$ "}] | ["B"] | null | Given $$\left| {2\widehat u \times 3\widehat v} \right| = 1$$
<br><br>and $$\theta $$ is acute angle between $$\widehat u$$
<br><br>and $$\widehat v,\,\,\left| {\widehat u} \right| = 1,\,\,\left| {\widehat v} \right| = 1\,\,\,$$
<br><br>$$ \Rightarrow \,\,\,6\left| {\widehat u} \right|\left| {\widehat v} \right|\left... | mcq | aieee-2007 | 8,734 |
c5QQOLp2BK9AMeWV | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = 2\widehat i + \widehat j -2 \widehat k$$ and $$\overrightarrow b = \widehat i + \widehat j$$.
<br/><br>Let $$\overrightarrow c $$ be a vector such that $$\left| {\overrightarrow c - \overrightarrow a } \right| = 3$$,
<br/><br>$$\left| {\left( {\overrightarrow a \times \overrightarrow b } \... | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "$${1 \\over 8}$$"}, {"identifier": "D", "content": "$${{25} \\over 8}$$"}] | ["A"] | null | Given:
<br><br>$$\overrightarrow a = 2\widehat i + \widehat j - 2\widehat k,\,\,\overrightarrow b = \widehat i + \widehat j$$
<br><br>$$ \Rightarrow $$ $$\left| {\overrightarrow a } \right| = 3$$
<br><br>$$ \therefore $$ $$\overrightarrow a \times \overrightarrow b = 2\widehat i - 2\widehat j ... | mcq | jee-main-2017-offline | 8,736 |
Pf8T6ywBp4CA8ZLZEdLkh | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | The area (in sq. units) of the parallelogram whose diagonals are along the vectors $$8\widehat i - 6\widehat j$$ and $$3\widehat i + 4\widehat j - 12\widehat k,$$ is : | [{"identifier": "A", "content": "26"}, {"identifier": "B", "content": "65"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "52"}] | ["B"] | null | When diagonal $${\overrightarrow {{d_1}} }$$ and $${\overrightarrow {{d_2}} }$$ are given of a parallelogram then the area of parallelogram = $${1 \over 2}\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$$
<br><br>Given, $${\overrightarrow {{d_1}} }$$ = 8$$\widehat i$$ $$-$$ 6$$\widehat j$$ + ... | mcq | jee-main-2017-online-8th-april-morning-slot | 8,737 |
vNlAtSKfwwTlji9dN29Ja | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If the vector $$\overrightarrow b = 3\widehat j + 4\widehat k$$ is written as the
sum of a vector $$\overrightarrow {{b_1}} ,$$ paralel to $$\overrightarrow a = \widehat i + \widehat j$$ and a vector $$\overrightarrow {{b_2}} ,$$ perpendicular to $$\overrightarrow a ,$$ then $$\overrightarrow {{b_1}} \times \over... | [{"identifier": "A", "content": "$$ - 3\\widehat i + 3\\widehat j - 9\\widehat k$$"}, {"identifier": "B", "content": "$$6\\widehat i - 6\\widehat j + {9 \\over 2}\\widehat k$$ "}, {"identifier": "C", "content": "$$ - 6\\widehat i + 6\\widehat j - {9 \\over 2}\\widehat k$$"}, {"identifier": "D", "content": "$$3\\widehat... | ["B"] | null | $$\overrightarrow {{b_1}} = {{\left( {\overrightarrow {{b_1}} .\overrightarrow a } \right)\widehat a} \over 1}$$
<br><br>= $$\left\{ {{{\left( {3\widehat j + 4\widehat k} \right).\left( {\widehat i + \widehat j} \right)} \over {\sqrt 2 }}} \right\}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} ... | mcq | jee-main-2017-online-9th-april-morning-slot | 8,738 |
MeeggVCHIAFtgLpDiij1E | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If $$\overrightarrow a ,\,\,\overrightarrow b ,$$ and $$\overrightarrow C $$ are unit vectors such that $$\overrightarrow a + 2\overrightarrow b + 2\overrightarrow c = \overrightarrow 0 ,$$ then $$\left| {\overrightarrow a \times \overrightarrow c } \right|$$ is equal to : | [{"identifier": "A", "content": "$${{\\sqrt {15} } \\over 4}$$ "}, {"identifier": "B", "content": "$${{1} \\over {4}}$$"}, {"identifier": "C", "content": "$${{15} \\over {16}}$$"}, {"identifier": "D", "content": "$${{\\sqrt {15} } \\over 16}$$"}] | ["A"] | null | Given, <br><br>
$$\overrightarrow a + 2\overrightarrow b + 2\overrightarrow c = \overrightarrow 0 $$<br><br/>
$$ \Rightarrow $$ $$\overrightarrow a + 2\overrightarrow c = - 2\overrightarrow b $$<br>
<br>
Squaring both sides,<br><br/>
$${\left| {\overrightarrow a } \right|^2} + 4\overrightarrow a .\overrightarrow... | mcq | jee-main-2018-online-15th-april-morning-slot | 8,739 |
t3jFwvsaHi92dFMbV9vrP | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow c = \widehat j - \widehat k$$ and a vector $$\overrightarrow b $$ be such that $$\overrightarrow a \times \overrightarrow b = \overrightarrow c $$ and $$\overrightarrow a .\overrightarrow b = 3.$$ Then $$\left| {\overrightarrow b } \rig... | [{"identifier": "A", "content": "$${{11} \\over 3}$$"}, {"identifier": "B", "content": "$${{11} \\over {\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$$\\sqrt {{{11} \\over 3}} $$"}, {"identifier": "D", "content": "$${{\\sqrt {11} } \\over 3}$$"}] | ["C"] | null | $$ \because $$ $$\overrightarrow a $$ $$=$$ $$\widehat i + \widehat j + \widehat k \Rightarrow \left| {\overrightarrow a } \right| = \sqrt 3 $$
<br><br>& $$\overrightarrow c = \widehat j - \widehat k \Rightarrow \left| {\overrightarrow c } \right|\sqrt 2 $$
<br><br>Now, $$\overrightarrow a $$ $$ \... | mcq | jee-main-2018-online-16th-april-morning-slot | 8,740 |
kz4wkHygFPPLV412753rsa0w2w9jx65kz0q | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = 3\widehat i + 2\widehat j + 2\widehat k$$ and $$\overrightarrow b = \widehat i + 2\widehat j - 2\widehat k$$ be two vectors. If a vector perpendicular to both the vectors
$$\overrightarrow a + \overrightarrow b $$ and $$\overrightarrow a - \overrightarrow b $$ has the magnitude 12 then one... | [{"identifier": "A", "content": "$$4\\left( {2\\widehat i - 2\\widehat j - \\widehat k} \\right)$$"}, {"identifier": "B", "content": "$$4\\left( { - 2\\widehat i - 2\\widehat j + \\widehat k} \\right)$$"}, {"identifier": "C", "content": "$$4\\left( {2\\widehat i + 2\\widehat j + \\widehat k} \\right)$$"}, {"identifier"... | ["A"] | null | Required vector is $\overrightarrow r$ = $$\lambda \left( {\left( {\overline a + \overline b } \right) \times \left( {\overline a - \overline b } \right)} \right)$$<br><br>
$$ \Rightarrow \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
4 & 4 & 0 \cr
2 & 0 & 4 \cr ... | mcq | jee-main-2019-online-12th-april-morning-slot | 8,742 |
bOQOFNGP3tT7x6szeD3rsa0w2w9jx2gbzy3 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | The distance of the point having position vector $$ - \widehat i + 2\widehat j + 6\widehat k$$
from the straight line passing through the point
(2, 3, – 4) and parallel to the vector, $$6\widehat i + 3\widehat j - 4\widehat k$$ is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "$$2\\sqrt {13} $$"}, {"identifier": "D", "content": "$$4\\sqrt 3 $$"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264693/exam_images/n0zytcbrpcd3kgpqtlsb.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263784/exam_images/axuiomv56i9jmm3qd8t0.webp"><source media="(max-wid... | mcq | jee-main-2019-online-10th-april-evening-slot | 8,743 |
3h3PJW4wWUVQxcoR3jRCO | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow \alpha = 3\widehat i + \widehat j$$ and $$\overrightarrow \beta = 2\widehat i - \widehat j + 3 \widehat k$$
. If $$\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}} $$,
where $${\overrightarrow \beta _1}$$
is parallel to $$\overrightarrow \alpha $$ and $$\o... | [{"identifier": "A", "content": "$$ 3\\widehat i - 9\\widehat j - 5\\widehat k$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$($$ - 3\\widehat i + 9\\widehat j + 5\\widehat k$$)"}, {"identifier": "C", "content": "$$ - 3\\widehat i + 9\\widehat j + 5\\widehat k$$"}, {"identifier": "D", "content": "$${1 \\over 2}$$... | ["B"] | null | Given $$\overrightarrow \alpha = 3\widehat i + \widehat j$$<br><br>$$\overrightarrow \beta = 2\widehat i - \widehat j + 3 \widehat k$$
<br><br>$${\overrightarrow \beta _1}$$
is parallel to $$\overrightarrow \alpha $$
<br><br>$$ \therefore $$ $${\overrightarrow \beta _1}$$ = $$\lambda $$ $$\overrightarrow \alpha$... | mcq | jee-main-2019-online-9th-april-morning-slot | 8,744 |
CB4aJTXbdSnUgLtX7X7k9k2k5hiep0f | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i - 2\widehat j + \widehat k$$ and $$\overrightarrow b = \widehat i - \widehat j + \widehat k$$ be two
vectors. If $$\overrightarrow c $$ is a vector such that $$\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a $$ and $$\overrightarrow c .\o... | [{"identifier": "A", "content": "$$ - {1 \\over 2}$$"}, {"identifier": "B", "content": "$$ - {3 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "-1"}] | ["A"] | null | $$\overrightarrow a = \widehat i - 2\widehat j + \widehat k$$
<br><br>$$\overrightarrow b = \widehat i - \widehat j + \widehat k$$
<br><br>$$\left| {\overrightarrow a } \right|$$ = $$\sqrt 6 $$, $$\left| {\overrightarrow b } \right|$$ = $$\sqrt 3 $$
<br><br>and $${\overrightarrow a .\overrightarrow b }$$ = 4
<br><br>... | mcq | jee-main-2020-online-8th-january-evening-slot | 8,747 |
uxkbzvwhQhZu358eew7k9k2k5ki4pum | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be three vectors such that $$\left| {\overrightarrow a } \right| = \sqrt 3 $$,
$$\left| {\overrightarrow b } \right| = 5,\overrightarrow b .\overrightarrow c = 10$$ and the angle between $$\overrightarrow b $$ and $$\overrightarrow c $$
is ... | [] | null | 30 | Given $$\left| {\overrightarrow a } \right| = \sqrt 3 $$,
$$\left| {\overrightarrow b } \right| = 5$$
<br><br>Given $$\overrightarrow b .\overrightarrow c = 10$$
<br><br>And the angle between $$\overrightarrow b $$ and $$\overrightarrow c $$
is $${\pi \over 3}$$
<br><br>$$ \therefore $$ $$bc\cos {\pi \over 3}$$ = 1... | integer | jee-main-2020-online-9th-january-evening-slot | 8,748 |
I6tuy9NF399qWeyYVhjgy2xukezfmp9m | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let the position vectors of points 'A' and 'B' be
<br/>$$\widehat i + \widehat j + \widehat k$$ and $$2\widehat i + \widehat j + 3\widehat k$$, respectively. A point
'P' divides the line segment AB internally in the
ratio
$$\lambda $$ : 1 (
$$\lambda $$ > 0). If O is the origin and
<br/>$$\overrightarrow {OB} .\ove... | [] | null | 0.8 | Let, $$\overrightarrow a $$ = $$\widehat i + \widehat j + \widehat k$$
<br><br>and $$\overrightarrow b $$ = $$2\widehat i + \widehat j + 3\widehat k$$
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265164/exam_images/hoezbwyywsjaggdoi7yh.webp"><source media="(... | integer | jee-main-2020-online-2nd-september-evening-slot | 8,749 |
XDKiZVSmcBrvn7Fb1njgy2xukfal01ga | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If $$\overrightarrow a = 2\widehat i + \widehat j + 2\widehat k$$, then the value of<br/><br/>
$${\left| {\widehat i \times \left( {\overrightarrow a \times \widehat i} \right)} \right|^2} + {\left| {\widehat j \times \left( {\overrightarrow a \times \widehat j} \right)} \right|^2} + {\left| {\widehat k \times \lef... | [] | null | 18 | Let $$\overrightarrow a = x\widehat i + y\widehat j + z\widehat k$$<br><br>Now $$\widehat i \times \left( {\overrightarrow a \times \widehat i} \right) = \left( {\widehat i.\widehat i} \right)\overrightarrow a - \left( {\widehat i.\overrightarrow a } \right)\widehat i$$<br><br>= $$y\widehat j + z\widehat k$$<br><br>... | integer | jee-main-2020-online-4th-september-evening-slot | 8,750 |
RkTo4epEnfRC9GpYlw1kls5v75w | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i + 2\widehat j - \widehat k$$, $$\overrightarrow b = \widehat i - \widehat j$$ and $$\overrightarrow c = \widehat i - \widehat j - \widehat k$$ be three given vectors. If $$\overrightarrow r $$ is a vector such that $$\overrightarrow r \times \overrightarrow a = \overrightarrow ... | [] | null | 12 | <p>Given, $$\overrightarrow a = \widehat i + 2\widehat j - \widehat k$$,</p>
<p>$$\overrightarrow b = \widehat i - \widehat j$$,</p>
<p>$$\overrightarrow c = \widehat i - \widehat j - \widehat k$$</p>
<p>$$\overrightarrow r \times \overrightarrow a = \overrightarrow c \times \overrightarrow a $$</p>
<p>$$ \Righta... | integer | jee-main-2021-online-25th-february-morning-slot | 8,751 |
s7RlE6savu4SIIyjxr1klta1k13 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$$ and $$\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$$. If the area of the parallelogram whose adjacent sides are represented by the vectors $$\overrightarrow a $$ and $$\overrightarrow b $$ is $$8\sqrt 3 $$ square units, then ... | [] | null | 2 | $$\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$$<br><br>$$\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$$<br><br>Area of parallelogram = $$\left| {\overrightarrow a \times \overrightarrow b } \right|$$<br><br>$$ = \left| {(\widehat i + \alpha \widehat j + 3\widehat k) \times (3... | integer | jee-main-2021-online-25th-february-evening-slot | 8,752 |
hkfGcy1FV2nIdZCA6X1kmiw7pzg | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$ = $$\widehat i$$ + 2$$\widehat j$$ $$-$$ 3$$\widehat k$$ and $$\overrightarrow b = 2\widehat i$$ $$-$$ 3$$\widehat j$$ + 5$$\widehat k$$. If $$\overrightarrow r $$ $$\times$$ $$\overrightarrow a $$ = $$\overrightarrow b $$ $$\times$$ $$\overrightarrow r $$, <br/><br/>$$\overrightarrow r $$ .... | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "11"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "15"}] | ["D"] | null | Given $$\overrightarrow r $$ $$\times$$ $$\overrightarrow a $$ = $$\overrightarrow b $$ $$\times$$ $$\overrightarrow r $$
<br><br>$$ \Rightarrow $$ $$\overrightarrow r \times \overrightarrow a = - \overrightarrow r \times \overrightarrow b $$<br><br>$$\overrightarrow r \times (\overrightarrow a + \overrightarrow... | mcq | jee-main-2021-online-16th-march-evening-shift | 8,753 |
Y2uW5HKoyGplfqL1nL1kmm2snqv | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$ and $$\overrightarrow b $$ be two non-zero vectors perpendicular to each other and $$|\overrightarrow a | = |\overrightarrow b |$$. If $$|\overrightarrow a \times \overrightarrow b | = |\overrightarrow a |$$, then the angle between the vectors $$\left( {\overrightarrow a + \overrightarrow b... | [{"identifier": "A", "content": "$${\\sin ^{ - 1}}\\left( {{1 \\over {\\sqrt 6 }}} \\right)$$"}, {"identifier": "B", "content": "$${\\cos ^{ - 1}}\\left( {{1 \\over {\\sqrt 2 }}} \\right)$$"}, {"identifier": "C", "content": "$${\\sin ^{ - 1}}\\left( {{1 \\over {\\sqrt 3 }}} \\right)$$"}, {"identifier": "D", "content": ... | ["D"] | null | $$\overrightarrow a $$ is perpendicular to $$\overrightarrow b $$<br><br>$$ \therefore $$ $$\overrightarrow a $$ . $$\overrightarrow b $$ = 0<br><br>Given, | $$\overrightarrow a $$ $$\times$$ $$\overrightarrow b $$ | = | $$\overrightarrow a $$ |<br><br>and | $$\overrightarrow a $$ | = | $$\overrightarrow b $$ |<br><br>... | mcq | jee-main-2021-online-18th-march-evening-shift | 8,754 |
1krq0xpcz | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If the shortest distance between the lines $$\overrightarrow {{r_1}} = \alpha \widehat i + 2\widehat j + 2\widehat k + \lambda (\widehat i - 2\widehat j + 2\widehat k)$$, $$\lambda$$ $$\in$$ R, $$\alpha$$ > 0 and $$\overrightarrow {{r_2}} = - 4\widehat i - \widehat k + \mu (3\widehat i - 2\widehat j - 2\widehat k... | [] | null | 6 | If $$\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b $$ and $$\overrightarrow r = \overrightarrow c + \lambda \overrightarrow d $$ then shortest distance between two lines is <br><br>$$L = {{(\overrightarrow a - \overrightarrow c ).(\overrightarrow b \times \overrightarrow d )} \over {|b \times ... | integer | jee-main-2021-online-20th-july-morning-shift | 8,755 |
1krw2kdmg | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow p = 2\widehat i + 3\widehat j + \widehat k$$ and $$\overrightarrow q = \widehat i + 2\widehat j + \widehat k$$ be two vectors. If a vector $$\overrightarrow r = (\alpha \widehat i + \beta \widehat j + \gamma \widehat k)$$ is perpendicular to each of the vectors ($$(\overrightarrow p + \overrig... | [] | null | 3 | $$\overrightarrow p = 2\widehat i + 3\widehat j + \widehat k$$ (Given )<br><br>$$\overrightarrow q = \widehat i + 2\widehat j + \widehat k$$<br><br>Now, $$(\overrightarrow p + \overrightarrow q ) \times (\overrightarrow p - \overrightarrow q ) = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k}... | integer | jee-main-2021-online-25th-july-morning-shift | 8,756 |
1krzmy5aw | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If $$\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 5$$ and $$\left| {\overrightarrow a \times \overrightarrow b } \right|$$ = 8, then $$\left| {\overrightarrow a .\,\overrightarrow b } \right|$$ is equal to : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "5"}] | ["A"] | null | $$\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 5$$<br><br>$$\left| {\overrightarrow a \times \overrightarrow b } \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta = \pm 8$$<br><br>$$\sin \theta = \pm \,{4 \over 5}$$<br><br>$$\therefore$$ $$... | mcq | jee-main-2021-online-25th-july-evening-shift | 8,757 |
1ks05h77r | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i + \widehat j + 2\widehat k$$ and $$\overrightarrow b = - \widehat i + 2\widehat j + 3\widehat k$$. Then the vector product $$\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\left( {\overrightarrow a \times \left( {\left( {\overrightarrow a - \overrightar... | [{"identifier": "A", "content": "$$5(34\\widehat i - 5\\widehat j + 3\\widehat k)$$"}, {"identifier": "B", "content": "$$7(34\\widehat i - 5\\widehat j + 3\\widehat k)$$"}, {"identifier": "C", "content": "$$7(30\\widehat i - 5\\widehat j + 7\\widehat k)$$"}, {"identifier": "D", "content": "$$5(30\\widehat i - 5\\wideha... | ["B"] | null | $$\overrightarrow a = \widehat i + \widehat j + 2\widehat k$$<br><br>$$\overrightarrow b = - \widehat i + 2\widehat j + 3\widehat k$$<br><br>$$\overrightarrow a + \overrightarrow b = 3\widehat j + 5\widehat k;\overrightarrow a.\overrightarrow b = - 1 + 2 + 6 = 7$$<br><br>$$\left( {\left( {\overrightarrow a \tim... | mcq | jee-main-2021-online-27th-july-morning-shift | 8,758 |
1ks0bqpuy | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow b $$ and $$\overrightarrow c = \widehat j - \widehat k$$ be three vectors such that $$\overrightarrow a \times \overrightarrow b = \overrightarrow c $$ and $$\overrightarrow a \,.\,\overrightarrow b = 1$$. If the length of projection ve... | [] | null | 2 | $$\overrightarrow a \times \overrightarrow b = \overrightarrow c $$<br><br>Take Dot with $$\overrightarrow c $$<br><br>$$\left( {\overrightarrow a \times \overrightarrow b } \right).\,\overrightarrow c = {\left| {\overrightarrow c } \right|^2} = 2$$<br><br>Projection of $$\overrightarrow b $$ or $$\overrightarrow a... | integer | jee-main-2021-online-27th-july-morning-shift | 8,759 |
1kteolpch | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i + 5\widehat j + \alpha \widehat k$$, $$\overrightarrow b = \widehat i + 3\widehat j + \beta \widehat k$$ and $$\overrightarrow c = - \widehat i + 2\widehat j - 3\widehat k$$ be three vectors such that, $$\left| {\overrightarrow b \times \overrightarrow c } \right| = 5\sqrt 3 $$... | [] | null | 90 | Since, $$\overrightarrow a .\,\overrightarrow b = 0$$<br><br>$$1 + 15 + \alpha \beta = 0 \Rightarrow \alpha \beta = - 16$$ .... (1)<br><br>Also, <br><br>$${\left| {\overrightarrow b \, \times \overrightarrow c } \right|^2} = 75 \Rightarrow (10 + {\beta ^2})14 - {(5 - 3\beta )^2} = 75$$<br><br>$$\Rightarrow$$ 5$$\be... | integer | jee-main-2021-online-27th-august-morning-shift | 8,760 |
1l544t6p5 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a = \alpha \widehat i + 3\widehat j - \widehat k$$, $$\overrightarrow b = 3\widehat i - \beta \widehat j + 4\widehat k$$ and $$\overrightarrow c = \widehat i + 2\widehat j - 2\widehat k$$ where $$\alpha ,\,\beta \in R$$, be three vectors. If the projection of $$\overrightarrow a $$ on $$\ov... | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "6"}] | ["A"] | null | <p>$$\overrightarrow a = \alpha \widehat i + 3\widehat j - \widehat k$$</p>
<p>$$\overrightarrow b = 3\widehat i - \beta \widehat j + 4\widehat k$$</p>
<p>$$\overrightarrow c = \widehat i + 2\widehat j - 2\widehat k$$</p>
<p>Projection of $$\overrightarrow a $$ on $$\overrightarrow c $$ is</p>
<p>$${{\overrightarrow... | mcq | jee-main-2022-online-29th-june-morning-shift | 8,762 |
1l54tqgwd | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a = \widehat i - 2\widehat j + 3\widehat k$$, $$\overrightarrow b = \widehat i + \widehat j + \widehat k$$ and $$\overrightarrow c $$ be a vector such that $$\overrightarrow a + \left( {\overrightarrow b \times \overrightarrow c } \right) = \overrightarrow 0 $$ and $$\overrig... | [] | null | BONUS | $$
\vec{a} \cdot \vec{b}=(\hat{i}-2 \hat{j}+3 \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})=2
$$ ........(i)
<br/><br/>Given: $\vec{a}+(\vec{b} \times \vec{c})=0$
<br/><br/>$$
\Rightarrow \vec{a} \cdot \vec{b}=0
$$ ........(ii)
<br/><br/>Equation (i) and equation (ii) are contradicting. | integer | jee-main-2022-online-29th-june-evening-shift | 8,763 |
1l55iksny | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a = \alpha \widehat i + 2\widehat j - \widehat k$$ and $$\overrightarrow b = - 2\widehat i + \alpha \widehat j + \widehat k$$, where $$\alpha \in R$$. If the area of the parallelogram whose adjacent sides are represented by the vectors $$\overrightarrow a $$ and $$\overrightarrow b $$ is $$... | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "14"}] | ["D"] | null | <p>$$\overrightarrow a = \alpha \widehat i + 2\widehat j - \widehat k$$ and $$\overrightarrow b = - 2\widehat i + \alpha \widehat j + \widehat k$$</p>
<p>$$\therefore$$ $$\overrightarrow a \times \overrightarrow b = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
\alpha & 2 & { - 1} \cr
... | mcq | jee-main-2022-online-28th-june-evening-shift | 8,764 |
1l55ithsn | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a $$ be a vector which is perpendicular to the vector $$3\widehat i + {1 \over 2}\widehat j + 2\widehat k$$. If $$\overrightarrow a \times \left( {2\widehat i + \widehat k} \right) = 2\widehat i - 13\widehat j - 4\widehat k$$, then the projection of the vector $$\overrightarrow a $$ on the vec... | [{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$${5 \\over 3}$$"}, {"identifier": "D", "content": "$${7 \\over 3}$$"}] | ["C"] | null | <p>Let $$\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$$</p>
<p>and $$\overrightarrow a \,.\,\left( {3\widehat i - {1 \over 2}\widehat j + 2\widehat k} \right) = 0 \Rightarrow 3{a_1} + {{{a_2}} \over 2} + 2{a_3} = 0$$ ..... (i)</p>
<p>and $$\overrightarrow a \times (2\widehat i + \widehat k)... | mcq | jee-main-2022-online-28th-june-evening-shift | 8,765 |
1l567mmea | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>If $$\overrightarrow a = 2\widehat i + \widehat j + 3\widehat k$$, $$\overrightarrow b = 3\widehat i + 3\widehat j + \widehat k$$ and $$\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k$$ are coplanar vectors and $$\overrightarrow a \,.\,\overrightarrow c = 5$$, $$\overrightarrow b \bot \o... | [] | null | 150 | <p>$$2{C_1} + {C_2} + 3{C_3} = 5$$ ...... (i)</p>
<p>$$3{C_1} + 3{C_2} + {C_3} = 0$$ ...... (ii)</p>
<p>$$\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = \left| {\matrix{
2 & 1 & 3 \cr
3 & 3 & 1 \cr
{{C_1}} & {{C_2}} & {{C_3}} \cr
} } \right|$$</p>
<p>$$ = 2(3{C_3} - {C_2}) - ... | integer | jee-main-2022-online-28th-june-morning-shift | 8,766 |
1l57okaze | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a = \widehat i + \widehat j - \widehat k$$ and $$\overrightarrow c = 2\widehat i - 3\widehat j + 2\widehat k$$. Then the number of vectors $$\overrightarrow b $$ such that $$\overrightarrow b \times \overrightarrow c = \overrightarrow a $$ and $$|\overrightarrow b | \in $$ {1, 2, ........, ... | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}] | ["A"] | null | <p>$$\overrightarrow a = \widehat i + \widehat j - \widehat k$$</p>
<p>$$\overrightarrow c = 2\widehat i - 3\widehat j + 2\widehat k$$</p>
<p>Now, $$\overrightarrow b \times \overrightarrow c = \overrightarrow a $$</p>
<p>$$\overrightarrow c \,.\,(\overrightarrow b \times \overrightarrow c ) = \overrightarrow c \,... | mcq | jee-main-2022-online-27th-june-morning-shift | 8,768 |
1l5ajnmz4 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\theta$$ be the angle between the vectors $$\overrightarrow a $$ and $$\overrightarrow b $$, where $$|\overrightarrow a | = 4,$$ $$|\overrightarrow b | = 3$$ and $$\theta \in \left( {{\pi \over 4},{\pi \over 3}} \right)$$. Then $${\left| {\left( {\overrightarrow a - \overrightarrow b } \right) \times \left... | [] | null | 576 | <p>$${\left| {\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$$</p>
<p>$$ \Rightarrow {\left| {\overrightarrow a \times \overrightarrow a + \overrightarrow a \times \... | integer | jee-main-2022-online-25th-june-morning-shift | 8,770 |
1l5ban31s | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\widehat a$$ and $$\widehat b$$ be two unit vectors such that $$|(\widehat a + \widehat b) + 2(\widehat a \times \widehat b)| = 2$$. If $$\theta$$ $$\in$$ (0, $$\pi$$) is the angle between $$\widehat a$$ and $$\widehat b$$, then among the statements :</p>
<p>(S1) : $$2|\widehat a \times \widehat b| = |\widehat... | [{"identifier": "A", "content": "Only (S1) is true."}, {"identifier": "B", "content": "Only (S2) is true."}, {"identifier": "C", "content": "Both (S1) and (S2) are true."}, {"identifier": "D", "content": "Both (S1) and (S2) are false."}] | ["C"] | null | <p>$$\left| {\widehat a + \widehat b + 2(\widehat a \times \widehat b)} \right| = 2,\,\theta \in (0,\,\pi )$$</p>
<p>$$ \Rightarrow {\left| {\widehat a + \widehat b + 2(\widehat a \times \widehat b)} \right|^2} = 4$$</p>
<p>$$ \Rightarrow {\left| {\widehat a} \right|^2} + {\left| {\widehat b} \right|^2} + 4{\left| {\w... | mcq | jee-main-2022-online-24th-june-evening-shift | 8,771 |
1l5c1or9b | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\widehat a$$, $$\widehat b$$ be unit vectors. If $$\overrightarrow c $$ be a vector such that the angle between $$\widehat a$$ and $$\overrightarrow c $$ is $${\pi \over {12}}$$, and $$\widehat b = \overrightarrow c + 2\left( {\overrightarrow c \times \widehat a} \right)$$, then $${\left| {6\overrightarrow ... | [{"identifier": "A", "content": "$$6\\left( {3 - \\sqrt 3 } \\right)$$"}, {"identifier": "B", "content": "$$3 + \\sqrt 3 $$"}, {"identifier": "C", "content": "$$6\\left( {3 + \\sqrt 3 } \\right)$$"}, {"identifier": "D", "content": "$$6\\left( {\\sqrt 3 + 1} \\right)$$"}] | ["C"] | null | $\because \quad \hat{b}=\vec{c}+2(\vec{c} \times \hat{a})$
<br/><br/>
$$
\begin{aligned}
&\Rightarrow \hat{b} \cdot \vec{c}=|\vec{c}|^{2} \\\\
&\therefore \hat{b}-\vec{c}=2(\vec{c} \times \vec{a})
\end{aligned}
$$
<br/><br/>
$$
\begin{aligned}
&\Rightarrow|\hat{b}|^{2}+|\vec{c}|^{2}-2 \hat{b} \cdot \vec{c}=4|\vec{c}|... | mcq | jee-main-2022-online-24th-june-morning-shift | 8,772 |
1l6dwjgp3 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\mathrm{ABC}$$ be a triangle such that $$\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{c}},|\overrightarrow{\mathrm{a}}|=6 \sqrt{2},|\overrightarrow{\mathrm{b}}|=2 \sqrt{3}$$ and $$\vec{b}... | [{"identifier": "A", "content": "both (S1) and (S2) are true"}, {"identifier": "B", "content": "only (S1) is true"}, {"identifier": "C", "content": "only (S2) is true"}, {"identifier": "D", "content": "both (S1) and (S2) are false"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l97sapyl/779dd912-c32e-40a3-8f62-d0f26f1743e1/b6acf9d0-4b5b-11ed-bfde-e1cb3fafe700/file-1l97sapym.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l97sapyl/779dd912-c32e-40a3-8f62-d0f26f1743e1/b6acf9d0-4b5b-11ed-bfde-e1cb3fafe700/fi... | mcq | jee-main-2022-online-25th-july-morning-shift | 8,773 |
1l6f2wm5r | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$$ and let $$\vec{b}$$ be a vector such that $$\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$$ and $$\vec{a} \cdot \vec{b}=3$$. Then the projection of $$\vec{b}$$ on the vector $$\vec{a}-\vec{b}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{2}{\\sqrt{21}}$$"}, {"identifier": "B", "content": "$$2 \\sqrt{\\frac{3}{7}}$$"}, {"identifier": "C", "content": "$$\n\\frac{2}{3} \\sqrt{\\frac{7}{3}}\n$$"}, {"identifier": "D", "content": "$$\\frac{2}{3}$$"}] | ["A"] | null | <p>$$\overrightarrow a = \widehat i - \widehat j + 2\widehat k$$</p>
<p>$$\overrightarrow a \times \overrightarrow b = 2\widehat i - \widehat k$$</p>
<p>$$\overrightarrow a \,.\,\overrightarrow b = 3$$</p>
<p>$$|\overrightarrow a \times \overrightarrow b {|^2} + |\overrightarrow a \,.\,\overrightarrow b {|^2} = |\... | mcq | jee-main-2022-online-25th-july-evening-shift | 8,774 |
1l6giq99g | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow{\mathrm{a}}=\alpha \hat{i}+\hat{j}-\hat{k}$$ and $$\overrightarrow{\mathrm{b}}=2 \hat{i}+\hat{j}-\alpha \hat{k}, \alpha>0$$. If the projection of $$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$$ on the vector $$-\hat{i}+2 \hat{j}-2 \hat{k}$$ is 30, then $$\alpha$$ is equal ... | [{"identifier": "A", "content": "$$\\frac{15}{2}$$"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "$$\\frac{13}{2}$$"}, {"identifier": "D", "content": "7"}] | ["D"] | null | <p>Given : $$\overrightarrow a = (\alpha ,1, - 1)$$ and $$\overrightarrow b = (2,1, - \alpha )$$</p>
<p>$$\overrightarrow c = \overrightarrow a \times \overrightarrow b = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
\alpha & 1 & { - 1} \cr
2 & 1 & { - \alpha } \cr
} } \right|$$... | mcq | jee-main-2022-online-26th-july-morning-shift | 8,775 |
1l6jc2mfl | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}$$ and $$\vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k}$$ be two vectors, such that $$\vec{a} \times \vec{b}=-\hat{i}+9 \hat{j}+12 \hat{k}$$. Then the projection of $$\vec{b}-2 \vec{a}$$ on $$\vec{b}+\vec{a}$$ is equal to :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$\\frac{39}{5}$$"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "$$\\frac{46}{5}$$"}] | ["D"] | null | <p>$$\overrightarrow a = \alpha \widehat i + \widehat j + \beta \widehat k$$, $$\overrightarrow b = 3\widehat i - 5\widehat j + 4\widehat k$$</p>
<p>$$\overrightarrow a \times \overrightarrow b = - \widehat i + 9\widehat j + 12\widehat k$$</p>
<p>$$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \c... | mcq | jee-main-2022-online-27th-july-morning-shift | 8,776 |
1l6km5v7z | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a $$, $$\overrightarrow b $$, $$\overrightarrow c $$ be three non-coplanar vectors such that $$\overrightarrow a $$ $$\times$$ $$\overrightarrow b $$ = 4$$\overrightarrow c $$, $$\overrightarrow b $$ $$\times$$ $$\overrightarrow c $$ = 9$$\overrightarrow a $$ and $$\overrightarrow c $$ $$\times... | [] | null | 36 | <p>Given,</p>
<p>$$\overrightarrow a \times \overrightarrow b = 4\,.\,\overrightarrow c $$ ..... (i)</p>
<p>$$\overrightarrow b \times \overrightarrow c = 9\,.\,\overrightarrow a $$ ..... (ii)</p>
<p>$$\overrightarrow c \times \overrightarrow a = \alpha \,.\,\overrightarrow b $$ .... (iii)</p>
<p>Taking dot produ... | integer | jee-main-2022-online-27th-july-evening-shift | 8,777 |
1l6m5gnnw | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let a vector $$\vec{a}$$ has magnitude 9. Let a vector $$\vec{b}$$ be such that for every $$(x, y) \in \mathbf{R} \times \mathbf{R}-\{(0,0)\}$$, the vector $$(x \vec{a}+y \vec{b})$$ is perpendicular to the vector $$(6 y \vec{a}-18 x \vec{b})$$. Then the value of $$|\vec{a} \times \vec{b}|$$ is equal to :</p> | [{"identifier": "A", "content": "$$9 \\sqrt{3}$$"}, {"identifier": "B", "content": "$$27 \\sqrt{3}$$"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "81"}] | ["B"] | null | <p>$$\left( {x\overrightarrow a + y\overrightarrow b } \right).\left( {6y\overrightarrow a - 18x\overrightarrow b } \right) = 0$$</p>
<p>$$ \Rightarrow \left( {6xy|\overrightarrow a {|^2} - 18xy|\overrightarrow b {|^2}} \right) + \left( {6{y^2} - 18{x^2}} \right)\overrightarrow a .\overrightarrow b = 0$$</p>
<p>As g... | mcq | jee-main-2022-online-28th-july-morning-shift | 8,778 |
1l6p2fvoj | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\hat{a}$$ and $$\hat{b}$$ be two unit vectors such that the angle between them is $$\frac{\pi}{4}$$. If $$\theta$$ is the angle between the vectors $$(\hat{a}+\hat{b})$$ and $$(\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b}))$$, then the value of $$164 \,\cos ^{2} \theta$$ is equal to :</p> | [{"identifier": "A", "content": "$$90+27 \\sqrt{2}$$"}, {"identifier": "B", "content": "$$45+18 \\sqrt{2}$$"}, {"identifier": "C", "content": "$$90+3 \\sqrt{2}$$"}, {"identifier": "D", "content": "$$54+90 \\sqrt{2}$$"}] | ["A"] | null | <p>$$\widehat a\,.\,\widehat b = {1 \over {\sqrt 2 }}$$ and $$|\widehat a \times \widehat b| = {1 \over {\sqrt 2 }}$$</p>
<p>$${{\left( {\widehat a + \widehat b} \right)\,.\,\left( {\widehat a + 2\widehat b + 2\left( {\widehat a \times \widehat b} \right)} \right)} \over {\left| {\widehat a + \widehat b} \right|\left| ... | mcq | jee-main-2022-online-29th-july-morning-shift | 8,779 |
1l6rf9wxz | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}, \vec{b}, \vec{c}$$ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and
$$(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})+(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})+(\vec{c} \times \vec{a}) \cdot(\vec{a} ... | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "16"}, {"identifier": "D", "content": "18"}] | ["C"] | null | $|\vec{a}||\vec{b}||\vec{c}|=14$
<br/><br/>
$$
\begin{aligned}
& \vec{a} \wedge \vec{b}=\vec{b} \wedge \vec{c}=\vec{c} \wedge \vec{a}=\theta=\frac{2 \pi}{3} \\\\
& \vec{a} \cdot \vec{b}=-\frac{1}{2}|\vec{a}||\vec{b}| \\\\
& \vec{b} \cdot \vec{c}=-\frac{1}{2}|\vec{b}||\vec{c}| \\\\
& \vec{c} \cdot \vec{a}=-\frac{1}{2}|\... | mcq | jee-main-2022-online-29th-july-evening-shift | 8,780 |
1l6rg17yk | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors such that $$|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2}, \vec{a} \cdot \vec{b}=3$$ and $$|\vec{a} \times \vec{b}|^{2}=75$$. Then $$|\vec{a}|^{2}$$ is equal to __________.</p> | [] | null | 14 | $\because|\vec{a}+\dot{b}|^{2}=|\vec{a}|^{2}+2|b|^{2}$
<br/><br/>or $|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{a}|^{2}+2|\vec{b}|^{2}$
<br/><br/>$\therefore|\vec{b}|^{2}=6$
<br/><br/>Now $|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}$
<br/><br/>$$
75=|\vec{a}|^{... | integer | jee-main-2022-online-29th-july-evening-shift | 8,781 |
ldo7cuo5 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c}=5 \hat{i}-3 \hat{j}+3 \hat{k}$ be three vectors. If $\vec{r}$ is a vector such
that, $\vec{r} \times \vec{b}=\vec{c} \times \vec{b}$ and $\vec{r} \cdot \vec{a}=0$, then $25|\vec{r}|^{2}$ is equal to : | [{"identifier": "A", "content": "336"}, {"identifier": "B", "content": "449"}, {"identifier": "C", "content": "339"}, {"identifier": "D", "content": "560"}] | ["C"] | null | $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
<br/><br/>$\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
<br/><br/>$\overrightarrow{\mathrm{c}}=\hat{5 \mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
<br/><br/>$(\overrightarrow{\mathrm{r}}-\o... | mcq | jee-main-2023-online-31st-january-evening-shift | 8,782 |
ldoaq5ul | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that
<br/><br/>$|\vec{a}|=\sqrt{31}, 4|\vec{b}|=|\vec{c}|=2$ and $2(\vec{a} \times \vec{b})=3(\vec{c} \times \vec{a})$.
<br/><br/>If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2 \pi}{3}$, then $\left(\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}}\righ... | [] | null | 3 | $2(\vec{a} \times \vec{b})=3(\vec{c} \times \vec{a})$
<br/><br/>$\vec{a} \times(2 \vec{b}+3 \vec{c})=0$
<br/><br/>$$
\begin{aligned}
& \vec{a}=\lambda(2 \vec{b}+3 \vec{c}) \\\\
& |\vec{a}|^{2}=\lambda^{2}\left(4|b|^{2}+9|c|^{2}+12 \vec{b} \cdot \vec{c}\right) \\\\
& 31=31 \lambda^{2} \\\\
& \lambda=\pm 1 \\\\
& \vec{a... | integer | jee-main-2023-online-31st-january-evening-shift | 8,783 |
1ldooe6vv | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>$$A(2,6,2), B(-4,0, \lambda), C(2,3,-1)$$ and $$D(4,5,0),|\lambda| \leq 5$$ are the vertices of a quadrilateral $$A B C D$$. If its area is 18 square units, then $$5-6 \lambda$$ is equal to __________.</p> | [] | null | 11 | $$
\begin{aligned}
& \mathrm{A}(2,6,2) \quad \mathrm{B}(-4,0, \lambda), \mathrm{C}(2,3,-1) \mathrm{D}(4,5,0) \\\\
& \text { Area }=\frac{1}{2}|\overrightarrow{B D} \times \overrightarrow{A C}|=18 \\\\
& \overrightarrow{A C} \times \overrightarrow{B D}=\left|\begin{array}{ccc}
\hat{i} & j & k \\\\
0 & -3 & -3 \\\\
8 & 5... | integer | jee-main-2023-online-1st-february-morning-shift | 8,784 |
1ldptp6vk | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors such that $$|\vec{a}|=\sqrt{14},|\vec{b}|=\sqrt{6}$$ and $$|\vec{a} \times \vec{b}|=\sqrt{48}$$. Then $$(\vec{a} \cdot \vec{b})^{2}$$ is equal to ___________.</p> | [] | null | 36 | $|\vec{a}|=\sqrt{14},|\vec{b}|=\sqrt{6}$ and $|\vec{a} \times \vec{b}|=\sqrt{48}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow |\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2} \\\\
& \Rightarrow 48+(\vec{a} \cdot \vec{b})^{2}=6 \times 14 \\\\
& \Rightarrow (\vec{a} \cdot \vec{b})^{2}=8... | integer | jee-main-2023-online-31st-january-morning-shift | 8,785 |
ldqx819b | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $\vec{a}$ and $\vec{b}$ be two vectors, Let $|\vec{a}|=1,|\vec{b}|=4$ and $\vec{a} \cdot \vec{b}=2$. If $\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b}$, then the value of $\vec{b} \cdot \vec{c}$ is : | [{"identifier": "A", "content": "$-48$"}, {"identifier": "B", "content": "$-60$"}, {"identifier": "C", "content": "$-84$"}, {"identifier": "D", "content": "$-24$"}] | ["A"] | null | <p>$$\overrightarrow b .\overrightarrow c = \overrightarrow b .(2\overline a \times \overrightarrow b ) - 3\overline b .\overrightarrow b $$</p>
<p>$$ = 0 - 3|\overline b {|^2} = - 48$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift | 8,786 |
1ldr7kwam | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let a unit vector $$\widehat{O P}$$ make angles $$\alpha, \beta, \gamma$$ with the positive directions of the co-ordinate axes $$\mathrm{OX}$$, $$\mathrm{OY}, \mathrm{OZ}$$ respectively, where $$\beta \in\left(0, \frac{\pi}{2}\right)$$. If $$\widehat{\mathrm{OP}}$$ is perpendicular to the plane through points $$(1,2... | [{"identifier": "A", "content": "$$\\alpha \\in\\left(\\frac{\\pi}{2}, \\pi\\right)$$ and $$\\gamma \\in\\left(\\frac{\\pi}{2}, \\pi\\right)$$"}, {"identifier": "B", "content": "$$\\alpha \\in\\left(0, \\frac{\\pi}{2}\\right)$$ and $$\\gamma \\in\\left(\\frac{\\pi}{2}, \\pi\\right)$$"}, {"identifier": "C", "content": "... | ["A"] | null | <p>Let $$A \equiv (1,2,3),B \equiv (2,3,4),C \equiv (1,5,7)$$</p>
<p>$$\overrightarrow n = \overrightarrow {AB} \times \overrightarrow {AC} = \left| {\matrix{
i & j & k \cr
1 & 1 & 1 \cr
0 & 3 & 4 \cr
} } \right|$$</p>
<p>$$ = \widehat i - 4\widehat j + 3\widehat k$$</p>
<p>$$\widehat {OP} = {{ \pm ... | mcq | jee-main-2023-online-30th-january-morning-shift | 8,787 |
1ldseppqy | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>If $$\overrightarrow a = \widehat i + 2\widehat k,\overrightarrow b = \widehat i + \widehat j + \widehat k,\overrightarrow c = 7\widehat i - 3\widehat j + 4\widehat k,\overrightarrow r \times \overrightarrow b + \overrightarrow b \times \overrightarrow c = \overrightarrow 0 $$ and $$\overrightarrow r \,.\,\ov... | [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "30"}, {"identifier": "C", "content": "34"}, {"identifier": "D", "content": "32"}] | ["C"] | null | <p>$$(\overrightarrow r - \overrightarrow c ) \times \overrightarrow b = 0$$</p>
<p>$$\overrightarrow r = \lambda \overrightarrow b + \overrightarrow c $$</p>
<p>$$ \Rightarrow \lambda \overrightarrow b \,.\,\overrightarrow a + \overrightarrow c \,.\,\overrightarrow a = 0$$</p>
<p>$$ \Rightarrow \lambda (3) + (7 ... | mcq | jee-main-2023-online-29th-january-evening-shift | 8,788 |
1ldsfbgwo | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a = 4\widehat i + 3\widehat j$$ and $$\overrightarrow b = 3\widehat i - 4\widehat j + 5\widehat k$$. If $$\overrightarrow c $$ is a vector such that $$\overrightarrow c .\left( {\overrightarrow a \times \overrightarrow b } \right) + 25 = 0,\overrightarrow c \,.(\widehat i + \widehat j + \wid... | [{"identifier": "A", "content": "$$\\frac{3}{\\sqrt2}$$"}, {"identifier": "B", "content": "$$\\frac{1}{\\sqrt2}$$"}, {"identifier": "C", "content": "$$\\frac{1}{5}$$"}, {"identifier": "D", "content": "$$\\frac{5}{\\sqrt2}$$"}] | ["D"] | null | <p>$$[\matrix{
{\overrightarrow c } & {\overrightarrow a } & {\overrightarrow b } \cr
} ] = - 25$$</p>
<p>Let $$\overrightarrow c = l\widehat i + n\widehat j + n\widehat k$$</p>
<p>$$\left| {\matrix{
l & m & n \cr
4 & 3 & 0 \cr
3 & { - 4} & 5 \cr
} } \right| = - 25$$</p>
<p>$$ \Rightarrow 3l ... | mcq | jee-main-2023-online-29th-january-evening-shift | 8,789 |
1ldwxj5lf | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a = \widehat i + 2\widehat j + \lambda \widehat k,\overrightarrow b = 3\widehat i - 5\widehat j - \lambda \widehat k,\overrightarrow a \,.\,\overrightarrow c = 7,2\overrightarrow b \,.\,\overrightarrow c + 43 = 0,\overrightarrow a \times \overrightarrow c = \overrightarrow b \times \over... | [] | null | 8 | $$
\begin{aligned}
& \vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}, \vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}, \vec{a} \cdot \vec{c}=7 \\\\
& \vec{a} \times \vec{c}-\vec{b} \times \vec{c}=\overrightarrow{0} \\\\
& (\vec{a}-\vec{b}) \times \vec{c}=0 \Rightarrow(\vec{a}-\vec{b}) \text { is paralleled to } \vec{c} \\\\
& \v... | integer | jee-main-2023-online-24th-january-evening-shift | 8,790 |
1lgow35tq | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$|\vec{a}|=2,|\vec{b}|=3$$ and the angle between the vectors $$\vec{a}$$ and $$\vec{b}$$ be $$\frac{\pi}{4}$$. Then $$|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "441"}, {"identifier": "B", "content": "482"}, {"identifier": "C", "content": "841"}, {"identifier": "D", "content": "882"}] | ["D"] | null | $$
\begin{aligned}
& |\vec{a}|=2 \\\\
& |\vec{b}|=3 \\\\
& \vec{a} \cdot \vec{b}=\frac{\pi}{4}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& |(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2 \\\\
& = |-3 \vec{a} \times \vec{b}+4 \vec{b} \times \vec{a}|^2 \\\\
& = |-3 \vec{a} \times \vec{b}-4 \vec{a} \times \vec{b}|... | mcq | jee-main-2023-online-13th-april-evening-shift | 8,791 |
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