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values | question stringlengths 17 24.5k | options stringlengths 2 4.26k | correct_option stringclasses 6
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lv5grw9p | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | <p>Let $$P(x, y, z)$$ be a point in the first octant, whose projection in the $$x y$$-plane is the point $$Q$$. Let $$O P=\gamma$$; the angle between $$O Q$$ and the positive $$x$$-axis be $$\theta$$; and the angle between $$O P$$ and the positive $$z$$-axis be $$\phi$$, where $$O$$ is the origin. Then the distance of ... | [{"identifier": "A", "content": "$$\\gamma \\sqrt{1-\\sin ^2 \\phi \\cos ^2 \\theta}$$\n"}, {"identifier": "B", "content": "$$\\gamma \\sqrt{1+\\cos ^2 \\theta \\sin ^2 \\phi}$$\n"}, {"identifier": "C", "content": "$$\\gamma \\sqrt{1+\\cos ^2 \\phi \\sin ^2 \\theta}$$\n"}, {"identifier": "D", "content": "$$\\gamma \\sq... | ["A"] | null | <p>$$\begin{aligned}
& \overrightarrow{O P}=x \hat{i}+y \hat{j}+z \hat{k} \\
& \overrightarrow{O Q}=x \hat{i}+y \hat{j} \\
& |O P|=\gamma=\sqrt{x^2+y^2+z^2} \\
& \cos \theta=\frac{x}{\sqrt{x^2+y^2}} \Rightarrow \cos ^2 \theta=\frac{x^2}{\gamma^2-z^2}=\frac{x^2}{\gamma^2-\gamma^2 \cos ^2 \phi} \\
& \cos \phi=\frac{z}{\s... | mcq | jee-main-2024-online-8th-april-morning-shift | 4,330 |
lv7v4fxr | maths | 3d-geometry | direction-cosines-and-direction-ratios-of-a-line | <p>If the line $$\frac{2-x}{3}=\frac{3 y-2}{4 \lambda+1}=4-z$$ makes a right angle with the line $$\frac{x+3}{3 \mu}=\frac{1-2 y}{6}=\frac{5-z}{7}$$, then $$4 \lambda+9 \mu$$ is equal to :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "13"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "6"}] | ["D"] | null | <p>$$\begin{aligned}
& L_1: \frac{x-2}{(-3)}=\frac{y-\frac{2}{3}}{\left(\frac{4 \lambda+1}{3}\right)}=\frac{}{(-1)} \\
& L_2: \frac{x+3}{3 \mu}=\frac{y-\frac{1}{2}}{-3}=\frac{z-5}{-7} \\
& \because L_1 \perp L_2 \\
& \Rightarrow(-3)(3 \mu)+\left(\frac{4 \lambda+1}{3}\right)(-3)+(-1)(-7)=0 \\
& -9 \mu-4 \lambda-1+7=0 \\... | mcq | jee-main-2024-online-5th-april-morning-shift | 4,331 |
hW9Uid5rVNiUGULz | maths | 3d-geometry | lines-and-plane | A plane which passes through the point $$(3,2,0)$$ and the line
<br/><br>$${{x - 4} \over 1} = {{y - 7} \over 5} = {{z - 4} \over 4}$$ is :</br> | [{"identifier": "A", "content": "$$x-y+z=1$$"}, {"identifier": "B", "content": "$$x+y+z=5$$ "}, {"identifier": "C", "content": "$$x+2y-z=1$$ "}, {"identifier": "D", "content": "$$2x-y+z=5$$"}] | ["A"] | null | As the point $$\left( {3,2,0} \right)$$ lies on the given line
<br><br>$${{x - 4} \over 1} = {{y - 7} \over 5} = {{z - 4} \over 4}$$
<br><br>$$\therefore$$ There can be infinite many planes passing through this line. But here out of the four options only first option is satisfied by the coordinates of both the points ... | mcq | aieee-2002 | 4,332 |
EbGhqY1JBAk9FShb | maths | 3d-geometry | lines-and-plane | If the angel $$\theta $$ between the line $${{x + 1} \over 1} = {{y - 1} \over 2} = {{z - 2} \over 2}$$ and
<br/><br/>the plane $$2x - y + \sqrt \lambda \,\,z + 4 = 0$$ is such that $$\sin \,\,\theta = {1 \over 3}$$ then value of $$\lambda $$ is : | [{"identifier": "A", "content": "$${5 \\over 3}$$"}, {"identifier": "B", "content": "$${-3 \\over 5}$$"}, {"identifier": "C", "content": "$${3 \\over 4}$$"}, {"identifier": "D", "content": "$${-4 \\over 3}$$"}] | ["A"] | null | If $$\theta $$ is the angle between line and plane then $$\left( {{\pi \over 2} - 0} \right)$$
<br><br>is the angle between line and normal to plane given by
<br><br>$$\cos \left( {{\pi \over 2} - 0} \right) = {{\left( {\widehat i + 2\widehat j + 2\widehat k} \right).\left( {2\widehat i - \widehat j + \sqrt \lambda... | mcq | aieee-2005 | 4,333 |
nTl94Fj4X6KBIr3k | maths | 3d-geometry | lines-and-plane | If the plane $$2ax-3ay+4az+6=0$$ passes through the midpoint of the line joining the centres of the spheres
<br/><br>$${x^2} + {y^2} + {z^2} + 6x - 8y - 2z = 13$$ and
<br/><br>$${x^2} + {y^2} + {z^2} - 10x + 4y - 2z = 8$$ then a equals :</br></br> | [{"identifier": "A", "content": "$$-1$$"}, {"identifier": "B", "content": "$$1$$"}, {"identifier": "C", "content": "$$-2$$ "}, {"identifier": "D", "content": "$$2$$"}] | ["C"] | null | Centers of given spheres are $$\left( { - 3,4,1} \right)$$ and $$\left( {5, - 2,1} \right).$$
<br><br>Mid point of centers is $$\left( {1,1,1} \right).$$
<br><br>Satisfying this in the equation of plane, we get
<br><br>$$2a - 3a + 4a + 6 = 0$$
<br/><br/>$$ \Rightarrow a = - 2$$ | mcq | aieee-2005 | 4,334 |
LrYeTFfhmmgV6ZGq | maths | 3d-geometry | lines-and-plane | The distance between the line
<br/><br>$$\overrightarrow r = 2\widehat i - 2\widehat j + 3\widehat k + \lambda \left( {i - j + 4k} \right),$$ and the plane
<br/><br>$$\overrightarrow r .\left( {\widehat i + 5\widehat j + \widehat k} \right) = 5$$ is </br></br> | [{"identifier": "A", "content": "$${{10} \\over 9}$$ "}, {"identifier": "B", "content": "$${{10} \\over {3\\sqrt 3 }}$$ "}, {"identifier": "C", "content": "$${{3} \\over 10}$$"}, {"identifier": "D", "content": "$${{10} \\over 3}$$"}] | ["B"] | null | A point on lines is $$\left( {2, - 2,3} \right)$$ its perpendicular distance
<br><br>from the plane $$x + 5y + z - 5 = 0$$ is
<br><br>$$ = \left| {{{2 - 10 + 3 - 5} \over {\sqrt {1 + 25 + 1} }}} \right| = {{10} \over {3\sqrt 3 }}$$ | mcq | aieee-2005 | 4,335 |
UvC5NyuvIjBty2fQ | maths | 3d-geometry | lines-and-plane | Let the line $$\,\,\,\,\,$$ $${{x - 2} \over 3} = {{y - 1} \over { - 5}} = {{z + 2} \over 2}$$ lie in the plane $$\,\,\,\,\,$$ $$x + 3y - \alpha z + \beta = 0.$$ Then $$\left( {\alpha ,\beta } \right)$$ equals | [{"identifier": "A", "content": "$$(-6,7)$$ "}, {"identifier": "B", "content": "$$(5,-15)$$ "}, {"identifier": "C", "content": "$$(-5,5)$$ "}, {"identifier": "D", "content": "$$(6, -17)$$ "}] | ["A"] | null | As the line $${{x - 2} \over 3} = {{y - 1} \over { - 5}} = {{z + 2} \over 2}$$ lie in the plane
<br><br>$$x + 3y - \alpha z + \beta = 0$$
<br><br>$$\therefore$$ $$Pt\left( {2,1, - 2} \right)$$ lies on the plane
<br><br>i.e. $$2 + 3 + 2\alpha + \beta = 0$$
<br><br>or $$\,\,\,\,2\alpha + \beta + 5 = 0\,\,\,\,\,\,\,... | mcq | aieee-2009 | 4,337 |
eeoIaVQuoj4uNE1A | maths | 3d-geometry | lines-and-plane | If the angle between the line $$x = {{y - 1} \over 2} = {{z - 3} \over \lambda }$$ and the plane
<br/><br>$$x+2y+3z=4$$ is $${\cos ^{ - 1}}\left( {\sqrt {{5 \over {14}}} } \right),$$ then $$\lambda $$ equals :</br> | [{"identifier": "A", "content": "$${3 \\over 2}$$"}, {"identifier": "B", "content": "$${2 \\over 5}$$"}, {"identifier": "C", "content": "$${5 \\over 3}$$"}, {"identifier": "D", "content": "$${2 \\over 3}$$"}] | ["D"] | null | If $$\theta $$ be the angle between the given line and plane, then
<br><br>$$\sin \theta = {{1 \times 1 + 2 \times 2 + \lambda \times 3} \over {\sqrt {{1^2} + {2^2} + {\lambda ^2}} .\sqrt {{1^2} + {2^2} + {3^2}} }}$$
<br><br>$$ = {{5 + 3\lambda } \over {\sqrt {14} .\sqrt {5 + {\lambda ^2}} }}$$
<br><br>But it is gi... | mcq | aieee-2011 | 4,338 |
3n9s47wXddiAH8vV | maths | 3d-geometry | lines-and-plane | The distance of the point $$(1, 0, 2)$$ from the point of intersection of the line $${{x - 2} \over 3} = {{y + 1} \over 4} = {{z - 2} \over {12}}$$ and the plane $$x - y + z = 16,$$ is : | [{"identifier": "A", "content": "$$3\\sqrt {21} $$ "}, {"identifier": "B", "content": "$$13$$"}, {"identifier": "C", "content": "$$2\\sqrt {14} $$"}, {"identifier": "D", "content": "$$8$$"}] | ["B"] | null | General point on given line $$ \equiv P\left( {3r + 2,4r - 1,12r + 2} \right)$$
<br><br>Point $$P$$ must satisfy equation of plane
<br><br>$$\left( {3r + 2} \right) - \left( {4r - 1} \right) + \left( {12r + 2} \right) = 16$$
<br><br>$$11r + 5 = 16$$
<br><br>$$r=1$$
<br><br>$$P\left( {3 \times 1 + 2,4 \times 1 - 1,12 ... | mcq | jee-main-2015-offline | 4,341 |
hxfsFTp3VBUn9E8D | maths | 3d-geometry | lines-and-plane | If the line, $${{x - 3} \over 2} = {{y + 2} \over { - 1}} = {{z + 4} \over 3}\,$$ lies in the planes, $$lx+my-z=9,$$ then $${l^2} + {m^2}$$ is equal to : | [{"identifier": "A", "content": "$$5$$ "}, {"identifier": "B", "content": "$$2$$ "}, {"identifier": "C", "content": "$$26$$"}, {"identifier": "D", "content": "$$18$$"}] | ["B"] | null | Line lies in the plane $$ \Rightarrow \left( {3, - 2, - 4} \right)$$ lie in the plane
<br><br>$$ \Rightarrow 3\ell - 2m + 4 = 9$$ or $$3\ell - 2m = 5....\left( 1 \right)$$
<br><br>Also, $$\ell ,$$ $$m, - 1$$ are dr's of line perpendicular to plane
<br><br>and $$2, - 1,3$$ are dr's of line lying in the plane
<br><b... | mcq | jee-main-2016-offline | 4,342 |
u2D3Iv4j5JXG4jCJ | maths | 3d-geometry | lines-and-plane | If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line,
<br/><br/>$${x \over 1} = {y \over 4} = {z \over 5}$$ is Q, then PQ is equal to:
| [{"identifier": "A", "content": "$$2\\sqrt {42} $$"}, {"identifier": "B", "content": "$$\\sqrt {42} $$"}, {"identifier": "C", "content": "$$6\\sqrt 5 $$"}, {"identifier": "D", "content": "$$3\\sqrt 5 $$"}] | ["A"] | null | Equation of line PQ is $${{x - 1} \over 1} = {{y + 2} \over 4} = {{z - 3} \over 5}$$
<br><br>Let F be ($$\lambda $$ + 1, 4$$\lambda $$ $$-$$ 2, 5$$\lambda $$ + 3)
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265539/exam_images/j1l2cyuk0ctnijsktmb6.webp" style="max-width: 100%; height: auto;... | mcq | jee-main-2017-offline | 4,345 |
Xs9FqdMNdzzpW5zC | maths | 3d-geometry | lines-and-plane | The distance of the point (1, 3, – 7) from the plane passing through the point (1, –1, – 1), having normal
perpendicular to both the lines
<br/><br/>$${{x - 1} \over 1} = {{y + 2} \over { - 2}} = {{z - 4} \over 3}$$
<br/><br>and
<br/><br/>$${{x - 2} \over 2} = {{y + 1} \over { - 1}} = {{z + 7} \over { - 1}}$$ is :</br> | [{"identifier": "A", "content": "$${{10} \\over {\\sqrt {83} }}$$"}, {"identifier": "B", "content": "$${{5} \\over {\\sqrt {83} }}$$"}, {"identifier": "C", "content": "$${{10} \\over {\\sqrt {74} }}$$"}, {"identifier": "D", "content": "$${{20} \\over {\\sqrt {74} }}$$"}] | ["A"] | null | Let the plane be
<br><br>a(x $$-$$ 1) + b(y + 1) + c (z + 1) = 0
<br><br>Normal vector
<br><br>$$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & { - 2} & 3 \cr
2 & { - 1} & { - 1} \cr
} } \right| = 5\widehat i + 7\widehat j + 3\widehat k$$
<br><br>So plane i... | mcq | jee-main-2017-offline | 4,346 |
qeI2oXKiRz35ggv6LQjaF | maths | 3d-geometry | lines-and-plane | The line of intersection of the planes $$\overrightarrow r .\left( {3\widehat i - \widehat j + \widehat k} \right) = 1\,\,$$ and
<br/>$$\overrightarrow r .\left( {\widehat i + 4\widehat j - 2\widehat k} \right) = 2,$$ is : | [{"identifier": "A", "content": "$${{x - {4 \\over 7}} \\over { - 2}} = {y \\over 7} = {{z - {5 \\over 7}} \\over {13}}$$"}, {"identifier": "B", "content": "$${{x - {4 \\over 7}} \\over 2} = {y \\over { - 7}} = {{z + {5 \\over 7}} \\over {13}}$$"}, {"identifier": "C", "content": "$${{x - {6 \\over {13}}} \\over 2} = {{... | ["C"] | null | $$\overrightarrow n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $$
<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
3 & { - 1} & 1 \cr
1 & 4 & { - 2} \cr
} } \right| = \widehat i\left( { - 2} \right) - \w... | mcq | jee-main-2017-online-8th-april-morning-slot | 4,348 |
TvlgxK9cJHYSO6FsebHnW | maths | 3d-geometry | lines-and-plane | If the line, $${{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + \lambda } \over { - 2}}$$ lies in the plane, 2x−4y+3z=2, then the shortest distance between this line and the line, $${{x - 1} \over {12}} = {y \over 9} = {z \over 4}$$ is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "3"}] | ["C"] | null | Point (3, $$-$$ 2, $$-$$ $$\lambda $$) on p line 2x $$-$$ 4y + 3z $$-$$ 2 $$=$$ 0
<br><br>$$=$$ 6 + 8 $$-$$ 3$$\lambda $$ $$-$$ 2 = 0
<br><br>$$=$$ 3$$\lambda $$ $$=$$ 12
<br><br><b>$$\lambda $$ $$=$$ 4</b>
<br><br>Now,
<br><br>$${{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + 4} \over { - 2}} = {k_1}$$ &... | mcq | jee-main-2017-online-9th-april-morning-slot | 4,349 |
Hby8suwc7u2mMH1O | maths | 3d-geometry | lines-and-plane | The length of the projection of the line segment joining the points (5, -1, 4) and (4, -1, 3) on the plane,
x + y + z = 7 is : | [{"identifier": "A", "content": "$$\\sqrt {{2 \\over 3}} $$"}, {"identifier": "B", "content": "$${2 \\over {\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["A"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266605/exam_images/sruywneq0jjxnahol5cq.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - 3D Geometry Question 286 English Explanation">
<br><br>PQ is the projection of line segment AB on the plane x + y + z = 7
<br>... | mcq | jee-main-2018-offline | 4,350 |
IHWRS21qK99uEo6wdYXzx | maths | 3d-geometry | lines-and-plane | An angle between the plane, x + y + z = 5 and the line of intersection of the planes, 3x + 4y + z $$-$$ 1 = 0 and 5x + 8y + 2z + 14 =0, is : | [{"identifier": "A", "content": "$${\\sin ^{ - 1}}\\left( {\\sqrt {{\\raise0.5ex\\hbox{$\\scriptstyle 3$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle {17}$}}} } \\right)$$ "}, {"identifier": "B", "content": "$${\\cos ^{ - 1}}\\left( {\\sqrt {{\\raise0.5ex\\hbox{$\\scriptstyle 3$}\n\\kern-0.1em/\\ker... | ["A"] | null | Normal to $$3x + 4y + z = 1$$ is $$3\widehat i + 4\widehat j + \widehat k$$
<br><br>Normal to $$5x + 8y + 2z =$$ $$ - 14$$ is $$5\widehat i + 8\widehat j + 2\widehat k$$
<br><br>The line of intersection of the planes is perpendicular to both normals, so, direction ratios of the intersection line are direc... | mcq | jee-main-2018-online-15th-april-morning-slot | 4,351 |
KhXrS1fo8yYoLcyUzbXUs | maths | 3d-geometry | lines-and-plane | The equation of the line passing through (–4, 3, 1), parallel
<br/><br>to the plane x + 2y – z – 5 = 0 and intersecting
<br/><br>the line $${{x + 1} \over { - 3}} = {{y - 3} \over 2} = {{z - 2} \over { - 1}}$$ is :</br></br> | [{"identifier": "A", "content": "$${{x + 4} \\over 3} = {{y - 3} \\over {-1}} = {{z - 1} \\over 1}$$"}, {"identifier": "B", "content": "$${{x + 4} \\over 1} = {{y - 3} \\over {1}} = {{z - 1} \\over 3}$$"}, {"identifier": "C", "content": "$${{x + 4} \\over -1} = {{y - 3} \\over {1}} = {{z - 1} \\over 1}$$"}, {"identifie... | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264981/exam_images/ehep4l3pw7lbyi0lsu7d.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Morning Slot Mathematics - 3D Geometry Question 268 English Explanation">
<br><br>Th... | mcq | jee-main-2019-online-9th-january-morning-slot | 4,352 |
xFbNYocl8Ychr6fBoL3rsa0w2w9jxayhpcy | maths | 3d-geometry | lines-and-plane | The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines
<br/>$$\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \lambda \left( {\widehat i + 2\widehat j - \widehat k} \right)$$ and $$\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \mu \left( { - \... | [{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 3 }}$$"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$${\\sqrt 3 }$$"}] | ["D"] | null | Vector of the plane is <br><br>
$$\left| {\matrix{
{\hat i} & {\hat j} & {\hat k} \cr
1 & 2 & { - 1} \cr
{ - 1} & 1 & { - 2} \cr
} } \right| = - 3\hat i + 3\hat j + 3\hat k$$<br><br>
Now equation of plane is <br><br>
$$ - 3x + 3y + 3z = c$$<br><br>
(1, 1, 0) will satisfy the pl... | mcq | jee-main-2019-online-12th-april-evening-slot | 4,353 |
klTd3ShmDde4vluBEB3rsa0w2w9jx6gsul6 | maths | 3d-geometry | lines-and-plane | If the line $${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 1} \over { - 1}}$$
intersects the plane 2x + 3y – z + 13 = 0 at a point P and the plane
3x + y + 4z = 16 at a point Q, then PQ is equal to : | [{"identifier": "A", "content": "$$2\\sqrt 7 $$"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "$$2\\sqrt {14} $$"}, {"identifier": "D", "content": "$$\\sqrt {14} $$"}] | ["C"] | null | $${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 1} \over { - 1}} = \lambda $$<br><br>
$$A(3\lambda + 2,2\lambda - 1, - \lambda + 1)$$ line on 2x + 3y -z + 13 = 0<br><br>
$$ \Rightarrow 2(3\lambda + 2) + 3(2\lambda - 1) - ( - \lambda + 1) + 13 = 0$$<br><br>
$$ \Rightarrow 13\lambda + 13 = 0 \Rightarrow \lambda =... | mcq | jee-main-2019-online-12th-april-morning-slot | 4,354 |
kvFqafTbB4SGPDJlq03rsa0w2w9jx2h1x8t | maths | 3d-geometry | lines-and-plane | A perpendicular is drawn from a point on the line $${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {z \over 1}$$ to the plane x + y + z = 3 such that the
foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are : | [{"identifier": "A", "content": "(4, 0, \u2013 1)"}, {"identifier": "B", "content": "(2, 0, 1)"}, {"identifier": "C", "content": "(1, 0, 2)"}, {"identifier": "D", "content": "(\u2013 1, 0, 4)"}] | ["B"] | null | $${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {z \over 1} = \lambda $$<br><br>
Let a point P on the line is <br><br>
(2$$\lambda $$ + 1, – $$\lambda $$ –1, + $$\lambda $$)<br><br>
Foot of $${ \bot ^r}Q$$ is given by<br><br>
$${{x - 2\lambda - 1} \over 1} = {{y + \lambda + 1} \over 1} = {{z - \lambda } \over 1} = - ... | mcq | jee-main-2019-online-10th-april-evening-slot | 4,355 |
sG8EkXlzDW1GqEPMdLzTF | maths | 3d-geometry | lines-and-plane | The equation of a plane containing the line of
intersection of the planes 2x – y – 4 = 0 and
y + 2z – 4 = 0 and passing through the point
(1, 1, 0) is : | [{"identifier": "A", "content": "x \u2013 3y \u2013 2z = \u20132"}, {"identifier": "B", "content": "2x \u2013 z = 2"}, {"identifier": "C", "content": "x \u2013 y \u2013 z = 0"}, {"identifier": "D", "content": "x + 3y + z = 4"}] | ["C"] | null | The equation of any plane passing through the intersection of the planes 2x – y – 4 = 0 and
y + 2z – 4 = 0 is :
<br><br>(2x – y – 4) + $$\lambda $$(y + 2z – 4) = 0 ........(1)
<br><br>As this plane passes through (1, 1, 0) then this point satisfy the equation (1).
<br><br>$$ \therefore $$ (2 – 1 – 4) + $$\lambda $$(1 +... | mcq | jee-main-2019-online-8th-april-morning-slot | 4,357 |
XPWSJ5jLrtku5PYAmM6MV | maths | 3d-geometry | lines-and-plane | The vector equation of the plane through the line
of intersection of the planes x + y + z = 1 and 2x
+ 3y+ 4z = 5 which is perpendicular to the plane
x – y + z = 0 is : | [{"identifier": "A", "content": "$$\\mathop r\\limits^ \\to \\times \\left( {\\mathop i\\limits^ \\wedge - \\mathop k\\limits^ \\wedge } \\right) - 2 = 0$$"}, {"identifier": "B", "content": "$$\\mathop r\\limits^ \\to . \\left( {\\mathop i\\limits^ \\wedge + \\mathop k\\limits^ \\wedge } \\right) + 2 = 0$$"},... | ["C"] | null | Given,
<br/><br/>P<sub>1</sub> : x + y + z = 1
<br><br>P<sub>1</sub> : 2x
+ 3y + 4z = 5
<br><br>Equation of the plane passing through the line
of intersection of the plane P<sub>1</sub> and P<sub>2</sub> is :
<br><br>P<sub>1</sub> + $$\lambda $$P<sub>2</sub> = 0
<br><br>$$ \Rightarrow $$ (x + y + z –1) + $$\lambda $$(2... | mcq | jee-main-2019-online-8th-april-evening-slot | 4,358 |
INBHKvgfzgvl3QVAjDdYB | maths | 3d-geometry | lines-and-plane | If an angle between the line, $${{x + 1} \over 2} = {{y - 2} \over 1} = {{z - 3} \over { - 2}}$$ and the plane, $$x - 2y - kz = 3$$ is $${\cos ^{ - 1}}\left( {{{2\sqrt 2 } \over 3}} \right),$$ then a value of k is : | [{"identifier": "A", "content": "$$\\sqrt {{3 \\over 5}} $$"}, {"identifier": "B", "content": "$$ - {5 \\over 2}$$"}, {"identifier": "C", "content": "$$ - {3 \\over 2}$$"}, {"identifier": "D", "content": "$$\\sqrt {{5 \\over 3}} $$"}] | ["D"] | null | DR's of line are 2, 1, $$-$$2
<br><br>normal vector of plane is $$\widehat i$$ $$-$$ 2$$\widehat j$$ $$-$$ k$$\widehat k$$
<br><br>sin$$\alpha $$ = $${{\left( {2\widehat i + \widehat j - 2\widehat k} \right).\left( {\widehat i - 2\widehat j - k\widehat k} \right)} \over {3\sqrt {1 + 4 + {k^2}} }}$$
<br><br>sin $$\alpha... | mcq | jee-main-2019-online-12th-january-evening-slot | 4,359 |
j3rxTjeh5YBNG4T1B1C5X | maths | 3d-geometry | lines-and-plane | The plane containing the line $${{x - 3} \over 2} = {{y + 2} \over { - 1}} = {{z - 1} \over 3}$$ and also containing its projection on the plane 2x + 3y $$-$$ z = 5, contains which one of the following points ? | [{"identifier": "A", "content": "($$-$$ 2, 2, 2)"}, {"identifier": "B", "content": "(2, 2, 0)"}, {"identifier": "C", "content": "(2, 0, $$-$$ 2)"}, {"identifier": "D", "content": "(0, $$-$$ 2, 2)"}] | ["C"] | null | The normal vector of required plane
<br><br>$$ = \left( {2\widehat i - \widehat j + 3\widehat k} \right) \times \left( {2\widehat i + 3\widehat j - \widehat k} \right)$$
<br><br>$$ = - 8\widehat i + 8\widehat j + 8\widehat k$$
<br><br>So, direction ratio of normal is $$\left( { - 1,1,1} \right)$$
<br><br>So required p... | mcq | jee-main-2019-online-11th-january-morning-slot | 4,361 |
kGwIvanvC36NRfACAH6nl | maths | 3d-geometry | lines-and-plane | On which of the following lines lies the point of intersection of the line, $${{x - 4} \over 2} = {{y - 5} \over 2} = {{z - 3} \over 1}$$ and the plane,
x + y + z = 2 ? | [{"identifier": "A", "content": "$${{x - 4} \\over 1} = {{y - 5} \\over 1} = {{z - 5} \\over { - 1}}$$"}, {"identifier": "B", "content": "$${{x - 2} \\over 2} = {{y - 3} \\over 2} = {{z + 3} \\over 3}$$"}, {"identifier": "C", "content": "$${{x - 1} \\over 1} = {{y - 3} \\over 2} = {{z + 4} \\over { - 5}}$$"}, {"identif... | ["C"] | null | General point on the given line is
<br><br>x = 2$$\lambda $$ + 4
<br><br>y = 2$$\lambda $$ + 5
<br><br>z = $$\lambda $$ + 3
<br><br>Solving with plane,
<br><br>2$$\lambda $$ + 4 + 2$$\lambda $$ + 5 + $$\lambda $$ + 3 = 2
<br><br>5$$\lambda $$ + 12 = 2
<br><br>5$$\lambda $$ = $$-$$ 10
<br><br>$$\lambda $$ = $$-$$ 2 | mcq | jee-main-2019-online-10th-january-evening-slot | 4,362 |
X9jCkBVS5S2zXHqiFo7k9k2k5ki7wil | maths | 3d-geometry | lines-and-plane | If the distance between the plane,
23x – 10y – 2z + 48 = 0 and the plane<br/><br/>
containing the lines
$${{x + 1} \over 2} = {{y - 3} \over 4} = {{z + 1} \over 3}$$<br/><br/> and
$${{x + 3} \over 2} = {{y + 2} \over 6} = {{z - 1} \over \lambda }\left( {\lambda \in R} \right)$$<br/><br/> is equal to
$${k \over {\sqrt ... | [] | null | 3 | Required distance = perpendicular distance of plane 23x – 10y – 2z + 48 = 0 either from (–1, 3, –1) or (–3, –2, 1)
<br><br>$$ \Rightarrow $$ $$\left| {{{ - 23 - 30 + 2 + 48} \over {\sqrt {{{\left( {23} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 2 \right)}^2}} }}} \right|$$ = $${k \over {\sqrt {633} }}$$
<br><br... | integer | jee-main-2020-online-9th-january-evening-slot | 4,364 |
z3jkKUxU8nDZA2wXVQjgy2xukg3b9f5y | maths | 3d-geometry | lines-and-plane | A plane P meets the coordinate axes at A, B
and C respectively. The centroid of $$\Delta $$ABC is
given to be (1, 1, 2). Then the equation of the
line through this centroid and perpendicular to
the plane P is : | [{"identifier": "A", "content": "$${{x - 1} \\over 1} = {{y - 1} \\over 1} = {{z - 2} \\over 2}$$"}, {"identifier": "B", "content": "$${{x - 1} \\over 2} = {{y - 1} \\over 1} = {{z - 2} \\over 1}$$"}, {"identifier": "C", "content": "$${{x - 1} \\over 2} = {{y - 1} \\over 2} = {{z - 2} \\over 1}$$"}, {"identifier": "D",... | ["C"] | null | Let, Equation of plane is
<br><br>$${x \over a} + {y \over b} + {z \over c}$$ = 1
<br><br>A = ($$a$$, 0, 0) B
= (0, b, 0), C
= (0, 0, c)
<br><br>$$ \therefore $$ Centroid = $$\left( {{a \over 3},{b \over 3},{c \over 3}} \right)$$ = (1, 1, 2)
<br><br>$$ \Rightarrow $$ $$a$$ = 3, b = 3, c = 6
<br><br>$$ \therefore $$ Pla... | mcq | jee-main-2020-online-6th-september-evening-slot | 4,365 |
Qojv10KpxgJcV2JxWHjgy2xukfuuxqrg | maths | 3d-geometry | lines-and-plane | The shortest distance between the lines
<br/><br>$${{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}$$ <br/><br>and x + y + z + 1 = 0, 2x – y + z
+ 3 = 0 is :</br></br> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt 3 }}$$"}] | ["D"] | null | Plane through line of intersection is
<br><br>x + y + z + 1 + $$\lambda $$ (2x –y + z + 3) = 0
<br><br>It should be parallel to given line $${{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}$$
<br><br>$$ \therefore $$ 0(1 + 2$$\lambda $$) - 1(1 - $$\lambda $$) + 1(1 + $$\lambda $$) = 0 $$ \Rightarrow $$ $$\lambd... | mcq | jee-main-2020-online-6th-september-morning-slot | 4,366 |
u8By5wccby20xeA0tgjgy2xukfqch09n | maths | 3d-geometry | lines-and-plane | If for some $$\alpha $$ $$ \in $$ R, the lines
<br/><br/>L<sub>1</sub> : $${{x + 1} \over 2} = {{y - 2} \over { - 1}} = {{z - 1} \over 1}$$ and
<br/><br>L<sub>2</sub> : $${{x + 2} \over \alpha } = {{y + 1} \over {5 - \alpha }} = {{z + 1} \over 1}$$ are coplanar,
<br/><br/>then the line L<sub>2</sub>
passes through the... | [{"identifier": "A", "content": "(10, 2, 2)"}, {"identifier": "B", "content": "(2, \u201310, \u20132)"}, {"identifier": "C", "content": "(10, \u20132, \u20132)"}, {"identifier": "D", "content": "(\u20132, 10, 2)"}] | ["B"] | null | L<sub>1</sub> : $${{x + 1} \over 2} = {{y - 2} \over { - 1}} = {{z - 1} \over 1}$$ and
<br><br>L<sub>2</sub> : $${{x + 2} \over \alpha } = {{y + 1} \over {5 - \alpha }} = {{z + 1} \over 1}$$ are coplanar.
<br><br>$$ \therefore $$ $$\left| {\matrix{
1 & 3 & 2 \cr
2 & { - 1} & 1 \cr
\alpha &... | mcq | jee-main-2020-online-5th-september-evening-slot | 4,367 |
b7yr7A2nAMnX9Hf8Fojgy2xukfagymx7 | maths | 3d-geometry | lines-and-plane | The distance of the point (1, –2, 3) from<br/><br> the plane x – y + z = 5 measured parallel to <br/><br>the line $${x \over 2} = {y \over 3} = {z \over { - 6}}$$ is :</br></br> | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$${1 \\over 7}$$"}, {"identifier": "D", "content": "$${7 \\over 5}$$"}] | ["B"] | null | Equation of line parallel to $${x \over 2} = {y \over 3} = {z \over { - 6}}$$ passes through $$(1, - 2,3)$$ is<br><br>$${{x - 1} \over 2} = {{y + 2} \over 3} = {{z - 3} \over { - 6}} = r$$<br><br>$$x = 2r + 1$$<br><br>$$y = 3r - 2$$, <br><br>$$z = - 6r + 3$$
<br><br>A point on whole line = (2r + 1, 3r – 2, – 6r + 3).
... | mcq | jee-main-2020-online-4th-september-evening-slot | 4,368 |
ZyHrTWGrvXewytqQX0jgy2xukezm71el | maths | 3d-geometry | lines-and-plane | The foot of the perpendicular drawn from the
point (4, 2, 3) to the line joining the points
(1, –2, 3) and (1, 1, 0) lies on the plane : | [{"identifier": "A", "content": "x \u2013 2y + z = 1"}, {"identifier": "B", "content": "x + 2y \u2013 z = 1"}, {"identifier": "C", "content": "x \u2013 y \u2013 2y = 1"}, {"identifier": "D", "content": "2x + y \u2013 z = 1"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267567/exam_images/staijxnqv8p4honzsnr6.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 3rd September Morning Slot Mathematics - 3D Geometry Question 228 English Explanation">
<br><br>Eq... | mcq | jee-main-2020-online-3rd-september-morning-slot | 4,369 |
Qarw3NWSaAza5ekKgYjgy2xukezbvldh | maths | 3d-geometry | lines-and-plane | A plane passing through the point (3, 1, 1)
contains two lines whose direction ratios are 1,
–2, 2 and 2, 3, –1 respectively. If this plane also
passes through the point ($$\alpha $$, –3, 5), then
$$\alpha $$ is
equal to: | [{"identifier": "A", "content": "-10"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "-5"}] | ["C"] | null | As normal is perpendicular to both the lines so normal vector to the plane is<br><br>
$$\overrightarrow n = \left( {\widehat i - 2\widehat j + 2\widehat k} \right) \times \left( {2\widehat i + 3\widehat j - \widehat k} \right)$$<br><br>
$$\overrightarrow n = \left| {\matrix{
{\widehat i} & {\widehat j} & {... | mcq | jee-main-2020-online-2nd-september-evening-slot | 4,370 |
mKqODJ2TViU4wWfYdSjgy2xukewrehxd | maths | 3d-geometry | lines-and-plane | The plane passing through the points (1, 2, 1),
<br/>(2, 1, 2) and parallel to the line, 2x = 3y, z = 1
<br/>also passes through the point : | [{"identifier": "A", "content": "(0, 6, \u20132)"}, {"identifier": "B", "content": "(\u20132, 0, 1)"}, {"identifier": "C", "content": "(0, \u20136, 2)"}, {"identifier": "D", "content": "(2, 0 \u20131)"}] | ["B"] | null | Equation of plane passing through (2, 1, 2)<br><br>a(x $$-$$ 2) + b(y $$-$$ 1) + c(z $$-$$ 2) = 0 ......(1)<br><br>As point (1, 2, 1) also passes through the plane, so it satisfy the equation, <br><br>a(1 $$-$$ 2) + b(2 $$-$$ 1) + c(1 $$-$$ 2) = 0<br><br>$$ \Rightarrow $$ $$-$$a + b $$-$$ c = 0 ....(2)<br><br>Given lin... | mcq | jee-main-2020-online-2nd-september-morning-slot | 4,371 |
1t1lV3VVbYNVWnJEu8jgy2xukf49l5v7 | maths | 3d-geometry | lines-and-plane | Let a plane P contain two lines
<br/>$$\overrightarrow r = \widehat i + \lambda \left( {\widehat i + \widehat j} \right)$$, $$\lambda \in R$$ and
<br/>$$\overrightarrow r = - \widehat j + \mu \left( {\widehat j - \widehat k} \right)$$, $$\mu \in R$$
<br/>If Q($$\alpha $$, $$\beta $$, $$\gamma $$) is the foot of th... | [] | null | 5 | Given lines,<br><br>$$\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$$ parallel to $$(\widehat i + \widehat j)$$<br><br>Let, $$\overrightarrow {{n_1}} = (\widehat i + \widehat j)$$<br><br>and $$\overrightarrow r = - \widehat j + \mu (\widehat j - \widehat k)$$ parallel to $$(\widehat j - \wideha... | integer | jee-main-2020-online-3rd-september-evening-slot | 4,372 |
a2z3tAA7ldpoqgXxjx1klrep90f | maths | 3d-geometry | lines-and-plane | The distance of the point (1, 1, 9) from the point of intersection of the line
$${{x - 3} \over 1} = {{y - 4} \over 2} = {{z - 5} \over 2}$$
and the plane x + y + z = 17 is : | [{"identifier": "A", "content": "$$19\\sqrt 2 $$"}, {"identifier": "B", "content": "$$2\\sqrt {19} $$"}, {"identifier": "C", "content": "38"}, {"identifier": "D", "content": "$$\\sqrt {38} $$"}] | ["D"] | null | Given, P(1, 1, 9).<br/><br/>Equation of plane x + y + z = 17<br/><br/>Equation of line $$\Rightarrow$$ $${{x - 3} \over 1} = {{y - 4} \over 2} = {{z - 5} \over 2}$$<br/><br/>$$\Rightarrow$$ $${{x - 3} \over 1} = {{y - 4} \over 2} = {{z - 5} \over 2} = \lambda $$ (let)<br/><br/>$$\Rightarrow$$ x = $$\lambda$$ + 3; y = 2... | mcq | jee-main-2021-online-24th-february-morning-slot | 4,373 |
wENOHw7UshvkPbCF2x1klrlu7yu | maths | 3d-geometry | lines-and-plane | The vector equation of the plane passing through the intersection<br/><br/> of the planes $$\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1$$ and $$\overrightarrow r .\left( {\widehat i - 2\widehat j} \right) = - 2$$, and the point (1, 0, 2) is : | [{"identifier": "A", "content": "$$\\overrightarrow r .\\left( {\\widehat i + 7\\widehat j + 3\\widehat k} \\right) = {7 \\over 3}$$"}, {"identifier": "B", "content": "$$\\overrightarrow r .\\left( {\\widehat i + 7\\widehat j + 3\\widehat k} \\right) = 7$$"}, {"identifier": "C", "content": "$$\\overrightarrow r .\\left... | ["B"] | null | Given, point (1, 0, 2)<br/><br/>Equation of plane = <br/><br/>$$\overrightarrow r\,.\,(\widehat i + \widehat j + \widehat k) = 1$$ and $$\overrightarrow r\,.\,(\widehat i - 2\widehat j) = - 2$$<br/><br/>Equation of plane passing through the intersection of given planes is<br/><br/>$$[\overrightarrow r\,.\,(\widehat i ... | mcq | jee-main-2021-online-24th-february-evening-slot | 4,374 |
oAhZZg0ACHs4c6hAhW1kluhipw0 | maths | 3d-geometry | lines-and-plane | Let ($$\lambda$$, 2, 1) be a point on the plane which passes through the point (4, $$-$$2, 2). If the plane is perpendicular to the line joining the points ($$-$$2, $$-$$21, 29) and ($$-$$1, $$-$$16, 23), then $${\left( {{\lambda \over {11}}} \right)^2} - {{4\lambda } \over {11}} - 4$$ is equal to __________. | [] | null | 8 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265776/exam_images/vghtupk7erbaxkhau8et.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th February Morning Shift Mathematics - 3D Geometry Question 209 English Explanation"><br><br>$$... | integer | jee-main-2021-online-26th-february-morning-slot | 4,375 |
0syG6wUdgkgrwz9vew1kluxx2hr | maths | 3d-geometry | lines-and-plane | Let L be a line obtained from the intersection of two planes x + 2y + z = 6 and y + 2z = 4. If point P($$\alpha$$, $$\beta$$, $$\gamma$$) is the foot of perpendicular from (3, 2, 1) on L, then the <br/>value of 21($$\alpha$$ + $$\beta$$ + $$\gamma$$) equals : | [{"identifier": "A", "content": "102"}, {"identifier": "B", "content": "142"}, {"identifier": "C", "content": "136"}, {"identifier": "D", "content": "68"}] | ["A"] | null | Dr's of line $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 2 & 1 \cr
0 & 1 & 2 \cr
} } \right| = 3\widehat i - 2\widehat j + \widehat k$$<br><br>Dr/s : - (3, $$-$$2, 1)<br><br>Points on the line ($$-$$2, 4, 0)<br><br>Equation of the line $${{x + 2} \over ... | mcq | jee-main-2021-online-26th-february-evening-slot | 4,376 |
CkOToyCiyoKHvpwLR91kmhx32x5 | maths | 3d-geometry | lines-and-plane | Let P be a plane lx + my + nz = 0 containing <br/><br/>the line, $${{1 - x} \over 1} = {{y + 4} \over 2} = {{z + 2} \over 3}$$. If plane P divides the line segment AB joining <br/><br/>points A($$-$$3, $$-$$6, 1) and B(2, 4, $$-$$3) in ratio k : 1 then the value of k is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "1.5"}, {"identifier": "D", "content": "4"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264908/exam_images/td5ojeuzdtmnzkretbhj.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Morning Shift Mathematics - 3D Geometry Question 205 English Explanation">
<br>Line lie... | mcq | jee-main-2021-online-16th-march-morning-shift | 4,377 |
jTuoAqCOlT0ueBNhrI1kmizayck | maths | 3d-geometry | lines-and-plane | If the distance of the point (1, $$-$$2, 3) from the plane x + 2y $$-$$ 3z + 10 = 0 measured parallel to the line, $${{x - 1} \over 3} = {{2 - y} \over m} = {{z + 3} \over 1}$$ is $$\sqrt {{7 \over 2}} $$, then the value of |m| is equal to _________. | [] | null | 2 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267379/exam_images/vmb11b1oodts8ory1wgg.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Evening Shift Mathematics - 3D Geometry Question 202 English Explanation">
<br>Given li... | integer | jee-main-2021-online-16th-march-evening-shift | 4,378 |
oMM8M7sXfZX0uLWycA1kmjbpapm | maths | 3d-geometry | lines-and-plane | If the equation of the plane passing through the line of intersection of the planes 2x $$-$$ 7y + 4z $$-$$ 3 = 0, 3x $$-$$ 5y + 4z + 11 = 0 and the point ($$-$$2, 1, 3) is ax + by + cz $$-$$ 7 = 0, then the value of 2a + b + c $$-$$ 7 is ____________. | [] | null | 4 | Equation of plane can be written using family of planes : P<sub>1</sub> + $$\lambda$$P<sub>2</sub> = 0<br><br>(2x $$-$$ 7y + 4z $$-$$ 3) + $$\lambda$$ (3x $$-$$ 5y + 4z + 11) = 0<br><br>It passes through ($$-$$2, 1, 3)<br><br>$$ \therefore $$ ($$-$$4 + 7 + 12 $$-$$ 3) + $$\lambda$$ ($$-$$6 $$-$$ 5 + 12 + 11) = 0<br><br... | integer | jee-main-2021-online-17th-march-morning-shift | 4,379 |
1krq0kpif | maths | 3d-geometry | lines-and-plane | Let P be a plane passing through the points (1, 0, 1), (1, $$-$$2, 1) and (0, 1, $$-$$2). Let a vector $$\overrightarrow a = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$$ be such that $$\overrightarrow a $$ is parallel to the plane P, perpendicular to $$(\widehat i + 2\widehat j + 3\widehat k)$$ and $$\ov... | [] | null | 81 | Equation of plane :<br><br>$$\left| {\matrix{
{x - 1} & {y - 0} & {z - 1} \cr
{1 - 1} & 2 & {1 - 1} \cr
{1 - 0} & {0 - 1} & {1 + 2} \cr
} } \right| = 0$$<br><br>$$ \Rightarrow 3x - z - 2 = 0$$<br><br>$$\overrightarrow a = \alpha \widehat i + \beta \widehat j + \gamma \widehat k... | integer | jee-main-2021-online-20th-july-morning-shift | 4,382 |
1krrtutnt | maths | 3d-geometry | lines-and-plane | Consider the line L given by the equation <br/><br/>$${{x - 3} \over 2} = {{y - 1} \over 1} = {{z - 2} \over 1}$$. <br/><br/>Let Q be the mirror image of the point (2, 3, $$-$$1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points i... | [{"identifier": "A", "content": "($$-$$1, 1, 2)"}, {"identifier": "B", "content": "(1, 1, 1)"}, {"identifier": "C", "content": "(1, 1, 2)"}, {"identifier": "D", "content": "(1, 2, 2)"}] | ["D"] | null | Plane p is $${ \bot ^r}$$ to line $${{x - 3} \over 2} = {{y - 1} \over 1} = {{z - 2} \over 1}$$ & passes through pt. (2, 3) equation of plane p <br><br>2(x $$-$$ 2) + 1(y $$-$$ 3) + 1 (z + 1) = 0<br><br>2x + y + z $$-$$ 6 = 0<br><br>Point (1, 2, 2) satisfies above equation | mcq | jee-main-2021-online-20th-july-evening-shift | 4,383 |
1krtbltik | maths | 3d-geometry | lines-and-plane | Let L be the line of intersection of planes $$\overrightarrow r .(\widehat i - \widehat j + 2\widehat k) = 2$$ and $$\overrightarrow r .(2\widehat i + \widehat j - \widehat k) = 2$$. If $$P(\alpha ,\beta ,\gamma )$$ is the foot of perpendicular on L from the point (1, 2, 0), then the value of $$35(\alpha + \beta + \g... | [{"identifier": "A", "content": "101"}, {"identifier": "B", "content": "119"}, {"identifier": "C", "content": "143"}, {"identifier": "D", "content": "134"}] | ["B"] | null | $${P_1}:x - y + 2z = 2$$<br><br>$${P_2}:2x + y - 3 = 2$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265409/exam_images/gzyjdvs460f1b9cskjhd.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 22th July Evening Shift Mathematic... | mcq | jee-main-2021-online-22th-july-evening-shift | 4,384 |
1krw1mzhz | maths | 3d-geometry | lines-and-plane | Let the foot of perpendicular from a point P(1, 2, $$-$$1) to the straight line $$L:{x \over 1} = {y \over 0} = {z \over { - 1}}$$ be N. Let a line be drawn from P parallel to the plane x + y + 2z = 0 which meets L at point Q. If $$\alpha$$ is the acute angle between the lines PN and PQ, then cos$$\alpha$$ is equal to ... | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 5 }}$$"}, {"identifier": "B", "content": "$${{\\sqrt 3 } \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${1 \\over {2\\sqrt 3 }}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264384/exam_images/ockxunbxex6ruzroagy0.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th July Morning Shift Mathematics - 3D Geometry Question 186 English Explanation 1"><br>$$\overr... | mcq | jee-main-2021-online-25th-july-morning-shift | 4,385 |
1krxgs6qs | maths | 3d-geometry | lines-and-plane | For real numbers $$\alpha$$ and $$\beta$$ $$\ne$$ 0, if the point of intersection of the straight lines<br/><br/>$${{x - \alpha } \over 1} = {{y - 1} \over 2} = {{z - 1} \over 3}$$ and $${{x - 4} \over \beta } = {{y - 6} \over 3} = {{z - 7} \over 3}$$, lies on the plane x + 2y $$-$$ z = 8, then $$\alpha$$ $$-$$ $$\beta... | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "7"}] | ["D"] | null | First line is ($$\phi$$ + $$\alpha$$, 2$$\phi$$ + 1, 3$$\phi$$ + 1)<br><br>and second line is (q$$\beta$$ + 4, 3q + 6, 3q + 7)<br><br>For intersection<br><br>$$\phi$$ + $$\alpha$$ = q$$\beta$$ + 4 ...... (i)<br><br>2$$\phi$$ + 1 = 3q + 6 .... (ii)<br><br>3$$\phi$$ + 1 = 3q + 7 ...... (iii)<br><br>for (ii) & (iii) $... | mcq | jee-main-2021-online-27th-july-evening-shift | 4,386 |
1kryflta1 | maths | 3d-geometry | lines-and-plane | The distance of the point P(3, 4, 4) from the point of intersection of the line joining the points. Q(3, $$-$$4, $$-$$5) and R(2, $$-$$3, 1) and the plane 2x + y + z = 7, is equal to ______________. | [] | null | 7 | $$\overrightarrow {QR} : - {{x - 3} \over 1} = {{y + 4} \over { - 1}} = {{z + 5} \over { - 6}} = r$$<br><br>$$ \Rightarrow (x,y,z) \equiv (r + 3, - r - 4, - 6r - 5)$$<br><br>Now, satisfying it in the given plane.<br><br>We get r = $$-$$2<br><br>so, required point of intersection is T(1, $$-$$2, 7).<br><br>Hence, PT = 7... | integer | jee-main-2021-online-27th-july-evening-shift | 4,387 |
1krzrobee | maths | 3d-geometry | lines-and-plane | If the lines $${{x - k} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$ and <br/>$${{x + 1} \over 3} = {{y + 2} \over 2} = {{z + 3} \over 1}$$ are co-planar, then the value of k is _____________. | [] | null | 1 | $$\left| {\matrix{
{k + 1} & 4 & 6 \cr
1 & 2 & 3 \cr
3 & 2 & 1 \cr
} } \right| = 0$$<br><br>$$ \Rightarrow $$ $$(k + 1)[2 - 6] - 4[1 - 9] + 6[2 - 6] = 0$$<br><br>$$ \Rightarrow $$ $$k = 1$$ | integer | jee-main-2021-online-25th-july-evening-shift | 4,388 |
1ks0cxriw | maths | 3d-geometry | lines-and-plane | Let a plane P pass through the point (3, 7, $$-$$7) and contain the line, $${{x - 2} \over { - 3}} = {{y - 3} \over 2} = {{z + 2} \over 1}$$. If distance of the plane P from the origin is d, then d<sup>2</sup> is equal to ______________. | [] | null | 3 | $$\overrightarrow {BA} = (\widehat i + 4\widehat j - 5\widehat k)$$<br><br>$$\overrightarrow {BA} \times \overrightarrow l = \overrightarrow n = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
{ - 3} & 2 & 1 \cr
1 & 4 & { - 5} \cr
} } \right|$$<br><br>$$a\wid... | integer | jee-main-2021-online-27th-july-morning-shift | 4,389 |
1ktbf6smu | maths | 3d-geometry | lines-and-plane | A plane P contains the line $$x + 2y + 3z + 1 = 0 = x - y - z - 6$$, and is perpendicular to the plane $$ - 2x + y + z + 8 = 0$$. Then which of the following points lies on P? | [{"identifier": "A", "content": "($$-$$1, 1, 2)"}, {"identifier": "B", "content": "(0, 1, 1)"}, {"identifier": "C", "content": "(1, 0, 1)"}, {"identifier": "D", "content": "(2, $$-$$1, 1)"}] | ["B"] | null | Equation of plane P can be assumed as<br><br>P : x + 2y + 3z + 1 + $$\lambda$$ (x $$-$$ y $$-$$ z $$-$$ 6) = 0<br><br>$$\Rightarrow$$ P : (1 + $$\lambda$$)x + (2 $$-$$ $$\lambda$$)y + (3 $$-$$ $$\lambda$$)z + 1 $$-$$ 6$$\lambda$$ = 0<br><br>$$ \Rightarrow {\overrightarrow n _1} = (1 + \lambda )\widehat i + (2 - \lambda... | mcq | jee-main-2021-online-26th-august-morning-shift | 4,390 |
1ktbi6k7h | maths | 3d-geometry | lines-and-plane | Let the line L be the projection of the line $${{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 4} \over 2}$$ in the plane x $$-$$ 2y $$-$$ z = 3. If d is the distance of the point (0, 0, 6) from L, then d<sup>2</sup> is equal to _______________. | [] | null | 26 | To find the projection let's find the foot of perpendicular from $(1,3$,
4) to plane $x-2 y-z=3$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lfg2ca35/20453aa3-5bf8-445c-837d-c8b6d848626f/8ecab100-c6b2-11ed-b4b3-b306e87ca523/file-1lfg2ca36.png?format=png" data-orsrc="https://app-content.cdn.... | integer | jee-main-2021-online-26th-august-morning-shift | 4,391 |
1ktd44pm7 | maths | 3d-geometry | lines-and-plane | Let Q be the foot of the perpendicular from the point P(7, $$-$$2, 13) on the plane containing the lines $${{x + 1} \over 6} = {{y - 1} \over 7} = {{z - 3} \over 8}$$ and $${{x - 1} \over 3} = {{y - 2} \over 5} = {{z - 3} \over 7}$$. Then (PQ)<sup>2</sup>, is equal to ___________. | [] | null | 96 | Containing the line $$\left| {\matrix{
{x + 1} & {y - 1} & {z - 3} \cr
6 & 7 & 8 \cr
3 & 5 & 7 \cr
} } \right| = 0$$<br><br>$$9(x + 1) - 18(y - 1) + 9(z - 3) = 0$$<br><br>$$x - 2y + z = 0$$<br><br>$$PQ = \left| {{{7 + 4 + 13} \over {\sqrt 6 }}} \right| = 4\sqrt 6 $$<br><br>$$P{Q... | integer | jee-main-2021-online-26th-august-evening-shift | 4,392 |
1kteihgh9 | maths | 3d-geometry | lines-and-plane | The distance of the point (1, $$-$$2, 3) from the plane x $$-$$ y + z = 5 measured parallel to a line, whose direction ratios are 2, 3, $$-$$6 is : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263282/exam_images/vaartuqi2uokdkrcurcw.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Morning Shift Mathematics - 3D Geometry Question 175 English Explanation"><br><br>$$(1... | mcq | jee-main-2021-online-27th-august-morning-shift | 4,393 |
1ktekeym0 | maths | 3d-geometry | lines-and-plane | Equation of a plane at a distance $$\sqrt {{2 \over {21}}} $$ from the origin, which contains the line of intersection of the planes x $$-$$ y $$-$$ z $$-$$ 1 = 0 and 2x + y $$-$$ 3z + 4 = 0, is : | [{"identifier": "A", "content": "$$3x - y - 5z + 2 = 0$$"}, {"identifier": "B", "content": "$$3x - 4z + 3 = 0$$"}, {"identifier": "C", "content": "$$ - x + 2y + 2z - 3 = 0$$"}, {"identifier": "D", "content": "$$4x - y - 5z + 2 = 0$$"}] | ["D"] | null | Required equation of plane<br><br>$${P_1} + \lambda {P_2} = 0$$<br><br>$$(x - y - z - 1) + \lambda (2x + y - 3z + 4) = 0$$<br><br>Given that its dist. From origin is $${2 \over {\sqrt {21} }}$$<br><br>Thus, $${{|4\lambda - 1|} \over {\sqrt {{{(2\lambda + 1)}^2} + {{(\lambda - 1)}^2} + {{( - 3\lambda - 1)}^2}} }} = ... | mcq | jee-main-2021-online-27th-august-morning-shift | 4,394 |
1ktfzryxs | maths | 3d-geometry | lines-and-plane | The equation of the plane passing through the line of intersection of the planes $$\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1$$ and $$\overrightarrow r .\left( {2\widehat i + 3\widehat j - \widehat k} \right) + 4 = 0$$ and parallel to the x-axis is : | [{"identifier": "A", "content": "$$\\overrightarrow r .\\left( {\\widehat j - 3\\widehat k} \\right) + 6 = 0$$"}, {"identifier": "B", "content": "$$\\overrightarrow r .\\left( {\\widehat i + 3\\widehat k} \\right) + 6 = 0$$"}, {"identifier": "C", "content": "$$\\overrightarrow r .\\left( {\\widehat i - 3\\widehat k} \\... | ["A"] | null | Equation of planes are<br><br>$$\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) - 1 = 0 \Rightarrow x + y + z - 1 = 0$$<br><br>and $$\overrightarrow r .\left( {2\widehat i + 3\widehat j - \widehat k} \right) + 4 = 0 \Rightarrow 2x + 3y - z + 4 = 0$$<br><br>equation of planes through line of int... | mcq | jee-main-2021-online-27th-august-evening-shift | 4,395 |
1ktgobl7e | maths | 3d-geometry | lines-and-plane | Let S be the mirror image of the point Q(1, 3, 4) with respect to the plane 2x $$-$$ y + z + 3 = 0 and let R(3, 5, $$\gamma$$) be a point of this plane. Then the square of the length of the line segment SR is ___________. | [] | null | 72 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266720/exam_images/h71pebht6whdybpiqe1o.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Evening Shift Mathematics - 3D Geometry Question 173 English Explanation"> <br><br>Sin... | integer | jee-main-2021-online-27th-august-evening-shift | 4,396 |
1ktiom3vk | maths | 3d-geometry | lines-and-plane | Let the equation of the plane, that passes through the point (1, 4, $$-$$3) and contains the line of intersection of the <br/>planes 3x $$-$$ 2y + 4z $$-$$ 7 = 0 <br/>and x + 5y $$-$$ 2z + 9 = 0, be <br/>$$\alpha$$x + $$\beta$$y + $$\gamma$$z + 3 = 0, then $$\alpha$$ + $$\beta$$ + $$\gamma$$ is equal to : | [{"identifier": "A", "content": "$$-$$23"}, {"identifier": "B", "content": "$$-$$15"}, {"identifier": "C", "content": "23"}, {"identifier": "D", "content": "15"}] | ["A"] | null | 3x $$-$$ 2y + 4z $$-$$ 7 + $$\lambda$$(x + 5y $$-$$ 2z + 9) = 0<br><br>(3 + $$\lambda$$)x + (5$$\lambda$$ $$-$$ 2)y + (4 $$-$$ 2$$\lambda$$)z + 9$$\lambda$$ $$-$$ 7 = 0<br><br>passing through (1, 4, $$-$$3)<br><br>$$\Rightarrow$$ 3 + $$\lambda$$ + 20$$\lambda$$ $$-$$ 8 $$-$$ 12 + 6$$\lambda$$ + 9$$\lambda$$ $$-$$ 7 = 0... | mcq | jee-main-2021-online-31st-august-morning-shift | 4,397 |
1ktis4alt | maths | 3d-geometry | lines-and-plane | The square of the distance of the point of intersection <br/><br/>of the line $${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6}$$ and the plane $$2x - y + z = 6$$ from the point ($$-$$1, $$-$$1, 2) is __________. | [] | null | 61 | $${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6} = \lambda $$<br><br>$$x = 2\lambda + 1,y = 3\lambda + 2,z = 6\lambda - 1$$<br><br>for point of intersection of line & plane<br><br>$$2(2\lambda + 1) - (3\lambda + 2) + (6\lambda - 1) = 6$$<br><br>$$7\lambda = 7 \Rightarrow \lambda = 1$$<br><br>poin... | integer | jee-main-2021-online-31st-august-morning-shift | 4,398 |
1ktk5cco4 | maths | 3d-geometry | lines-and-plane | The distance of the point ($$-$$1, 2, $$-$$2) from the line of intersection of the planes 2x + 3y + 2z = 0 and x $$-$$ 2y + z = 0 is : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${5 \\over 2}$$"}, {"identifier": "C", "content": "$${{\\sqrt {42} } \\over 2}$$"}, {"identifier": "D", "content": "$${{\\sqrt {34} } \\over 2}$$"}] | ["D"] | null | P<sub>1</sub> : 2x + 3y + 2z = 0<br><br>$$\Rightarrow$$ $${\overrightarrow n _1} = 2\widehat i + 3\widehat j + 2\widehat k$$<br><br>P<sub>2</sub> : x $$-$$ 2y + z = 0<br><br>$$\Rightarrow$$ $${\overrightarrow n _2} = \widehat i - 2\widehat j + \widehat k$$<br><br>Direction vector of line L which is line of intersection... | mcq | jee-main-2021-online-31st-august-evening-shift | 4,399 |
1ktkdi4ed | maths | 3d-geometry | lines-and-plane | Suppose, the line $${{x - 2} \over \alpha } = {{y - 2} \over { - 5}} = {{z + 2} \over 2}$$ lies on the plane $$x + 3y - 2z + \beta = 0$$. Then $$(\alpha + \beta )$$ is equal to _______. | [] | null | 7 | <p>Given equation of line</p>
<p>$${{x - 2} \over \alpha } = {{y - 2} \over { - 5}} = {{z + 2} \over 2}$$ ...... (i)</p>
<p>and plane x + 3y $$-$$ 2z + $$\beta$$ = 0 ...... (ii)</p>
<p>Line (i) passes through (2, 2, $$-$$2)</p>
<p>which lies on plane (ii).</p>
<p>$$\therefore$$ 2 + 6 + 4 + $$\beta$$ = 0 $$\Rightarrow$$... | integer | jee-main-2021-online-31st-august-evening-shift | 4,400 |
1l546e82f | maths | 3d-geometry | lines-and-plane | <p>Let $${P_1}:\overrightarrow r \,.\,\left( {2\widehat i + \widehat j - 3\widehat k} \right) = 4$$ be a plane. Let P<sub>2</sub> be another plane which passes through the points (2, $$-$$3, 2), (2, $$-$$2, $$-$$3) and (1, $$-$$4, 2). If the direction ratios of the line of intersection of P<sub>1</sub> and P<sub>2</sub... | [] | null | 28 | <p>Direction ratio of normal to $${P_1} \equiv < 2,1, - 3 > $$</p>
<p>and that of $${P_2} \equiv \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
0 & 1 & { - 5} \cr
{ - 1} & { - 2} & 5 \cr
} } \right| = - 5\widehat i - \widehat j( - 5) + \widehat k(1)$$</p>
<p>i.e. $$ < - 5,5,1 > $$</... | integer | jee-main-2022-online-29th-june-morning-shift | 4,401 |
1l54bcrlo | maths | 3d-geometry | lines-and-plane | <p>Let $${{x - 2} \over 3} = {{y + 1} \over { - 2}} = {{z + 3} \over { - 1}}$$ lie on the plane $$px - qy + z = 5$$, for some p, q $$\in$$ R. The shortest distance of the plane from the origin is :</p> | [{"identifier": "A", "content": "$$\\sqrt {{3 \\over {109}}} $$"}, {"identifier": "B", "content": "$$\\sqrt {{5 \\over {142}}} $$"}, {"identifier": "C", "content": "$${5 \\over {\\sqrt {71} }}$$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt {142} }}$$"}] | ["B"] | null | $\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1}=\lambda$
<br/><br/>
$(3 \lambda+2,-2 \lambda-1,-\lambda-3)$ lies on plane $p x-q y+z=5$ <br/><br/>$p(3 \lambda+2)-q(-2 \lambda-1)+(-\lambda-3)=5$
<br/><br/>
$\lambda(3 p+2 q-1)+(2 p+q-8)=0$
<br/><br/>
$3 p+2 q-1=0\} p=15$
<br/><br/>
$2 p+q-8=0\} q=-22$
<br/><br/>
Equation of... | mcq | jee-main-2022-online-29th-june-evening-shift | 4,402 |
1l54t8ew5 | maths | 3d-geometry | lines-and-plane | <p>Let Q be the mirror image of the point P(1, 2, 1) with respect to the plane x + 2y + 2z = 16. Let T be a plane passing through the point Q and contains the line $$\overrightarrow r = - \widehat k + \lambda \left( {\widehat i + \widehat j + 2\widehat k} \right),\,\lambda \in R$$. Then, which of the following point... | [{"identifier": "A", "content": "(2, 1, 0)"}, {"identifier": "B", "content": "(1, 2, 1)"}, {"identifier": "C", "content": "(1, 2, 2)"}, {"identifier": "D", "content": "(1, 3, 2)"}] | ["B"] | null | $P(1,2,1)$ image in plane $x+2 y+2 z=16$
<br><br>
$$
\begin{aligned}
& \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=\frac{-2(1+2 \times 2+2 \times 1-16)}{1^{2}+2^{2}+2^{2}} \\\\
& \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=2 \\\\
& Q(3,6,5) \\\\
& \vec{r}=-\hat{k}+\lambda(\hat{i}+\hat{j}+2 \hat{k})
\end{ali... | mcq | jee-main-2022-online-29th-june-evening-shift | 4,403 |
1l566xtof | maths | 3d-geometry | lines-and-plane | <p>If two distinct point Q, R lie on the line of intersection of the planes $$ - x + 2y - z = 0$$ and $$3x - 5y + 2z = 0$$ and $$PQ = PR = \sqrt {18} $$ where the point P is (1, $$-$$2, 3), then the area of the triangle PQR is equal to :</p> | [{"identifier": "A", "content": "$${2 \\over 3}\\sqrt {38} $$"}, {"identifier": "B", "content": "$${4 \\over 3}\\sqrt {38} $$"}, {"identifier": "C", "content": "$${8 \\over 3}\\sqrt {38} $$"}, {"identifier": "D", "content": "$$\\sqrt {{{152} \\over 3}} $$"}] | ["B"] | null | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5obyhm8/67807093-6d4d-453b-a541-6801368b15df/45fa9b00-0544-11ed-987f-3938cfc0f7f1/file-1l5obyhm9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5obyhm8/67807093-6d4d-453b-a541-6801368b15df/45fa9b00-0544-11ed-987f-3938cfc0f7f... | mcq | jee-main-2022-online-28th-june-morning-shift | 4,404 |
1l58a1l78 | maths | 3d-geometry | lines-and-plane | <p>Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x $$-$$ 3y + 5z = 8. If the mirror image of the point $$\left( {2, - {1 \over 2},2} \right)$$ in the rotated plane is B(a, b, c), then :</p> | [{"identifier": "A", "content": "$${a \\over 8} = {b \\over 5} = {c \\over { - 4}}$$"}, {"identifier": "B", "content": "$${a \\over 4} = {b \\over 5} = {c \\over { - 2}}$$"}, {"identifier": "C", "content": "$${a \\over 8} = {b \\over { - 5}} = {c \\over 4}$$"}, {"identifier": "D", "content": "$${a \\over 4} = {b \\over... | ["A"] | null | <p>Consider the equation of plane,</p>
<p>$$P:(2x + 3y + z + 20) + \lambda (x - 3y + 5z - 8) = 0$$</p>
<p>$$P:(2 + \lambda )x + 3(3 - 3\lambda )y + 1(1 + 5\lambda )z + (20 - 8\lambda ) = 0$$</p>
<p>$$\because$$ Plane P is perpendicular to $$2x + 3y + z + 20 = 0$$</p>
<p>So, $$4 + 2\lambda + 9 - 9\lambda + 1 + 5\lambd... | mcq | jee-main-2022-online-26th-june-morning-shift | 4,407 |
1l5aiu9gk | maths | 3d-geometry | lines-and-plane | <p>Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a line L passing through (1, $$-$$1, $$-$$1), parallel to the line PQ meets the plane S at R, then QR<sup>2</sup> is equal to :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "11"}] | ["B"] | null | <p>As L is parallel to PQ d.r.s of S is <1, 1, 1></p>
<p>$$\therefore$$ $$L \equiv {{x - 1} \over 1} = {{y + 1} \over 1} = {{z + 1} \over 1}$$</p>
<p>Point of intersection of L and S be $$\lambda$$</p>
<p>$$ \Rightarrow (\lambda + 1) + (\lambda - 1) + (\lambda - 1) = S$$</p>
<p>$$ \Rightarrow \lambda = 2$$</p>
<p>$... | mcq | jee-main-2022-online-25th-june-morning-shift | 4,409 |
1l5ajvyuf | maths | 3d-geometry | lines-and-plane | <p>Let the lines</p>
<p>$${L_1}:\overrightarrow r = \lambda \left( {\widehat i + 2\widehat j + 3\widehat k} \right),\,\lambda \in R$$</p>
<p>$${L_2}:\overrightarrow r = \left( {\widehat i + 3\widehat j + \widehat k} \right) + \mu \left( {\widehat i + \widehat j + 5\widehat k} \right);\,\mu \in R$$,</p>
<p>intersect... | [] | null | 5 | <p>As plane is parallel to both the lines we have d.r's of normal to the plane as <7, $$-$$2, $$-$$1></p>
<p>$$\left( {from\,\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 2 & 3 \cr
1 & 1 & 5 \cr
} } \right| = 7\widehat i - \widehat j(2) + \widehat k( - 1)} \right)$$</p>
<p>Also po... | integer | jee-main-2022-online-25th-june-morning-shift | 4,410 |
1l6dwfnc0 | maths | 3d-geometry | lines-and-plane | <p>Let $$\mathrm{P}$$ be the plane containing the straight line $$\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$$ and perpendicular to the plane containing the straight lines $$\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$$ and $$\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$$. If $$\mathrm{d}$$ is the distance of $$\mathrm{P}$$ from the p... | [{"identifier": "A", "content": "$$\\frac{147}{2}$$"}, {"identifier": "B", "content": "96"}, {"identifier": "C", "content": "$$\\frac{32}{3}$$"}, {"identifier": "D", "content": "54"}] | ["C"] | null | Let $\langle a, b, c\rangle$ be direction ratios of plane containing
<br/><br/>
$$
\begin{aligned}
&\text { lines } \frac{x}{2}=\frac{y}{3}=\frac{z}{5} \text { and } \frac{x}{3}=\frac{y}{7}=\frac{z}{8} . \\\\
&\therefore \quad 2 a+3 b+5 c=0 \quad \ldots \text { (i) } \\\\
&\text { and } 3 a+7 b+8 c=0 \quad \ldots \text... | mcq | jee-main-2022-online-25th-july-morning-shift | 4,412 |
1l6dxjgo4 | maths | 3d-geometry | lines-and-plane | <p>The line of shortest distance between the lines $$\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}$$ and $$\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}$$ makes an angle of $$\cos ^{-1}\left(\sqrt{\frac{2}{27}}\right)$$ with the plane $$\mathrm{P}: \mathrm{a} x-y-z=0$$, $$(a>0)$$. If the image of the point $$(1,1,-5)$$ in the... | [] | null | 3 | DR's of line of shortest distance<br/><br/>
$$
\left|\begin{array}{lll}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
0 & 1 & 1 \\
2 & 2 & 1
\end{array}\right|=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}
$$<br/><br/>
angle between line and plane is $\cos ^{-1} \sqrt{\frac{2}{27}}=\alpha$<br/><br... | integer | jee-main-2022-online-25th-july-morning-shift | 4,413 |
1l6hzmtfd | maths | 3d-geometry | lines-and-plane | <p>The largest value of $$a$$, for which the perpendicular distance of the plane containing the lines $$
\vec{r}=(\hat{i}+\hat{j})+\lambda(\hat{i}+a \hat{j}-\hat{k})$$ and $$\vec{r}=(\hat{i}+\hat{j})+\mu(-\hat{i}+\hat{j}-a \hat{k})$$ from the point $$(2,1,4)$$ is $$\sqrt{3}$$, is _________.</p> | [] | null | 2 | <p>Normal to plane $$ = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & a & { - 1} \cr
{ - 1} & 1 & { - a} \cr
} } \right|$$</p>
<p>$$ = \widehat i(1 - {a^2}) - \widehat j( - a - 1) + \widehat k(1 + a)$$</p>
<p>$$ = (1 - a)\widehat i + \widehat j + \widehat k$$</p>
<p>$$\therefore$$... | integer | jee-main-2022-online-26th-july-evening-shift | 4,414 |
1l6hzyvwf | maths | 3d-geometry | lines-and-plane | <p>The plane passing through the line $$L: l x-y+3(1-l) z=1, x+2 y-z=2$$ and perpendicular to the plane $$3 x+2 y+z=6$$ is $$3 x-8 y+7 z=4$$. If $$\theta$$ is the acute angle between the line $$L$$ and the $$y$$-axis, then $$415 \cos ^{2} \theta$$ is equal to _____________.</p> | [] | null | 125 | <p>$$L:lx - y + 3(1 - l)z = 1$$, $$x + 2y - z = 2$$ and plane containing the line $$p:3x - 8y + 7z = 4$$</p>
<p>Let $$\overrightarrow n $$ be the vector parallel to L.</p>
<p>then $$\overrightarrow n = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
l & { - 1} & {3(1 - l)} \cr
1 & 2 & { - ... | integer | jee-main-2022-online-26th-july-evening-shift | 4,415 |
1l6kkntgj | maths | 3d-geometry | lines-and-plane | <p>If the line of intersection of the planes $$a x+b y=3$$ and $$a x+b y+c z=0$$, a $$>0$$ makes an angle $$30^{\circ}$$ with the plane $$y-z+2=0$$, then the direction cosines of the line are :</p> | [{"identifier": "A", "content": "$$\\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}}, 0$$"}, {"identifier": "B", "content": "$$\\frac{1}{\\sqrt{2}}, \\pm \\,\\frac{1}{\\sqrt{2}}, 0$$"}, {"identifier": "C", "content": "$$\\frac{1}{\\sqrt{5}},-\\frac{2}{\\sqrt{5}}, 0$$"}, {"identifier": "D", "content": "$$\\frac{1}{2},-\\frac{\... | ["B"] | null | <p>$${P_1}:ax + by + 0z = 3$$, normal vector : $${\overrightarrow n _1} = (a,b,0)$$</p>
<p>$${P_2}:ax + by + cz = 0$$, normal vector : $${\overrightarrow n _2} = (a,b,c)$$</p>
<p>Vector parallel to the line of intersection $$ = {\overrightarrow n _1} \times {\overrightarrow n _2}$$</p>
<p>$${\overrightarrow n _1} \time... | mcq | jee-main-2022-online-27th-july-evening-shift | 4,417 |
1l6nnkfa5 | maths | 3d-geometry | lines-and-plane | <p>Let the lines <br/><br/>$$\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}$$ and <br/><br/>$$\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}$$ be coplanar <br/><br/>and $$\mathrm{P}$$ be the plane containing these two lines. <br/><br/>Then which of the following points does <b>NOT</b> lie on P?</p> | [{"identifier": "A", "content": "$$(0,-2,-2)$$"}, {"identifier": "B", "content": "$$(-5,0,-1)$$"}, {"identifier": "C", "content": "$$(3,-1,0)$$"}, {"identifier": "D", "content": "$$(0,4,5)$$"}] | ["D"] | null | <p>$${L_1}:{{x - 1} \over \lambda } = {{y - 2} \over 1} = {{z - 3} \over 2}$$,</p>
<p>through a point $${\overrightarrow a _1} \equiv (1,2,3)$$</p>
<p>parallel to $${\overrightarrow b _1} \equiv (\lambda ,1,2)$$</p>
<p>$${L_2}:{{x + 26} \over { - 2}} = {{y + 18} \over 3} = {{z + 28} \over \lambda }$$</p>
<p>through a p... | mcq | jee-main-2022-online-28th-july-evening-shift | 4,418 |
1l6p39yy3 | maths | 3d-geometry | lines-and-plane | <p>Let a line with direction ratios $$a,-4 a,-7$$ be perpendicular to the lines with direction ratios $$3,-1,2 b$$ and $$b, a,-2$$. If the point of intersection of the line $$\frac{x+1}{a^{2}+b^{2}}=\frac{y-2}{a^{2}-b^{2}}=\frac{z}{1}$$ and the plane $$x-y+z=0$$ is $$(\alpha, \beta, \gamma)$$, then $$\alpha+\beta+\gamm... | [] | null | 10 | <p>Given $$a\,.\,3 + ( - 4a)( - 1) + ( - 7)2b = 0$$ ...... (1)</p>
<p>and $$ab - 4{a^2} + 14 = 0$$ ....... (2)</p>
<p>$$ \Rightarrow {a^2} = 4$$ and $${b^2} = 1$$</p>
<p>$$\therefore$$ $$L \equiv {{x + 1} \over 5} = {{y - 2} \over 3} = {z \over 1} = \lambda $$ (say)</p>
<p>$$\Rightarrow$$ General point on line is $$(5\... | integer | jee-main-2022-online-29th-july-morning-shift | 4,420 |
1ldo4up2x | maths | 3d-geometry | lines-and-plane | <p>Let the plane P pass through the intersection of the planes $$2x+3y-z=2$$ and $$x+2y+3z=6$$, and be perpendicular to the plane $$2x+y-z+1=0$$. If d is the distance of P from the point ($$-$$7, 1, 1), then $$\mathrm{d^{2}}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{250}{83}$$"}, {"identifier": "B", "content": "$$\\frac{250}{82}$$"}, {"identifier": "C", "content": "$$\\frac{15}{53}$$"}, {"identifier": "D", "content": "$$\\frac{25}{83}$$"}] | ["A"] | null | Plane $P$, is passing through intersection of the two planes, so,
<br/><br/>$$
\begin{aligned}
& 2 x+3 y-z-2+\lambda(x+2 y+3 z-6)=0 \\\\
& x(2+\lambda)+y(3+2 \lambda)+z(3 \lambda-1)-2-6 \lambda=0
\end{aligned}
$$
<br/><br/>It is perpendicular with plane, $2 x+y-2+1=0$
<br/><br/>So, $(2+\lambda) 2+(3+2 \lambda) 1+(3 \la... | mcq | jee-main-2023-online-1st-february-evening-shift | 4,421 |
1ldo7avkk | maths | 3d-geometry | lines-and-plane | <p>The point of intersection $$\mathrm{C}$$ of the plane $$8 x+y+2 z=0$$ and the line joining the points $$\mathrm{A}(-3,-6,1)$$ and $$\mathrm{B}(2,4,-3)$$ divides the line segment $$\mathrm{AB}$$ internally in the ratio $$\mathrm{k}: 1$$. If $$\mathrm{a}, \mathrm{b}, \mathrm{c}(|\mathrm{a}|,|\mathrm{b}|,|\mathrm{c}|$$... | [] | null | 10 | Plane: $8 x+y+2 z=0$
<br><br>Given line $\mathrm{AB}: \frac{\mathrm{x}-2}{5}=\frac{\mathrm{y}-4}{10}=\frac{\mathrm{z}+3}{-4}=\lambda$
<br><br>Any point on line $(5 \lambda+2,10 \lambda+4,-4 \lambda-3)$
<br><br>Point of intersection of line and plane
<br><br>$$
\begin{aligned}
& 8(5 \lambda+2)+10 \lambda+4-8 \lambda... | integer | jee-main-2023-online-1st-february-evening-shift | 4,422 |
1ldo7ffl8 | maths | 3d-geometry | lines-and-plane | <p>Let $$\alpha x+\beta y+\gamma z=1$$ be the equation of a plane passing through the point $$(3,-2,5)$$ and perpendicular to the line joining the points $$(1,2,3)$$ and $$(-2,3,5)$$. Then the value of $$\alpha \beta y$$ is equal to _____________.</p> | [] | null | 6 | Plane :
<br/><br/>$$
a(x-3)+b(y+2)+c(z-5)=0
$$
<br/><br/>Dr's of plane : $3 \hat{i}-\hat{j}-2 \hat{k}$
<br/><br/>$$
\begin{aligned}
& <3,-1,-2> \\\\
& P: 3(x-3)-1(y+2)-2(z-5)=0 \\\\
& 3 x-9-y-2-2 z+10=0 \\\\
& 3 x-y-2 z=1 \\\\
& \therefore \alpha=3, \beta=-1, \gamma=-2 \\\\
& \Rightarrow \alpha \beta \gamma=6
\end{alig... | integer | jee-main-2023-online-1st-february-evening-shift | 4,423 |
ldo7laxw | maths | 3d-geometry | lines-and-plane | Let the plane $\mathrm{P}: 8 x+\alpha_{1} y+\alpha_{2} z+12=0$ be parallel to<br/><br/> the line $\mathrm{L}: \frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$. If the
intercept of $\mathrm{P}$<br/><br/> on the $y$-axis is 1 , then the distance between $\mathrm{P}$ and $\mathrm{L}$ is : | [{"identifier": "A", "content": "$\\frac{6}{\\sqrt{14}}$"}, {"identifier": "B", "content": "$\\sqrt{14}$"}, {"identifier": "C", "content": "$\\sqrt{\\frac{2}{7}}$"}, {"identifier": "D", "content": "$\\sqrt{\\frac{7}{2}}$"}] | ["B"] | null | P: $8 x+\alpha_{1} \mathrm{y}+\alpha_{2} \mathrm{z}+12=0$
<br/><br/>L: $\frac{\mathrm{x}+2}{2}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}+4}{5}$
<br/><br/>$\because \mathrm{P}$ is parallel to $\mathrm{L}$
<br/><br/>$\Rightarrow 8(2)+\alpha_{1}(3)+5\left(\alpha_{2}\right)=0$
<br/><br/>$\Rightarrow 3 \alpha_{1}+5\left(\... | mcq | jee-main-2023-online-31st-january-evening-shift | 4,424 |
ldo9cplr | maths | 3d-geometry | lines-and-plane | The foot of perpendicular from the origin $\mathrm{O}$ to a plane $\mathrm{P}$ which meets the co-ordinate axes at the points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ is $(2, \mathrm{a}, 4), \mathrm{a} \in \mathrm{N}$. If the volume of the tetrahedron $\mathrm{OABC}$ is 144 unit$^{3}$, then which of the following points is... | [{"identifier": "A", "content": "$(3,0,4)$"}, {"identifier": "B", "content": "$(0,6,3)$"}, {"identifier": "C", "content": "$(0,4,4)$"}, {"identifier": "D", "content": "$(2,2,4)$"}] | ["A"] | null | Equation of Plane:
<br/><br/>$$
\begin{aligned}
& (2 \hat{i}+a \hat{j}+4 \hat{\mathrm{k}}) \cdot[(\mathrm{x}-2) \hat{\mathrm{i}}+(\mathrm{y}-\mathrm{a}) \hat{\mathrm{j}}+(\mathrm{z}-4) \hat{\mathrm{k}}]=0 \\\\
& \Rightarrow 2 \mathrm{x}+\mathrm{ay}+4 \mathrm{z}=20+\mathrm{a}^{2} \\\\
& \Rightarrow \mathrm{A} \equiv\l... | mcq | jee-main-2023-online-31st-january-evening-shift | 4,425 |
1ldptgy8w | maths | 3d-geometry | lines-and-plane | <p>Let the line $$L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}$$ intersect the plane $$2 x+y+3 z=16$$ at the point
$$P$$. Let the point $$Q$$ be the foot of perpendicular from the point $$R(1,-1,-3)$$ on the line $$L$$. If $$\alpha$$ is the area of triangle $$P Q R$$, then $$\alpha^{2}$$ is equal to __________.</p> | [] | null | 180 | $\quad L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}=r$ (say)
<br/><br/>Let $P \equiv\left(2 r_{1}+1,-r_{1}, r_{1}+3\right)$
<br/><br/>$P$ lies on $2 x+y+3 z=16$
<br/><br/>$\therefore 2\left(2 r_{1}+1\right)+\left(-r_{1}-1\right)+3\left(r_{1}+3\right)=16$
<br/><br/>$r_{1}=1$
<br/><br/>$P \equiv(3,-2,4)$
<br/><br/... | integer | jee-main-2023-online-31st-january-morning-shift | 4,427 |
ldqv1su4 | maths | 3d-geometry | lines-and-plane | A vector $\vec{v}$ in the first octant is inclined to the $x$-axis at $60^{\circ}$, to the $y$-axis at 45 and to the $z$-axis at an acute angle. If a plane passing through the points $(\sqrt{2},-1,1)$ and $(a, b, c)$, is normal to $\vec{v}$, then : | [{"identifier": "A", "content": "$a+b+\\sqrt{2} c=1$"}, {"identifier": "B", "content": "$\\sqrt{2} a+b+c=1$"}, {"identifier": "C", "content": "$\\sqrt{2} a-b+c=1$"}, {"identifier": "D", "content": "$a+\\sqrt{2} b+c=1$"}] | ["D"] | null | <p>$$l = {1 \over 2},m = {1 \over {\sqrt 2 }},n = \cos \theta $$</p>
<p>$${l^2} + {m^2} + {n^2} = 1$$</p>
<p>$$ \Rightarrow {1 \over 4} + {1 \over 2} + {n^2} = 1 \Rightarrow {n^2} = {1 \over 4} \Rightarrow n = \, + \,{1 \over 2}$$</p>
<p>$$\theta$$ is acute $$\therefore$$ $$n = {1 \over 2}$$</p>
<p>$$\therefore$$ $$\ov... | mcq | jee-main-2023-online-30th-january-evening-shift | 4,429 |
ldqvn3td | maths | 3d-geometry | lines-and-plane | If a plane passes through the points $(-1, k, 0),(2, k,-1),(1,1,2)$ and is parallel to the line $\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}$, then the value of $\frac{k^2+1}{(k-1)(k-2)}$ is : | [{"identifier": "A", "content": "$\\frac{17}{5}$"}, {"identifier": "B", "content": "$\\frac{6}{13}$"}, {"identifier": "C", "content": "$\\frac{13}{6}$"}, {"identifier": "D", "content": "$\\frac{5}{17}$"}] | ["C"] | null | <p>Let $$P \equiv ( - 1,k,0),Q \equiv (2,k, - 1)$$ & $$R(1,1,2)$$</p>
<p>$$\overrightarrow P R = 2\widehat i + (1 - k)\widehat j + 2\widehat k$$</p>
<p>& $$\overrightarrow Q R = - \widehat i + (1 - k)\widehat j + 3\widehat k$$</p>
<p>$$\therefore$$ Normal to plane will be</p>
<p>$$\left| {\matrix{
{\widehat i} & {\... | mcq | jee-main-2023-online-30th-january-evening-shift | 4,430 |
1ldr7gixq | maths | 3d-geometry | lines-and-plane | <p>The line $$l_1$$ passes through the point (2, 6, 2) and is perpendicular to the plane $$2x+y-2z=10$$. Then the shortest distance between the line $$l_1$$ and the line $$\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$$ is :</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "$$\\frac{19}{3}$$"}, {"identifier": "D", "content": "$$\\frac{13}{3}$$"}] | ["A"] | null | <p>Equation of $${l_1} = {{x - 2} \over 2} = {{y - 6} \over 1} = {{z - 2} \over { - 2}}$$</p>
<p>Shortest distance with $${{x + 1} \over 2} = {{y + 4} \over { - 3}} = {z \over 2}$$ is</p>
<p>S.d $$ = \left| {{{\matrix{
3 & {10} & 2 \cr
2 & 1 & { - 2} \cr
2 & { - 3} & 2 \cr
} } \over {\left| { - 4\wideh... | mcq | jee-main-2023-online-30th-january-morning-shift | 4,431 |
1ldr7v822 | maths | 3d-geometry | lines-and-plane | <p>If the equation of the plane passing through the point $$(1,1,2)$$ and perpendicular to the line $$x-3 y+ 2 z-1=0=4 x-y+z$$ is $$\mathrm{A} x+\mathrm{B} y+\mathrm{C} z=1$$, then $$140(\mathrm{C}-\mathrm{B}+\mathrm{A})$$ is equal to ___________.</p> | [] | null | 15 | <p>Line of intersection of the planes $$x - 3y + 2z - 1 = 0$$ and $$4x - y + z = 0$$ is normal $$(\overrightarrow n )$$ to the required plane.</p>
<p>$$\overrightarrow n = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & { - 3} & 2 \cr
4 & { - 1} & 1 \cr
} } \right| = - \widehat i ... | integer | jee-main-2023-online-30th-january-morning-shift | 4,432 |
1ldsesbom | maths | 3d-geometry | lines-and-plane | <p>The plane $$2x-y+z=4$$ intersects the line segment joining the points A ($$a,-2,4)$$ and B ($$2,b,-3)$$ at the point C in the ratio 2 : 1 and the distance of the point C from the origin is $$\sqrt5$$. If $$ab < 0$$ and P is the point $$(a-b,b,2b-a)$$ then CP$$^2$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{17}{3}$$"}, {"identifier": "B", "content": "$$\\frac{97}{3}$$"}, {"identifier": "C", "content": "$$\\frac{16}{3}$$"}, {"identifier": "D", "content": "$$\\frac{73}{3}$$"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1let6xptn/17e2a896-c42e-490f-a47a-4034b21a373c/808505a0-ba1e-11ed-b1c7-2f0d4a78b053/file-1let6xpto.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1let6xptn/17e2a896-c42e-490f-a47a-4034b21a373c/808505a0-ba1e-11ed-b1c7-2f0d4a78b053... | mcq | jee-main-2023-online-29th-january-evening-shift | 4,433 |
1ldsffwjf | maths | 3d-geometry | lines-and-plane | <p>If the lines $${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over 1}$$ and $${{x - a} \over 2} = {{y + 2} \over 3} = {{z - 3} \over 1}$$ intersect at the point P, then the distance of the point P from the plane $$z = a$$ is :</p> | [{"identifier": "A", "content": "28"}, {"identifier": "B", "content": "22"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "16"}] | ["A"] | null | <p>$${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over 1} = \lambda $$ (say)</p>
<p>& $${{x - a} \over 2} = {{y + 2} \over 3} = {{z - 3} \over 1} = \mu $$ (say)</p>
<p>$$\therefore$$ $$\lambda + 1 = 2\mu + a$$ ...... (i)</p>
<p>$$2\lambda + 2 = 3\mu - 2$$ ..... (ii)</p>
<p>$$\lambda - 3 = \mu + 3$$ .... (iii... | mcq | jee-main-2023-online-29th-january-evening-shift | 4,434 |
1ldww23tv | maths | 3d-geometry | lines-and-plane | <p>If the foot of the perpendicular drawn from (1, 9, 7) to the line passing through the point (3, 2, 1) and parallel to the planes $$x+2y+z=0$$ and $$3y-z=3$$ is ($$\alpha,\beta,\gamma$$), then $$\alpha+\beta+\gamma$$ is equal to :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$-$$1"}, {"identifier": "D", "content": "5"}] | ["D"] | null | Direction of line
<br/><br/>
$$
\begin{aligned}
\vec{b} & =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 1 \\
0 & 3 & -1
\end{array}\right| \\\\
& =\hat{i}(-5)-\hat{j}(-1)+\hat{k}(3) \\\\
& =-5 \hat{i}+\hat{j}+3 \hat{k}
\end{aligned}
$$
<br/><br/>
Equation of line
<br/><br/>
$$
\frac{x-3}{-5}=\frac{y-... | mcq | jee-main-2023-online-24th-january-evening-shift | 4,437 |
1ldwwhcmx | maths | 3d-geometry | lines-and-plane | <p>Let the plane containing the line of intersection of the planes <br/><br/>P<sub>1</sub> : $$x+(\lambda+4)y+z=1$$ and <br/><br/>P<sub>2</sub> : $$2x+y+z=2$$ <br/><br/>pass through the points (0, 1, 0) and (1, 0, 1). Then the distance of <br/><br/>the point (2$$\lambda,\lambda,-\lambda$$) from the plane P<sub>2</sub> ... | [{"identifier": "A", "content": "$$2\\sqrt6$$"}, {"identifier": "B", "content": "$$3\\sqrt6$$"}, {"identifier": "C", "content": "$$4\\sqrt6$$"}, {"identifier": "D", "content": "$$5\\sqrt6$$"}] | ["B"] | null | Equation of plane passing through point of intersection of $\mathrm{P} 1$ and $\mathrm{P} 2$<br/><br/>
$$
\begin{aligned}
& \mathrm{P}_1+\mathrm{kP}_2 = 0 \\\\
& (\mathrm{x}+(\lambda+4) \mathrm{y}+\mathrm{z}-1)+\mathrm{k}(2 \mathrm{x}+\mathrm{y}+\mathrm{z}-2)=0
\end{aligned}
$$<br/><br/>
Passing through $(0,1,0)$ and $... | mcq | jee-main-2023-online-24th-january-evening-shift | 4,438 |
1ldybfknv | maths | 3d-geometry | lines-and-plane | <p>The distance of the point ($$-1,9,-16$$) from the plane <br/><br/>$$2x+3y-z=5$$ measured parallel to the line <br/><br/>$${{x + 4} \over 3} = {{2 - y} \over 4} = {{z - 3} \over {12}}$$ is :</p> | [{"identifier": "A", "content": "13$$\\sqrt2$$"}, {"identifier": "B", "content": "26"}, {"identifier": "C", "content": "20$$\\sqrt2$$"}, {"identifier": "D", "content": "31"}] | ["B"] | null | Given, $${{x + 4} \over 3} = {{2 - y} \over 4} = {{z - 3} \over {12}}$$
<br><br>$$ \Rightarrow $$ $${{x + 4} \over 3} = {{y - 2} \over -4} = {{z - 3} \over {12}}$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leyorqpr/6043abe7-b1a8-4575-ada8-a5ec43b191a3/29ee5170-bd24-11ed-8df0-25e8b2ace386/... | mcq | jee-main-2023-online-24th-january-morning-shift | 4,439 |
lgnyni4l | maths | 3d-geometry | lines-and-plane | Let the plane $P$ contain the line $2 x+y-z-3=0=5 x-3 y+4 z+9$ and be<br/><br/> parallel to the line $\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}$. Then the distance of the point<br/><br/> $\mathrm{A}(8,-1,-19)$ from the plane $\mathrm{P}$ measured parallel to the line $\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}$<br/><b... | [] | null | 26 | Plane $\mathrm{P} \equiv \mathrm{P}_1+\lambda \mathrm{P}_2=0$<br><br>
$$
\begin{aligned}
& (2 x+y-z-3)+\lambda(5 x-3 y)+4 z+9)=0 \\\\
& (5 \lambda+2) x+(1-3 \lambda) y+(4 \lambda-1) z+9 \lambda-3=0 \\\\
& \overrightarrow{\mathrm{n}} \cdot \overrightarrow{\mathrm{b}}=0 \text { where } \overrightarrow{\mathrm... | integer | jee-main-2023-online-15th-april-morning-shift | 4,440 |
1lgowhqoz | maths | 3d-geometry | lines-and-plane | <p>The plane, passing through the points $$(0,-1,2)$$ and $$(-1,2,1)$$ and parallel to the line passing through $$(5,1,-7)$$ and $$(1,-1,-1)$$, also passes through the point :</p> | [{"identifier": "A", "content": "$$(0,5,-2)$$"}, {"identifier": "B", "content": "$$(2,0,1)$$"}, {"identifier": "C", "content": "$$(1,-2,1)$$"}, {"identifier": "D", "content": "$$(-2,5,0)$$"}] | ["D"] | null | <p>The first step is to find the normal vector to the desired plane. Since the plane is parallel to the line passing through the points (5, 1, -7) and (1, -1, -1), the direction vector of that line is also parallel to the plane. The direction vector is the difference between the coordinates of the two points, which is ... | mcq | jee-main-2023-online-13th-april-evening-shift | 4,441 |
1lgpyb2eu | maths | 3d-geometry | lines-and-plane | <p>The distance of the point $$(-1,2,3)$$ from the plane $$\vec{r} \cdot(\hat{i}-2 \hat{j}+3 \hat{k})=10$$ parallel to the line of the shortest distance between the lines $$\vec{r}=(\hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{k})$$ and $$\vec{r}=(2 \hat{i}-\hat{j})+\mu(\hat{i}-\hat{j}+\hat{k})$$ is :</p> | [{"identifier": "A", "content": "$$3 \\sqrt{6}$$"}, {"identifier": "B", "content": "$$3 \\sqrt{5}$$"}, {"identifier": "C", "content": "$$2 \\sqrt{6}$$"}, {"identifier": "D", "content": "$$2 \\sqrt{5}$$"}] | ["C"] | null | 1. Determine the line of shortest distance between the given two lines:
<br/><br/>Direction vector of line 1: $$\vec{d_1} = 2\hat{i} + \hat{k}$$
<br/><br/>Direction vector of line 2: $$\vec{d_2} = \hat{i} - \hat{j} + \hat{k}$$
<br/><br/>Now, let's find the cross product $$\vec{N} = \vec{d_1} \times \vec{d_2}$$
<br/... | mcq | jee-main-2023-online-13th-april-morning-shift | 4,442 |
1lgrgb0dt | maths | 3d-geometry | lines-and-plane | <p>Let the plane P: $$4 x-y+z=10$$ be rotated by an angle $$\frac{\pi}{2}$$ about its line of intersection with the plane $$x+y-z=4$$. If $$\alpha$$ is the distance of the point $$(2,3,-4)$$ from the new position of the plane $$\mathrm{P}$$, then $$35 \alpha$$ is equal to :</p> | [{"identifier": "A", "content": "126"}, {"identifier": "B", "content": "105"}, {"identifier": "C", "content": "85"}, {"identifier": "D", "content": "90"}] | ["A"] | null | Equation of plane after rotation :
<br/><br/>$$
\begin{aligned}
& (4 x-y+z-10)+\lambda(x+y-z-y)=0 \\\\
\Rightarrow & (4+\lambda) x+y(\lambda-1)+z(1-\lambda)-4 \lambda-10=0 \\\\
& \overrightarrow{n_1} \cdot \overrightarrow{n_2}=0 \\\\
\Rightarrow & (4+\lambda) 4+(\lambda-1)(-1)+(1-\lambda) 1=0 \\\\
\Rightarrow & 16+4 \l... | mcq | jee-main-2023-online-12th-april-morning-shift | 4,443 |
1lgsw3mg4 | maths | 3d-geometry | lines-and-plane | <p>Let the line $$l: x=\frac{1-y}{-2}=\frac{z-3}{\lambda}, \lambda \in \mathbb{R}$$ meet the plane $$P: x+2 y+3 z=4$$ at the point $$(\alpha, \beta, \gamma)$$. If the angle between the line $$l$$ and the plane $$P$$ is $$\cos ^{-1}\left(\sqrt{\frac{5}{14}}\right)$$, then $$\alpha+2 \beta+6 \gamma$$ is equal to ________... | [] | null | 11 | $L: \frac{x-0}{1}=\frac{y-1}{2}=\frac{z-3}{\lambda} $
<br/><br/>$ P: x+2 y+3 z=4$
<br/><br/>Vector parallel to line : $\langle 1,2, \lambda\rangle=\bar{b}$
<br/><br/>Normal vector to plane $P:<1,2,3\rangle=\bar{n}$
<br/><br/>Angle between plane and line is $\theta$
<br/><br/>Then, $\sin \theta=\frac{<1,2, \lambda>\cdot... | integer | jee-main-2023-online-11th-april-evening-shift | 4,444 |
1lgvqjg9j | maths | 3d-geometry | lines-and-plane | <p>Let the foot of perpendicular from the point $$\mathrm{A}(4,3,1)$$ on the plane $$\mathrm{P}: x-y+2 z+3=0$$ be N. If B$$(5, \alpha, \beta), \alpha, \beta \in \mathbb{Z}$$ is a point on plane P such that the area of the triangle ABN is $$3 \sqrt{2}$$, then $$\alpha^{2}+\beta^{2}+\alpha \beta$$ is equal to __________... | [] | null | 7 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnkifvys/3c8e033b-fa69-4b24-96b7-4df950fb761d/62d0e540-6786-11ee-8adf-57893cbbad41/file-6y3zli1lnkifvyt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnkifvys/3c8e033b-fa69-4b24-96b7-4df950fb761d/62d0e540-6786-11ee-8a... | integer | jee-main-2023-online-10th-april-evening-shift | 4,445 |
1lgxh5015 | maths | 3d-geometry | lines-and-plane | <p>Let two vertices of a triangle ABC be (2, 4, 6) and (0, $$-$$2, $$-$$5), and its centroid be (2, 1, $$-$$1). If the image of the third vertex in the plane $$x+2y+4z=11$$ is $$(\alpha,\beta,\gamma)$$, then $$\alpha\beta+\beta\gamma+\gamma\alpha$$ is equal to :</p> | [{"identifier": "A", "content": "72"}, {"identifier": "B", "content": "74"}, {"identifier": "C", "content": "76"}, {"identifier": "D", "content": "70"}] | ["B"] | null | Given that two vertex of a $\triangle A B C$ be $A(2,4,6)$ and <br><br>$B(0,-2,-5)$ and $G=$ centroid $=(2,1,-1)$
[Given]
<br><br>Let, the other vertex is $C(x, y, z)$
<br><br>According to the question,
<br><br>$$
\begin{aligned}
& \frac{2+0+x}{3}=2 \\\\
&\Rightarrow x =4
\end{aligned}
$$
<br><br>$$
\begin{ali... | mcq | jee-main-2023-online-10th-april-morning-shift | 4,446 |
1lgxsud7o | maths | 3d-geometry | lines-and-plane | <p>Let P be the point of intersection of the line $${{x + 3} \over 3} = {{y + 2} \over 1} = {{1 - z} \over 2}$$ and the plane $$x+y+z=2$$. If the distance of the point P from the plane $$3x - 4y + 12z = 32$$ is q, then q and 2q are the roots of the equation :</p> | [{"identifier": "A", "content": "$${x^2} + 18x - 72 = 0$$"}, {"identifier": "B", "content": "$${x^2} - 18x - 72 = 0$$"}, {"identifier": "C", "content": "$${x^2} + 18x + 72 = 0$$"}, {"identifier": "D", "content": "$${x^2} - 18x + 72 = 0$$"}] | ["D"] | null | Given, equation of line is
<br/><br/>$$
\begin{aligned}
& \frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}=k \\\\
& \therefore x=3 k-3, y=k-2, z=1-2 k
\end{aligned}
$$
<br/><br/>Since, given that $P \equiv(3 k-3, k-2,1-2 k)$ be the point of intersection of the given line and the plane $x+y+z=2$
<br/><br/>$$
\begin{aligned}
&\... | mcq | jee-main-2023-online-10th-april-morning-shift | 4,447 |
1lgylgam9 | maths | 3d-geometry | lines-and-plane | <p>For $$\mathrm{a}, \mathrm{b} \in \mathbb{Z}$$ and $$|\mathrm{a}-\mathrm{b}| \leq 10$$, let the angle between the plane $$\mathrm{P}: \mathrm{ax}+y-\mathrm{z}=\mathrm{b}$$ and the line $$l: x-1=\mathrm{a}-y=z+1$$ be $$\cos ^{-1}\left(\frac{1}{3}\right)$$. If the distance of the point $$(6,-6,4)$$ from the plane P is ... | [{"identifier": "A", "content": "48"}, {"identifier": "B", "content": "85"}, {"identifier": "C", "content": "32"}, {"identifier": "D", "content": "25"}] | ["C"] | null | We have, $\theta=\cos ^{-1} \frac{1}{3}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \cos \theta=\frac{1}{3} \\\\
& \therefore \sin \theta=\sqrt{1-\left(\frac{1}{3}\right)^2}=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}
\end{aligned}
$$
<br/><br/>The given plane line and are
<br/><br/>$$
a x+y-z=b $$
<br/><br/>$$ x-1=a-y=z+1... | mcq | jee-main-2023-online-8th-april-evening-shift | 4,448 |
1lh00mq7x | maths | 3d-geometry | lines-and-plane | <p>Let $$\lambda_{1}, \lambda_{2}$$ be the values of $$\lambda$$ for which the points $$\left(\frac{5}{2}, 1, \lambda\right)$$ and $$(-2,0,1)$$ are at equal distance from the plane $$2 x+3 y-6 z+7=0$$. If $$\lambda_{1} > \lambda_{2}$$, then the distance of the point $$\left(\lambda_{1}-\lambda_{2}, \lambda_{2}, \lam... | [] | null | 9 | Since $\left(\frac{5}{2}, 1, \lambda\right)$ and $(-2,0,1)$ are equidistant
<br/><br/>from plane $2 x+3 y-6 z+7=0$
<br/><br/>$$
\begin{aligned}
& \therefore\left|\frac{2\left(\frac{5}{2}\right)+3(1)-6(\lambda)+7}{\sqrt{2^2+3^2+6^2}}\right|=\left|\frac{2(-2)+3(0)-6(1)+7}{\sqrt{2^2+3^2+6^2}}\right| \\\\
& \Rightarrow|5+... | integer | jee-main-2023-online-8th-april-morning-shift | 4,450 |
1lh216te7 | maths | 3d-geometry | lines-and-plane | <p>If the equation of the plane passing through the line of intersection of the planes $$2 x-y+z=3,4 x-3 y+5 z+9=0$$ and parallel to the line $$\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}$$ is $$a x+b y+c z+6=0$$, then $$a+b+c$$ is equal to :</p> | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "14"}, {"identifier": "D", "content": "12"}] | ["C"] | null | Equation of plane intersection of two plane
<br/><br/>$$
\mathrm{P}_1+\lambda \mathrm{P}_2=0
$$
<br/><br/>$$
\mathrm{P}_1: 2 x-y+z=3, \text { and } \mathrm{P}_2: 4 x-3 y+5 z+9=0
$$
<br/><br/>Equation of any plane passing through the intersection of given planes is
<br/><br/>$(2 x-y+z-3)+\lambda(4 x-3 y+5 z+9)=0$
<br/>... | mcq | jee-main-2023-online-6th-april-morning-shift | 4,451 |
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