question_id stringlengths 8 35 | subject stringclasses 3
values | chapter stringclasses 90
values | topic stringclasses 459
values | question stringlengths 17 24.5k | options stringlengths 2 4.26k | correct_option stringclasses 6
values | answer stringclasses 460
values | explanation stringlengths 1 10.6k | question_type stringclasses 3
values | paper_id stringclasses 154
values | __index_level_0__ int64 2 13.4k |
|---|---|---|---|---|---|---|---|---|---|---|---|
1lsgafa13 | maths | 3d-geometry | lines-and-plane | <p>Let $$A(2,3,5)$$ and $$C(-3,4,-2)$$ be opposite vertices of a parallelogram $$A B C D$$. If the diagonal $$\overrightarrow{\mathrm{BD}}=\hat{i}+2 \hat{j}+3 \hat{k}$$, then the area of the parallelogram is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{1}{2} \\sqrt{410}$$\n"}, {"identifier": "B", "content": "$$\\frac{1}{2} \\sqrt{306}$$\n"}, {"identifier": "C", "content": "$$\\frac{1}{2} \\sqrt{586}$$\n"}, {"identifier": "D", "content": "$$\\frac{1}{2} \\sqrt{474}$$"}] | ["D"] | null | <p>$$\begin{aligned}
& \text { Area }=|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}}| \\
& =\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
5 & -1 & 7 \\
1 & 2 & 3
\end{array}\right| \\
& =\frac{1}{2}|-17 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}|=\fra... | mcq | jee-main-2024-online-30th-january-morning-shift | 4,453 |
lvc57bf8 | maths | 3d-geometry | lines-and-plane | <p>If $$A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1)$$ and $$D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$$ are the vertices of a quadrilateral $$A B C D$$, then its area is</p> | [{"identifier": "A", "content": "$$\\frac{4 \\sqrt{2}}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{2 \\sqrt{2}}{3}$$\n"}, {"identifier": "C", "content": "$$\\frac{5 \\sqrt{2}}{3}$$\n"}, {"identifier": "D", "content": "$$2 \\sqrt{2}$$"}] | ["A"] | null | <p>$$A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1), D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$$ are vertices of a quadrilateral</p>
<p>$$\begin{aligned}
& \overrightarrow{A C}=(2 \hat{i}+2 \hat{j}+\hat{k})-(3 \hat{i}+\hat{j}-\hat{k}) \\
& =-\hat{i}+\hat{j}+2 \hat{k} \\
& \overrigh... | mcq | jee-main-2024-online-6th-april-morning-shift | 4,454 |
lCoOxMgGdxQ2IcRT | maths | 3d-geometry | lines-in-space | The lines $${{x - 2} \over 1} = {{y - 3} \over 1} = {{z - 4} \over { - k}}$$ and $${{x - 1} \over k} = {{y - 4} \over 2} = {{z - 5} \over 1}$$ are coplanar if : | [{"identifier": "A", "content": "$$k=3$$ or $$-2$$"}, {"identifier": "B", "content": "$$k=0$$ or $$-1$$"}, {"identifier": "C", "content": "$$k=1$$ or $$-1$$"}, {"identifier": "D", "content": "$$k=0$$ or $$-3$$"}] | ["D"] | null | Coplanar if
<br/><br/>$$\left| {\matrix{
{{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr
{{l_1}} & {{m_1}} & {{n_1}} \cr
{{l_2}} & {{m_2}} & {{n_2}} \cr
} } \right| = 0$$
<br><br>$$\therefore$$ $$\left| {\matrix{
1 & { - 1} & { - 1} \cr
1 & 1 & { ... | mcq | aieee-2003 | 4,455 |
QB1hUjgch8hPRWNt | maths | 3d-geometry | lines-in-space | A line with direction cosines proportional to $$2,1,2$$ meets each of the lines $$x=y+a=z$$ and $$x+a=2y=2z$$ . The co-ordinates of each of the points of intersection are given by : | [{"identifier": "A", "content": "$$\\left( {2a,3a,3a} \\right),\\left( {2a,a,a} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {3a,2a,3a} \\right),\\left( {a,a,a} \\right)$$ "}, {"identifier": "C", "content": "$$\\left( {3a,2a,3a} \\right),\\left( {a,a,2a} \\right)$$ "}, {"identifier": "D", "content": "$$\\le... | ["B"] | null | Let a point on the line
<br><br>$$x = y + a = z$$ is $$\left( {\lambda ,\lambda - a,\lambda } \right)$$
<br><br>and a point on the line
<br><br>$$x + a = 2y = 2z$$ is $$\left( {\mu - a,{\mu \over 2},{\mu \over 2}} \right),$$
<br><br>then direction ratio of the line joining these points are
<br><br>$$\lambda -... | mcq | aieee-2004 | 4,456 |
4OZMVKi7SP8wxWVL | maths | 3d-geometry | lines-in-space | If the straight lines
<br/>$$x=1+s,y=-3$$$$ - \lambda s,$$ $$z = 1 + \lambda s$$ and $$x = {t \over 2},y = 1 + t,z = 2 - t,$$ with parameters $$s$$ and $$t$$ respectively, are co-planar, then $$\lambda $$ equals : | [{"identifier": "A", "content": "$$0$$ "}, {"identifier": "B", "content": "$$-1$$ "}, {"identifier": "C", "content": "$$ - {1 \\over 2}$$ "}, {"identifier": "D", "content": "$$-2$$ "}] | ["D"] | null | The given lines are
<br><br>$$x - 1 = {{y + 3} \over { - \lambda }} = {{z - 1} \over \lambda } = s.........\left( 1 \right)$$
<br><br>and $$2x = y - 1 = {{z - 2} \over { - 1}} = t..........\left( 2 \right)$$
<br><br>The lines are coplanar, if
<br><br>$$\left| {\matrix{
{0 - \left( { - 1} \right)} & { - 1 - 3} ... | mcq | aieee-2004 | 4,457 |
u6SMbm7CaVhVyKjm | maths | 3d-geometry | lines-in-space | The angle between the lines $$2x=3y=-z$$ and $$6x=-y=-4z$$ is : | [{"identifier": "A", "content": "$${0^ \\circ }$$ "}, {"identifier": "B", "content": "$${90^ \\circ }$$"}, {"identifier": "C", "content": "$${45^ \\circ }$$"}, {"identifier": "D", "content": "$${30^ \\circ }$$"}] | ["B"] | null | The given lines are $$2x = 3y = - z$$
<br><br>or $$\,\,\,\,\,\,\,\,\,\,{x \over 3} = {y \over 2} = {z \over { - 6}}$$ $$\,\,\,\left[ {} \right.$$ Dividing by $$6$$ $$\left. {} \right]$$
<br><br>and $$6x = - y = - 4z$$
<br><br>or $$\,\,\,\,\,\,\,\,\,{x \over 2} = {y \over { - 12}} = {z \over { - 3}}$$ $$\,\,\,\,\left... | mcq | aieee-2005 | 4,458 |
Z4uIG8y06yV4jPYd | maths | 3d-geometry | lines-in-space | The two lines $$x=ay+b, z=cy+d;$$ and $$x=a'y+b' ,$$ $$z=c'y+d'$$ are perpendicular to each other if : | [{"identifier": "A", "content": "$$aa'+cc'=-1$$"}, {"identifier": "B", "content": "$$aa'+cc'=1$$ "}, {"identifier": "C", "content": "$${a \\over {a'}} + {c \\over {c'}} = - 1$$ "}, {"identifier": "D", "content": "$${a \\over {a'}} + {c \\over {c'}} = 1$$"}] | ["A"] | null | Equation of lines
<br><br>$${{x - b} \over a} = {y \over 1} = {{z - d} \over c}$$
<br><br>$${{x - b'} \over {a'}} = {y \over 1} = {{z - d'} \over {c'}}$$
<br><br>Line are perpendicular
<br><br>$$ \Rightarrow aa' + 1 + cc' = 0$$ | mcq | aieee-2006 | 4,459 |
2cg5qE8eqUZZH2mZ | maths | 3d-geometry | lines-in-space | The line passing through the points $$(5,1,a)$$ and $$(3, b, 1)$$ crosses the $$yz$$-plane at the point $$\left( {0,{{17} \over 2}, - {{ - 13} \over 2}} \right)$$ . Then | [{"identifier": "A", "content": "$$a=2,$$ $$b=8$$"}, {"identifier": "B", "content": "$$a=4,$$ $$b=6$$ "}, {"identifier": "C", "content": "$$a=6,$$ $$b=4$$"}, {"identifier": "D", "content": "$$a=8,$$ $$b=2$$"}] | ["C"] | null | Equation of line through $$\left( {5,1,a} \right)$$ and
<br><br>$$\left( {3,b,1} \right)$$ is $${{x - 5} \over { - 2}} = {{y - 1} \over {b - 1}} = {{z - a} \over {1 - a}} = \lambda $$
<br><br>$$\therefore$$ Any point on this line is a
<br><br>$$\left[ { - 2\lambda + 5,\left( {b - 1} \right)\lambda + 1,\left( {1 - \... | mcq | aieee-2008 | 4,460 |
ynbPZFBIbyg94BX0 | maths | 3d-geometry | lines-in-space | If the straight lines $$\,\,\,\,\,$$ $$\,\,\,\,\,$$ $${{x - 1} \over k} = {{y - 2} \over 2} = {{z - 3} \over 3}$$ $$\,\,\,\,\,$$ and$$\,\,\,\,\,$$ $${{x - 2} \over 3} = {{y - 3} \over k} = {{z - 1} \over 2}$$ intersects at a point, then the integer $$k$$ is equal to | [{"identifier": "A", "content": "$$-5$$"}, {"identifier": "B", "content": "$$5$$"}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$$-2$$"}] | ["A"] | null | The two lines intersect if shortest distance between them is zero $$i.e.$$
<br><br>$${{\left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}} = 0$$
<br><br>$$ \Righ... | mcq | aieee-2008 | 4,461 |
7cGbYtvbBGU0bEmH | maths | 3d-geometry | lines-in-space | If the line $${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over 4}$$ and $${{x - 3} \over 1} = {{y - k} \over 2} = {z \over 1}$$ intersect, then $$k$$ is equal to : | [{"identifier": "A", "content": "$$-1$$ "}, {"identifier": "B", "content": "$${2 \\over 9}$$"}, {"identifier": "C", "content": "$${9 \\over 2}$$"}, {"identifier": "D", "content": "$$0$$ "}] | ["C"] | null | Given lines in vector form are
<br><br>$$\overrightarrow r = \left( {\widehat i - \overrightarrow j + \overrightarrow k } \right) + \lambda \left( {2\overrightarrow i + 3\overrightarrow j + 4\overrightarrow j } \right)$$
<br><br>and $$\overrightarrow r = \left( {3\widehat i + k\widehat j} \right) + \mu \left( {\... | mcq | aieee-2012 | 4,463 |
CXIqtyKExltEbk3B | maths | 3d-geometry | lines-in-space | If the lines $${{x - 2} \over 1} = {{y - 3} \over 1} = {{z - 4} \over { - k}}$$ and $${{x - 1} \over k} = {{y - 4} \over 2} = {{z - 5} \over 1}$$ are coplanar, then $$k$$ can have : | [{"identifier": "A", "content": "any value "}, {"identifier": "B", "content": "exactly one value"}, {"identifier": "C", "content": "exactly two values "}, {"identifier": "D", "content": "exactly three values"}] | ["C"] | null | Given lines will be coplanar
<br><br>If $$\,\,\,\,\left| {\matrix{
{ - 1} & 1 & 1 \cr
1 & 1 & { - k} \cr
k & 2 & 1 \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow - 1\left( {1 + 2k} \right) - \left( {1 + {k^2}} \right) + 1\left( {2 - k} \right) = 0$$
<br><br>$$ \Rightarrow k = 0,... | mcq | jee-main-2013-offline | 4,464 |
wufSe9GWqPvxj5A6ZtUib | maths | 3d-geometry | lines-in-space | If the angle between the lines, $${x \over 2} = {y \over 2} = {z \over 1}$$
<br/><br/>and $${{5 - x} \over { - 2}} = {{7y - 14} \over p} = {{z - 3} \over 4}\,\,$$ is $${\cos ^{ - 1}}\left( {{2 \over 3}} \right),$$ then p is equal to : | [{"identifier": "A", "content": "$${7 \\over 2}$$ "}, {"identifier": "B", "content": "$${2 \\over 7}$$"}, {"identifier": "C", "content": "$$-$$ $${7 \\over 4}$$"}, {"identifier": "D", "content": "$$-$$ $${4 \\over 7}$$"}] | ["A"] | null | Let $$\theta $$ be the angle between the two lines
<br><br>Here direction cosines of $${x \over 2}$$ = $${y \over 2}$$ = $${z \over 1}$$ are 2, 2, 1
<br><br>Also second line can be written as :
<br><br>$${{x - 5} \over 2} = {{y - 2} \over {{P \over 7}}} = {{z - 3} \over 4}$$
<br><br>$$ \therefore $$ its direc... | mcq | jee-main-2018-online-16th-april-morning-slot | 4,466 |
0mlTVuKdlxftvs27Ar3rsa0w2w9jwxk75cq | maths | 3d-geometry | lines-in-space | If the length of the perpendicular from the point ($$\beta $$, 0, $$\beta $$) ($$\beta $$ $$ \ne $$ 0) to the line,
<br/>$${x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}}$$ is $$\sqrt {{3 \over 2}} $$, then
$$\beta $$ is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "-2"}, {"identifier": "D", "content": "-1"}] | ["D"] | null | $${x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}} = p\,\,P\left( {\beta ,0,\beta } \right)$$<br><br>
any point on line A = (p, 1, – p – 1)<br><br>
Now, DR of AP $$ \equiv $$ < p – $$\beta $$, 1 – 0, – p – 1 – $$\beta $$ ><br><br>
Which is perpendicular to line so<br><br>
(p – $$\beta $$). 1 + 0.1 – 1(– p... | mcq | jee-main-2019-online-10th-april-morning-slot | 4,467 |
SGw6KlnNG3CtzIgUGr18hoxe66ijvwv9w1o | maths | 3d-geometry | lines-in-space | The vertices B and C of a $$\Delta $$ABC lie on the line,
<br/><br/>$${{x + 2} \over 3} = {{y - 1} \over 0} = {z \over 4}$$ such that BC = 5 units. <br/><br/>Then the
area (in sq. units) of this triangle, given that the
point A(1, –1, 2), is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "$$5\\sqrt {17} $$"}, {"identifier": "C", "content": "$$\\sqrt {34} $$"}, {"identifier": "D", "content": "$$2\\sqrt {34} $$"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265608/exam_images/ei4rpd9as7gipkdqbmus.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264422/exam_images/cu7zggkrbhyehugwdzx8.webp"><img src="https://res.c... | mcq | jee-main-2019-online-9th-april-evening-slot | 4,468 |
BUAbYW0qzI2qZL65Fu5S1 | maths | 3d-geometry | lines-in-space | The length of the perpendicular from the point
(2, –1, 4) on the straight line,
<br/><br/>$${{x + 3} \over {10}}$$= $${{y - 2} \over {-7}}$$ = $${{z} \over {1}}$$
is :
| [{"identifier": "A", "content": "less than 2"}, {"identifier": "B", "content": "greater than 4\n"}, {"identifier": "C", "content": "greater than 2 but less than 3"}, {"identifier": "D", "content": "greater than 3 but less than 4"}] | ["D"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266241/exam_images/fhdx0dohqlxs327d7ok9.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267798/exam_images/i5tycucchql2hd8qs3c1.webp" style="max-width: 100%;height: auto;display: block;margi... | mcq | jee-main-2019-online-8th-april-morning-slot | 4,469 |
H7fMwfkoz6SPD9jgNn7k9k2k5gznyfy | maths | 3d-geometry | lines-in-space | The shortest distance between the lines
<br/><br/>$${{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}$$ and
<br/><br/>$${{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}$$ is : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$${7 \\over 2}\\sqrt {30} $$"}, {"identifier": "C", "content": "$$3\\sqrt {30} $$"}, {"identifier": "D", "content": "$$2\\sqrt {30} $$"}] | ["C"] | null | $$\overrightarrow a $$
= < 3, 8, 3 >
<br><br>$$\overrightarrow b $$
= < – 3, – 7, 6 >
<br><br>$$\overrightarrow p $$
= < 3, – 1, 1 >
<br><br>$$\overrightarrow q $$
= < –3, 2, 4 >
<br><br>$$\overrightarrow p \times \overrightarrow q = \left| {\matrix{
{\widehat i} & {\widehat j} &... | mcq | jee-main-2020-online-8th-january-morning-slot | 4,472 |
HngSHjDPG5hOAlG07kjgy2xukfg75pwu | maths | 3d-geometry | lines-in-space | If (a, b, c) is the image of the point (1, 2, -3) in<br/><br/> the line $${{x + 1} \over 2} = {{y - 3} \over { - 2}} = {z \over { - 1}}$$, then a + b + c is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "-1"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267424/exam_images/kdcf95puxmjkf83bykfg.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 5th September Morning Slot Mathematics - 3D Geometry Question 223 English Explanation">
<br><br>Eq... | mcq | jee-main-2020-online-5th-september-morning-slot | 4,473 |
cy8c71WuyhUNtGF5jA1klrkomrx | maths | 3d-geometry | lines-in-space | Let a, b$$ \in $$R. If the mirror image of the point P(a, 6, 9) with respect to the line <br/><br/>$${{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}}$$ is (20, b, $$-$$a$$-$$9), then | a + b |, is equal to : | [{"identifier": "A", "content": "88"}, {"identifier": "B", "content": "90"}, {"identifier": "C", "content": "86"}, {"identifier": "D", "content": "84"}] | ["A"] | null | Given, P(a, 6, 9)<br/><br/>Equation of line $${{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}}$$<br/><br/>Image of point P with respect to line is point Q(20, b, $$-$$a $$-$$9)<br/><br/>Mid-point of P and Q = $$\left( {{{a + 20} \over 2},{{6 + b} \over 2},{{ - a} \over 2}} \right)$$<br/><br/>This point li... | mcq | jee-main-2021-online-24th-february-evening-slot | 4,474 |
Jp12LbScNdtH0VokX81klrmjwqw | maths | 3d-geometry | lines-in-space | Let $$\lambda$$ be an integer. If the shortest distance between the lines <br/><br/>x $$-$$ $$\lambda$$ = 2y $$-$$ 1 = $$-$$2z and x = y + 2$$\lambda$$ = z $$-$$ $$\lambda$$ is $${{\sqrt 7 } \over {2\sqrt 2 }}$$, then the value of | $$\lambda$$ | is _________. | [] | null | 1 | $${{x - \lambda } \over 1} = {{y - {1 \over 2}} \over {{1 \over 2}}} = {z \over { - {1 \over 2}}}$$<br><br>$${{x - \lambda } \over 2} = {{y - {1 \over 2}} \over 1} = {2 \over { - 1}}$$ ....... (1)<br><br>Point on line = $$\left( {\lambda ,{1 \over 2},0} \right)$$<br><br>$${x \over 1} = {{y + 2\lambda } \over 1} = {{z -... | integer | jee-main-2021-online-24th-february-evening-slot | 4,475 |
8zfZrqb7fcL0HgCiZX1klt9uh4s | maths | 3d-geometry | lines-in-space | A line 'l' passing through origin is perpendicular to the lines<br/><br/>$${l_1}:\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat k$$<br/><br/>$${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$$<br/><br/>If the co-ordinates of the point in the firs... | [] | null | 44 | $${l_1}:\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat k$$<br><br>$${l_1}:{{x - 3} \over 1} = {{y + 1} \over 2} = {{z - 4} \over 2} \Rightarrow $$ D.R. of $${l_1} = 1,2,2$$<br><br>$${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$$<br><br>$${l_2}... | integer | jee-main-2021-online-25th-february-evening-slot | 4,477 |
YtsOhjwJ46iodfXK1u1kmhvwg6d | maths | 3d-geometry | lines-in-space | Let the position vectors of two points P and Q be 3$$\widehat i$$ $$-$$ $$\widehat j$$ + 2$$\widehat k$$ and $$\widehat i$$ + 2$$\widehat j$$ $$-$$ 4$$\widehat k$$, respectively. Let R and S be two points such that the direction ratios of lines PR and QS are (4, $$-$$1, 2) and ($$-$$2, 1, $$-$$2), respectively. Let lin... | [{"identifier": "A", "content": "$$\\sqrt {171} $$"}, {"identifier": "B", "content": "$$\\sqrt {227} $$"}, {"identifier": "C", "content": "$$\\sqrt {482} $$"}, {"identifier": "D", "content": "$$\\sqrt {5} $$"}] | ["A"] | null | $$\overrightarrow p = 3\widehat i - \widehat j + 2\widehat k$$ & $$\overrightarrow Q = \widehat i + 2\widehat j - 4\widehat k$$<br><br>$${\overrightarrow v _{PR}} = (4, - 1,2)$$ & $${\overrightarrow v _{QS}} = ( - 2,1, - 2)$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265215/exam... | mcq | jee-main-2021-online-16th-march-morning-shift | 4,478 |
ASltiYdrFmn8DXdoyu1kmiwbg93 | maths | 3d-geometry | lines-in-space | If the foot of the perpendicular from point (4, 3, 8) on the line $${L_1}:{{x - a} \over l} = {{y - 2} \over 3} = {{z - b} \over 4}$$, <i>l</i> $$\ne$$ 0 is (3, 5, 7), then the shortest distance between the line L<sub>1</sub> and line $${L_2}:{{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 6 }}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$$\\sqrt {{2 \\over 3}} $$"}] | ["A"] | null | (3, 5, 7) lie on given line L<sub>1</sub><br><br>$${{3 - a} \over l} = {3 \over 3} = {{7 - b} \over 4}$$<br><br>$${{7 - b} \over 4} = 1 \Rightarrow b = 3$$<br><br>$${{3 - a} \over l} = 1 \Rightarrow 3 - a = l$$<br><br>M(4, 3, 8)<br><br>N(3, 5, 7)<br><br>DR'S of MN = (1, $$-$$2, 1)<br><br>MN $$ \bot $$ line L<sub>1</sub... | mcq | jee-main-2021-online-16th-march-evening-shift | 4,479 |
1krrpbbk7 | maths | 3d-geometry | lines-in-space | The lines x = ay $$-$$ 1 = z $$-$$ 2 and x = 3y $$-$$ 2 = bz $$-$$ 2, (ab $$\ne$$ 0) are coplanar, if : | [{"identifier": "A", "content": "b = 1, a$$\\in$$R $$-$$ {0}"}, {"identifier": "B", "content": "a = 1, b$$\\in$$R $$-$$ {0}"}, {"identifier": "C", "content": "a = 2, b = 2"}, {"identifier": "D", "content": "a = 2, b = 3"}] | ["A"] | null | Lines are $$x = ay - 1 = z - 2$$<br><br>$$\therefore$$ $${x \over 1} = {{y - {1 \over a}} \over {{1 \over a}}} = {{z - 2} \over 1}$$ .... (i)<br><br>and $$x = 3y - 2 = bz - 2$$<br><br>$$\therefore$$ $${x \over 1} = {{y - {2 \over 3}} \over {{1 \over 3}}} = {{z - {2 \over b}} \over {{1 \over b}}}$$ .... (ii)<br><br>$$\t... | mcq | jee-main-2021-online-20th-july-evening-shift | 4,480 |
1krti6m7i | maths | 3d-geometry | lines-in-space | If the shortest distance between the straight lines $$3(x - 1) = 6(y - 2) = 2(z - 1)$$ and $$4(x - 2) = 2(y - \lambda ) = (z - 3),\lambda \in R$$ is $${1 \over {\sqrt {38} }}$$, then the integral value of $$\lambda$$ is equal to : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "$$-$$1"}] | ["A"] | null | $${L_1}:{{(x - 1)} \over 2} = {{(y - 2)} \over 1} = {{(z - 1)} \over 3}\overrightarrow {{r_1}} = 2\widehat i + \widehat j + 3\widehat k$$<br><br>$${L_2}:{{(x - 2)} \over 1} = {{y - \lambda } \over 2} = {{z - 3} \over 4}\overrightarrow {{r_2}} = \widehat i + 2\widehat j + 4\widehat k$$<br><br><img src="https://res.clo... | mcq | jee-main-2021-online-22th-july-evening-shift | 4,481 |
1l55iy2cg | maths | 3d-geometry | lines-in-space | <p>Let the image of the point P(1, 2, 3) in the line $$L:{{x - 6} \over 3} = {{y - 1} \over 2} = {{z - 2} \over 3}$$ be Q. Let R ($$\alpha$$, $$\beta$$, $$\gamma$$) be a point that divides internally the line segment PQ in the ratio 1 : 3. Then the value of 22 ($$\alpha$$ + $$\beta$$ + $$\gamma$$) is equal to _________... | [] | null | 125 | <p>The point dividing PQ in the ratio 1 : 3 will be mid-point of P & foot of perpendicular from P on the line.</p>
<p>$$\therefore$$ Let a point on line be $$\lambda$$</p>
<p>$$ \Rightarrow {{x - 6} \over 3} = {{y - 1} \over 2} = {{z - 2} \over 3} = \lambda $$</p>
<p>$$ \Rightarrow P'(3\lambda + 6,\,2\lambda + 1,\,3\... | integer | jee-main-2022-online-28th-june-evening-shift | 4,483 |
1l56ra7cv | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines <br/><br/>$${{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$$ and $${{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$$, is :</p> | [{"identifier": "A", "content": "$${{18} \\over {\\sqrt 5 }}$$"}, {"identifier": "B", "content": "$${{22} \\over {3\\sqrt 5 }}$$"}, {"identifier": "C", "content": "$${{46} \\over {3\\sqrt 5 }}$$"}, {"identifier": "D", "content": "$$6\\sqrt 3 $$"}] | ["A"] | null | <p>$${L_1}:{{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$$</p>
<p>$${L_2}:{{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$$</p>
<p>Now, $$\overrightarrow p \times \overrightarrow q = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
2 & 3 & { - 1} \cr
2 & 1 & 3 \... | mcq | jee-main-2022-online-27th-june-evening-shift | 4,484 |
1l58a01dt | maths | 3d-geometry | lines-in-space | <p>If the two lines $${l_1}:{{x - 2} \over 3} = {{y + 1} \over {-2}},\,z = 2$$ and $${l_2}:{{x - 1} \over 1} = {{2y + 3} \over \alpha } = {{z + 5} \over 2}$$ are perpendicular, then an angle between the lines l<sub>2</sub> and $${l_3}:{{1 - x} \over 3} = {{2y - 1} \over { - 4}} = {z \over 4}$$ is :</p> | [{"identifier": "A", "content": "$${\\cos ^{ - 1}}\\left( {{{29} \\over 4}} \\right)$$"}, {"identifier": "B", "content": "$${\\sec ^{ - 1}}\\left( {{{29} \\over 4}} \\right)$$"}, {"identifier": "C", "content": "$${\\cos ^{ - 1}}\\left( {{2 \\over {29}}} \\right)$$"}, {"identifier": "D", "content": "$${\\cos ^{ - 1}}\\l... | ["B"] | null | <p>$$\because$$ L<sub>1</sub> and L<sub>2</sub> are perpendicular, so</p>
<p>$$3 \times 1 + ( - 2)\left( {{\alpha \over 2}} \right) + 0 \times 2 = 0$$</p>
<p>$$ \Rightarrow \alpha = 3$$</p>
<p>Now angle between l<sub>2</sub> and l<sub>3</sub>,</p>
<p>$$\cos \theta = {{1( - 3) + {\alpha \over 2}( - 2) + 2(4)} \over ... | mcq | jee-main-2022-online-26th-june-morning-shift | 4,485 |
1l58gby99 | maths | 3d-geometry | lines-in-space | <p>Let $$\overrightarrow a = \widehat i + \widehat j + 2\widehat k$$, $$\overrightarrow b = 2\widehat i - 3\widehat j + \widehat k$$ and $$\overrightarrow c = \widehat i - \widehat j + \widehat k$$ be three given vectors. Let $$\overrightarrow v $$ be a vector in the plane of $$\overrightarrow a $$ and $$\overrighta... | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "9"}] | ["D"] | null | <p>Let $$\overrightarrow v = {\lambda _1}\overrightarrow a + {\lambda _2}\overrightarrow b $$, where $${\lambda _1},\,{\lambda _2} \in R$$.</p>
<p>$$ = ({\lambda _1} + 2{\lambda _2})\widehat i + ({\lambda _1} - 3{\lambda _2})\widehat j + (2{\lambda _1} + {\lambda _2})\widehat k$$</p>
<p>$$\because$$ Projection of $$\... | mcq | jee-main-2022-online-26th-june-evening-shift | 4,486 |
1l5baeet6 | maths | 3d-geometry | lines-in-space | <p>If the shortest distance between the lines $${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over \lambda }$$ and $${{x - 2} \over 1} = {{y - 4} \over 4} = {{z - 5} \over 5}$$ is $${1 \over {\sqrt 3 }}$$, then the sum of all possible value of $$\lambda$$ is :</p> | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "15"}] | ["A"] | null | <p>Let $${\overrightarrow a _1} = \widehat i + 2\widehat j + 3\widehat k$$</p>
<p>$${\overrightarrow a _2} = 2\widehat i + 4\widehat j + 5\widehat k$$</p>
<p>$$\overrightarrow p = 2\widehat i + 3\widehat j + \lambda \widehat k,\,\overrightarrow q = \widehat i + 4\widehat j + 5\widehat k$$</p>
<p>$$\therefore$$ $$\ove... | mcq | jee-main-2022-online-24th-june-evening-shift | 4,488 |
1l5c2ach2 | maths | 3d-geometry | lines-in-space | <p>Let a line having direction ratios, 1, $$-$$4, 2 intersect the lines $${{x - 7} \over 3} = {{y - 1} \over { - 1}} = {{z + 2} \over 1}$$ and $${x \over 2} = {{y - 7} \over 3} = {z \over 1}$$ at the points A and B. Then (AB)<sup>2</sup> is equal to ___________.</p> | [] | null | 84 | Let $A(3 \lambda+7,-\lambda+1, \lambda-2)$ and $B(2 \mu, 3 \mu+7, \mu)$ <br/><br/>So, DR's of $A B \propto 3 \lambda-2 \mu+7,-(\lambda+3 \mu+6), \lambda-\mu$ $-2$
<br/><br/>
Clearly $\frac{3 \lambda-2 \mu+7}{1}=\frac{\lambda+3 \mu+6}{4}=\frac{\lambda-\mu-2}{2}$
<br/><br/>
$\Rightarrow 5 \lambda-3 \mu=-16$
<br/><br/>
An... | integer | jee-main-2022-online-24th-june-morning-shift | 4,489 |
1l5w1j95q | maths | 3d-geometry | lines-in-space | <p>Consider a triangle ABC whose vertices are A(0, $$\alpha$$, $$\alpha$$), B($$\alpha$$, 0, $$\alpha$$) and C($$\alpha$$, $$\alpha$$, 0), $$\alpha$$ > 0. Let D be a point moving on the line x + z $$-$$ 3 = 0 = y and G be the centroid of $$\Delta$$ABC. If the minimum length of GD is $$\sqrt {{{57} \over 2}} $$, then... | [] | null | 6 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l6dmv11z/765801b8-1a82-4486-8b88-3108e93f1dbd/8269d270-132e-11ed-941a-4dd6502f33e3/file-1l6dmv120.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l6dmv11z/765801b8-1a82-4486-8b88-3108e93f1dbd/8269d270-132e-11ed-941a-4dd6502f33e3... | integer | jee-main-2022-online-30th-june-morning-shift | 4,491 |
1l6f2sskl | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $$\frac{x+7}{-6}=\frac{y-6}{7}=z$$ and $$\frac{7-x}{2}=y-2=z-6$$ is :</p> | [{"identifier": "A", "content": "$$2 \\sqrt{29}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$\\sqrt{\\frac{37}{29}}$$"}, {"identifier": "D", "content": "$$\\frac{\\sqrt{29}}{2}$$"}] | ["A"] | null | <p>$${L_1}:{{x + 7} \over 6} = {{y - 6} \over 7} = {{z - 0} \over 1}$$</p>
<p>Any point on it $${\overrightarrow a _1}( - 7,6,0)$$ and L<sub>1</sub> is parallel to $${\overrightarrow b _1}( - 6,7,1)$$</p>
<p>$${L_2}:{{x - 7} \over { - 2}} = {{y - 2} \over 1} = {{z - 6} \over 1}$$</p>
<p>Any point on it $${\overrightarr... | mcq | jee-main-2022-online-25th-july-evening-shift | 4,492 |
1l6gio0a1 | maths | 3d-geometry | lines-in-space | <p>The length of the perpendicular from the point $$(1,-2,5)$$ on the line passing through $$(1,2,4)$$ and parallel to the line $$x+y-z=0=x-2 y+3 z-5$$ is :</p> | [{"identifier": "A", "content": "$$\\sqrt{\\frac{21}{2}}$$"}, {"identifier": "B", "content": "$$\\sqrt{\\frac{9}{2}}$$"}, {"identifier": "C", "content": "$$\\sqrt{\\frac{73}{2}}$$"}, {"identifier": "D", "content": "1"}] | ["A"] | null | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nbsbor/276c38c8-004d-4184-95dc-ab0e753739ef/40a706c0-2c4f-11ed-9dc0-a1792fcc650d/file-1l7nbsbos.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nbsbor/276c38c8-004d-4184-95dc-ab0e753739ef/40a706c0-2c4f-11ed-9dc0-a1792fcc650... | mcq | jee-main-2022-online-26th-july-morning-shift | 4,493 |
1l6gjpve9 | maths | 3d-geometry | lines-in-space | <p>Let $$\mathrm{Q}$$ and $$\mathrm{R}$$ be two points on the line $$\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}$$ at a distance $$\sqrt{26}$$ from the point $$P(4,2,7)$$. Then the square of the area of the triangle $$P Q R$$ is ___________.</p> | [] | null | 153 | <p>$$L:{{x + 1} \over 2} = {{y + 2} \over 3} = {{2 - 1} \over 2}$$</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nc1t7n/ab3c067b-907d-4d61-90d4-40c5dd219f33/4878de40-2c50-11ed-9dc0-a1792fcc650d/file-1l7nc1t7o.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nc1t7n/a... | integer | jee-main-2022-online-26th-july-morning-shift | 4,494 |
1ldomd3je | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines <br/><br/>$${{x - 5} \over 1} = {{y - 2} \over 2} = {{z - 4} \over { - 3}}$$ and <br/><br/>$${{x + 3} \over 1} = {{y + 5} \over 4} = {{z - 1} \over { - 5}}$$ is :</p> | [{"identifier": "A", "content": "$$7\\sqrt 3 $$"}, {"identifier": "B", "content": "$$5\\sqrt 3 $$"}, {"identifier": "C", "content": "$$4\\sqrt 3 $$"}, {"identifier": "D", "content": "$$6\\sqrt 3 $$"}] | ["D"] | null | $$
\begin{aligned}
& \mathrm{L}_1: \frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3} \\\\
& \overrightarrow{a_1}=5 \hat{i}+2 \hat{j}+4 \hat{k} \\\\
& \overrightarrow{r_1}=\hat{i}+2 \hat{j}-3 \hat{k} \\\\
& \mathrm{~L}_2: \frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5} \\\\
& \overrightarrow{a_2}=-3 \hat{i}-5 \hat{j}+\hat{k} \\\\
&... | mcq | jee-main-2023-online-1st-february-morning-shift | 4,496 |
ldr0kpgp | maths | 3d-geometry | lines-in-space | Let a line $L$ pass through the point $P(2,3,1)$ and be parallel to the line $x+3 y-2 z-2=0=x-y+2 z$. If the distance of $L$ from the point $(5,3,8)$ is $\alpha$, then $3 \alpha^2$ is equal to : | [] | null | 158 | <p>$$L:{{x - 2} \over 1} = {{y - 3} \over { - 1}} = {{z - 1} \over { - 1}} = \lambda $$</p>
<p>Any point on L can be taken as</p>
<p>$$B(\lambda + 2, - \lambda + 3, - \lambda + 1)$$</p>
<p>Let $$A(5,3,8)$$</p>
<p>So, $$AB\,.\,(\widehat i - \widehat j - \widehat k) = 0$$</p>
<p>$$[(\lambda - 3)\widehat i - \lambda \... | integer | jee-main-2023-online-30th-january-evening-shift | 4,498 |
1ldsfqrb6 | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $${{x - 1} \over 2} = {{y + 8} \over -7} = {{z - 4} \over 5}$$ and $${{x - 1} \over 2} = {{y - 2} \over 1} = {{z - 6} \over { - 3}}$$ is :</p> | [{"identifier": "A", "content": "$$2\\sqrt3$$"}, {"identifier": "B", "content": "$$3\\sqrt3$$"}, {"identifier": "C", "content": "$$4\\sqrt3$$"}, {"identifier": "D", "content": "$$5\\sqrt3$$"}] | ["C"] | null | <p>$${\overrightarrow r _1} = \widehat i - 8\widehat j + 4\widehat k$$</p>
<p>$${\overrightarrow r _2} = \widehat i + 2\widehat j + 6\widehat k$$</p>
<p>$$\overrightarrow a = 2\widehat i - 7\widehat j + 5\widehat k$$</p>
<p>$$\overrightarrow b = 2\widehat i + \widehat j - 3\widehat k$$</p>
<p>S.D. $$ = {{\left| {\mat... | mcq | jee-main-2023-online-29th-january-evening-shift | 4,499 |
1ldsx6ir2 | maths | 3d-geometry | lines-in-space | <p>Let the co-ordinates of one vertex of $$\Delta ABC$$ be $$A(0,2,\alpha)$$ and the other two vertices lie on the line $${{x + \alpha } \over 5} = {{y - 1} \over 2} = {{z + 4} \over 3}$$. For $$\alpha \in \mathbb{Z}$$, if the area of $$\Delta ABC$$ is 21 sq. units and the line segment $$BC$$ has length $$2\sqrt{21}$$ ... | [] | null | 9 | <p>A. $\left(\mathrm{O}_{1} 2, \alpha\right)$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lekt8kqs/c8535748-1d32-4d4a-8b11-6aa2b4c12893/a97e5c40-b582-11ed-b66a-d501f51666d6/file-1lekt8kqt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lekt8kqs/c8535748-1d32-4d4a-8b11... | integer | jee-main-2023-online-29th-january-morning-shift | 4,500 |
1ldu4b5nl | maths | 3d-geometry | lines-in-space | <p>The foot of perpendicular of the point (2, 0, 5) on the line $${{x + 1} \over 2} = {{y - 1} \over 5} = {{z + 1} \over { - 1}}$$ is ($$\alpha,\beta,\gamma$$). Then, which of the following is NOT correct?</p> | [{"identifier": "A", "content": "$$\\frac{\\alpha}{\\beta}=-8$$"}, {"identifier": "B", "content": "$$\\frac{\\alpha \\beta}{\\gamma}=\\frac{4}{15}$$"}, {"identifier": "C", "content": "$$\\frac{\\beta}{\\gamma}=-5$$"}, {"identifier": "D", "content": "$$\\frac{\\gamma}{\\alpha}=\\frac{5}{8}$$"}] | ["C"] | null | <p>$$
\mathrm{L}: \frac{\mathrm{x}+1}{2}=\frac{y-1}{5}=\frac{z+1}{-1}=\lambda \text { (let) }
$$</p><p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lef5vgns/9610cbeb-a0ae-4272-89c8-8cbf8a31b61b/5f9aceb0-b267-11ed-9d4d-b96eca78f2e5/file-1lef5vgnx.png?format=png" data-orsrc="https://app-content.cdn.ex... | mcq | jee-main-2023-online-25th-january-evening-shift | 4,501 |
1ldu52orl | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $$x+1=2y=-12z$$ and $$x=y+2=6z-6$$ is :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$$\\frac{5}{2}$$"}, {"identifier": "C", "content": "$$\\frac{3}{2}$$"}, {"identifier": "D", "content": "2"}] | ["D"] | null | $L_{1}: \frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{-1}{12}}$
<br/><br/>
$$
\begin{aligned}
& L_{2}: \frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}} \\\\
& \text { S.D }=\left|\frac{(-\hat{i}+2 \hat{j}-\hat{k}) \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})}{7}\right| \\\\
& =\left|\frac{-2-6-6}{7}\right|=2 \text { unit... | mcq | jee-main-2023-online-25th-january-evening-shift | 4,502 |
1ldu68afw | maths | 3d-geometry | lines-in-space | <p>If the shortest distance between the line joining the points (1, 2, 3) and (2, 3, 4), and the line $${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {{z - 2} \over 0}$$ is $$\alpha$$, then 28$$\alpha^2$$ is equal to ____________.</p> | [] | null | 18 | Points $(1,2,3)$ and $(2,3,4)$
<br/><br/>
$L_{1}: \frac{(x-1)}{1}=\frac{(y-2)}{1}=\frac{(2-3)}{1}$
<br/><br/>
$L_{2}: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-2}{0}$
<br/><br/>
$\vec{b}_{1}=\hat{i}+\hat{j}+\hat{k}$
<br/><br/>
$\vec{b}_{2}=2 \hat{i}-\hat{j}+0 \hat{k}$<br/><br/>
$$
\begin{aligned}
& \overrightarrow{a_{1}}-\... | integer | jee-main-2023-online-25th-january-evening-shift | 4,503 |
1ldv1rqg2 | maths | 3d-geometry | lines-in-space | <p>The distance of the point P(4, 6, $$-$$2) from the line passing through the point ($$-$$3, 2, 3) and parallel to a line with direction ratios 3, 3, $$-$$1 is equal to :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$$\\sqrt{14}$$"}, {"identifier": "C", "content": "$$\\sqrt6$$"}, {"identifier": "D", "content": "$$2\\sqrt3$$"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lebz24vh/fd89cb9d-00d8-44b0-a190-8609bc533a04/5f4f5dd0-b0a6-11ed-8017-e30f07067392/file-1lebz24vi.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lebz24vh/fd89cb9d-00d8-44b0-a190-8609bc533a04/5f4f5dd0-b0a6-11ed-8017-e30f07067392/fi... | mcq | jee-main-2023-online-25th-january-morning-shift | 4,504 |
1ldv1wbyd | maths | 3d-geometry | lines-in-space | <p>Consider the lines $$L_1$$ and $$L_2$$ given by</p>
<p>$${L_1}:{{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 2} \over 2}$$</p>
<p>$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$.</p>
<p>A line $$L_3$$ having direction ratios 1, $$-$$1, $$-$$2, intersects $$L_1$$ and $$L_2$$ at the points $$P$$ an... | [{"identifier": "A", "content": "$$4\\sqrt3$$"}, {"identifier": "B", "content": "$$2\\sqrt6$$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "$$3\\sqrt2$$"}] | ["B"] | null | Let,
<br/><br/>
$$
\begin{aligned}
& P \equiv(2 \lambda+1, \lambda+3,2 \lambda+2) \text { and } Q(\mu+2,2 \mu+2 \text {, } 3 \mu+3) \\\\
& \text { d.r's of } P Q \equiv<2 \lambda-\mu-1, \lambda-2 \mu+1,2 \lambda-3 \mu-1> \\\\
& \therefore \quad \frac{2 \lambda-\mu-1}{1}-\frac{\lambda-2 \mu-1}{-1}=\frac{2 \lambda-3 \mu-... | mcq | jee-main-2023-online-25th-january-morning-shift | 4,505 |
1ldwxg3ot | maths | 3d-geometry | lines-in-space | <p>If the shortest between the lines $${{x + \sqrt 6 } \over 2} = {{y - \sqrt 6 } \over 3} = {{z - \sqrt 6 } \over 4}$$ and $${{x - \lambda } \over 3} = {{y - 2\sqrt 6 } \over 4} = {{z + 2\sqrt 6 } \over 5}$$ is 6, then the square of sum of all possible values of $$\lambda$$ is :</p> | [] | null | 384 | Shortest distance between the lines<br/><br/> $\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$<br/><br/> $\frac{x-\lambda}{3}=\frac{y-2 \sqrt{6}}{4}=\frac{2+2 \sqrt{6}}{5}$ is 6<br/><br/>
Vector along line of shortest distance<br/><br/> $=\left|\begin{array}{lll}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 2... | integer | jee-main-2023-online-24th-january-evening-shift | 4,506 |
1ldybxlrv | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 6} \over 2}$$ and $${{x - 6} \over 3} = {{1 - y} \over 2} = {{z + 8} \over 0}$$ is equal to ________</p> | [] | null | 14 | <p>For $${L_1}:$$$${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 6} \over 2}$$</p>
<p>$$\therefore$$ Point on line is $$A(2,-1,6)$$</p>
<p>Parallel vector to this line is,</p>
<p>$$\overrightarrow p = 3\widehat i + 2\widehat j + 2\widehat k$$</p>
<p>For $${L_2}:$$$${{x - 6} \over 3} = {{y - 1} \over -2} = {{z + 8} \ov... | integer | jee-main-2023-online-24th-january-morning-shift | 4,507 |
1lgowajui | maths | 3d-geometry | lines-in-space | <p>The line, that is coplanar to the line $$\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$$, is :</p> | [{"identifier": "A", "content": "$$\\frac{x+1}{-1}=\\frac{y-2}{2}=\\frac{z-5}{4}$$"}, {"identifier": "B", "content": "$$\\frac{x+1}{-1}=\\frac{y-2}{2}=\\frac{z-5}{5}$$"}, {"identifier": "C", "content": "$$\\frac{x-1}{-1}=\\frac{y-2}{2}=\\frac{z-5}{5}$$"}, {"identifier": "D", "content": "$$\\frac{x+1}{1}=\\frac{y-2}{2}=... | ["B"] | null | <p>Given two lines:</p>
<p>$\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}$</p>
<p>and </p>
<p>$\frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}$</p>
<p>These lines are coplanar if the determinant of the matrix</p>
<p>$$
\begin{vmatrix}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a... | mcq | jee-main-2023-online-13th-april-evening-shift | 4,509 |
1lgrebnuy | maths | 3d-geometry | lines-in-space | <p>Let the lines $$l_{1}: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}$$ and $$l_{2}: 3 x+2 y+z-2=0=x-3 y+2 z-13$$ be coplanar. If the point $$\mathrm{P}(a, b, c)$$ on $$l_{1}$$ is nearest to the point $$\mathrm{Q}(-4,-3,2)$$, then $$|a|+|b|+|c|$$ is equal to</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "8"}] | ["C"] | null | $$
\begin{aligned}
& (3 \mathrm{x}+2 \mathrm{y}+\mathrm{z}-2)+\mu(\mathrm{x}-3 \mathrm{y}+2 \mathrm{z}-13)=0 \\\\
& 3(3+\mu)+1 \cdot(2-3 \mu)-2(1+2 \mu)=0 \\\\
& 9-4 \mu=0 \\\\
& \mu=\frac{9}{4} \\\\
& 4(-15-8+\alpha-2)+9(-5+12+2 \alpha-13)=0 \\\\
& -100+4 \alpha-54+18 \alpha=0 \\\\
& \Rightarrow \alpha=7 \\\\
& \text ... | mcq | jee-main-2023-online-12th-april-morning-shift | 4,510 |
1lguwwp0s | maths | 3d-geometry | lines-in-space | <p>Let a line $$l$$ pass through the origin and be perpendicular to the lines</p>
<p>$$l_{1}: \vec{r}=(\hat{\imath}-11 \hat{\jmath}-7 \hat{k})+\lambda(\hat{i}+2 \hat{\jmath}+3 \hat{k}), \lambda \in \mathbb{R}$$ and</p>
<p>$$l_{2}: \vec{r}=(-\hat{\imath}+\hat{\mathrm{k}})+\mu(2 \hat{\imath}+2 \hat{\jmath}+\hat{\mathrm{k... | [] | null | 5 | We have,<br/>
<p>$$l_{1}: \vec{r}=(\hat{\imath}-11 \hat{\jmath}-7 \hat{k})+\lambda(\hat{i}+2 \hat{\jmath}+3 \hat{k}), \lambda \in \mathbb{R}$$ and</p>
<p>$$l_{2}: \vec{r}=(-\hat{\imath}+\hat{\mathrm{k}})+\mu(2 \hat{\imath}+2 \hat{\jmath}+\hat{\mathrm{k}}), \mu \in \mathbb{R}$$.</p>
Let direction ratio of line $l$ be $a... | integer | jee-main-2023-online-11th-april-morning-shift | 4,511 |
1lgxgzye9 | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $${{x + 2} \over 1} = {y \over { - 2}} = {{z - 5} \over 2}$$ and $${{x - 4} \over 1} = {{y - 1} \over 2} = {{z + 3} \over 0}$$ is :</p> | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "9"}] | ["D"] | null | Given, the lines are
<br/><br/>$$
\begin{aligned}
\frac{x+2}{1} & =\frac{y}{-2}=\frac{z-5}{2} ~~~~..........(i)\\\\
\text { and } \frac{x-4}{1} & =\frac{y-1}{2}=\frac{z+3}{0} ~~~~..........(ii)
\end{aligned}
$$
<br/><br/>Formula for shortest distance between two skew-lines,
<br/><br/>$$
\begin{aligned}
S D & =\left|\fr... | mcq | jee-main-2023-online-10th-april-morning-shift | 4,512 |
1lgzydtfy | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $$\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$$ and $$\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$$ is :</p> | [{"identifier": "A", "content": "$$3 \\sqrt{6}$$"}, {"identifier": "B", "content": "$$6 \\sqrt{2}$$"}, {"identifier": "C", "content": "$$6 \\sqrt{3}$$"}, {"identifier": "D", "content": "$$2 \\sqrt{6}$$"}] | ["A"] | null | The given lines are
<br/><br/>$$\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$$ and $$\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$$
<br/><br/>$$
\begin{aligned}
& \text { So, } \vec{b}_1=4 \hat{i}+5 \hat{j}+3 \hat{k} \\\\
& \vec{b}_2=3 \hat{i}+4 \hat{j}+2 \hat{k} \\\\
& \vec{a}_1=4 \hat{i}-2 \hat{j}-3 \hat{k} \\\\
&\vec{a}_2... | mcq | jee-main-2023-online-8th-april-morning-shift | 4,513 |
lsamrrrt | maths | 3d-geometry | lines-in-space | If the mirror image of the point $P(3,4,9)$ in the line
<br/><br/>$\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}$ is $(\alpha, \beta, \gamma)$, then 14 $(\alpha+\beta+\gamma)$ is : | [{"identifier": "A", "content": "102"}, {"identifier": "B", "content": "138"}, {"identifier": "C", "content": "132"}, {"identifier": "D", "content": "108"}] | ["D"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsohd1pb/c1eafb9f-de3f-4c37-82e0-908506938702/622ee6f0-ccb2-11ee-aa98-13f456b8f7af/file-6y3zli1lsohd1pc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsohd1pb/c1eafb9f-de3f-4c37-82e0-908506938702/622ee6f0-ccb2-11ee-aa... | mcq | jee-main-2024-online-1st-february-evening-shift | 4,517 |
lsaptbc8 | maths | 3d-geometry | lines-in-space | If the shortest distance between the lines <br/><br/>$\frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$ is 1 , then the sum of all possible values of $\lambda$ is : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$2 \\sqrt{3}$"}, {"identifier": "C", "content": "$3 \\sqrt{3}$"}, {"identifier": "D", "content": "$-2 \\sqrt{3}$"}] | ["B"] | null | <p>Given the two lines:</p>
<p>$$ L_1: \frac{x-\lambda}{-2} = \frac{y-2}{1} = \frac{z-1}{1} $$</p>
<p>$$ L_2: \frac{x-\sqrt{3}}{1} = \frac{y-1}{-2} = \frac{z-2}{1} $$</p>
<p>We observe that these lines are not parallel as their directional vectors are not proportional. The directional vector for $L_1$ is $(-2,1,1)$ ... | mcq | jee-main-2024-online-1st-february-morning-shift | 4,518 |
lsaqawqo | maths | 3d-geometry | lines-in-space | Let the line of the shortest distance between the lines
<br/><br/>$$
\begin{aligned}
& \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) \text { and } \\\\
& \mathrm{L}_2: \overrightarrow{\mathrm{r}}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{... | [] | null | 21 | $\begin{array}{ll}L_1 \equiv \vec{r}=(1,2,3)+\lambda(1,-1,1) & \left(\vec{r}=\vec{a}_1+\lambda \vec{b}_1\right) \\\\ L_2 \equiv \vec{r}=(4,5,6)+\mu(1,1,-1) & \left(\vec{r}=\vec{a}_2+\lambda \vec{b}_2\right)\end{array}$
<br/><br/>$P \equiv(\lambda+1,-\lambda+2, \lambda+3)$
<br/><br/>$Q \equiv(\mu+4, \mu+5,-\mu+6)$
<br/>... | integer | jee-main-2024-online-1st-february-morning-shift | 4,519 |
lsbklvv1 | maths | 3d-geometry | lines-in-space | The distance, of the point $(7,-2,11)$ from the line <br/><br/>$\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is : | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "18"}, {"identifier": "C", "content": "21"}, {"identifier": "D", "content": "14"}] | ["D"] | null | <p>$$\mathrm{B}=(2 \lambda+7,-3 \lambda-2,6 \lambda+11)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1c0zyx/3ef0ec08-6ff8-4ca7-97e0-1c5c0fde9c3c/b5ee7090-d3c3-11ee-a50b-bb659a2e1d74/file-1lt1c0zyy.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1c0zyx/3ef0ec08-6f... | mcq | jee-main-2024-online-27th-january-morning-shift | 4,520 |
lsbkxlh7 | maths | 3d-geometry | lines-in-space | If the shortest distance between the lines <br/><br/>$\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$ is $\frac{6}{\sqrt{5}}$, then the sum of all possible values of $\lambda$ is : | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "8"}] | ["D"] | null | <p>$$\begin{aligned}
& \frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3} \\
& \frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}
\end{aligned}$$</p>
<p>the shortest distance between the lines</p>
<p>$$=\left|\frac{(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \cdot\left(\overrightarrow{\mathrm{d}_1} \times \overrighta... | mcq | jee-main-2024-online-27th-january-morning-shift | 4,521 |
jaoe38c1lscn35vz | maths | 3d-geometry | lines-in-space | <p>Let the image of the point $$(1,0,7)$$ in the line $$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$$ be the point $$(\alpha, \beta, \gamma)$$. Then which one of the following points lies on the line passing through $$(\alpha, \beta, \gamma)$$ and making angles $$\frac{2 \pi}{3}$$ and $$\frac{3 \pi}{4}$$ with $$y$$-axis an... | [{"identifier": "A", "content": "$$(1,-2,1+\\sqrt{2})$$\n"}, {"identifier": "B", "content": "$$(3,-4,3+2 \\sqrt{2})$$\n"}, {"identifier": "C", "content": "$$(3,4,3-2 \\sqrt{2})$$\n"}, {"identifier": "D", "content": "$$(1,2,1-\\sqrt{2})$$"}] | ["C"] | null | <p>$$\mathrm{L}_1=\frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}=\lambda$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1toimr/45988a05-1bb2-47f1-9316-43cb63fa6ccd/bf271320-d408-11ee-b9d5-0585032231f0/file-1lt1toimt.png?format=png" data-orsrc="https://app-content.cdn.exa... | mcq | jee-main-2024-online-27th-january-evening-shift | 4,522 |
jaoe38c1lscogowm | maths | 3d-geometry | lines-in-space | <p>The lines $$\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}$$ and $$\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}$$ intersect at the point $$P$$. If the distance of $$\mathrm{P}$$ from the line $$\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$$ is $$l$$, then $$14 l^2$$ is equal to __________.</p> | [] | null | 108 | <p>$$\begin{aligned}
& \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-7}{8}=\lambda \\
& \frac{\mathrm{x}+3}{4}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}+2}{1}=\mathrm{k} \\
& \Rightarrow \lambda+2=4 \mathrm{k}-3 \\
& -\lambda=3 \mathrm{k}-2 \\
& \Rightarrow \mathrm{k}=1, \lambda=-1 \\
... | integer | jee-main-2024-online-27th-january-evening-shift | 4,523 |
jaoe38c1lsd3h061 | maths | 3d-geometry | lines-in-space | <p>Let $$(\alpha, \beta, \gamma)$$ be the mirror image of the point $$(2,3,5)$$ in the line $$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$$. Then, $$2 \alpha+3 \beta+4 \gamma$$ is equal to</p> | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "31"}, {"identifier": "C", "content": "33"}, {"identifier": "D", "content": "34"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsjwr9eg/7ae9597b-3dbe-4745-92d2-f1eee5f94c62/ab25ed80-ca2e-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwr9eh.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsjwr9eg/7ae9597b-3dbe-4745-92d2-f1eee5f94c62/ab25ed80-ca2e-11ee... | mcq | jee-main-2024-online-31st-january-evening-shift | 4,524 |
jaoe38c1lsd4flem | maths | 3d-geometry | lines-in-space | <p>The shortest distance, between lines $$L_1$$ and $$L_2$$, where $$L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$$ and $$L_2$$ is the line, passing through the points $$\mathrm{A}(-4,4,3), \mathrm{B}(-1,6,3)$$ and perpendicular to the line $$\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$$, is</p> | [{"identifier": "A", "content": "$$\\frac{141}{\\sqrt{221}}$$\n"}, {"identifier": "B", "content": "$$\\frac{24}{\\sqrt{117}}$$\n"}, {"identifier": "C", "content": "$$\\frac{42}{\\sqrt{117}}$$\n"}, {"identifier": "D", "content": "$$\\frac{121}{\\sqrt{221}}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& \mathrm{L}_2=\frac{\mathrm{x}+4}{3}=\frac{\mathrm{y}-4}{2}=\frac{\mathrm{z}-3}{0} \\
& \therefore \mathrm{S} . \mathrm{D}=\frac{\left|\begin{array}{ccc}
\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\
2 & -3 & 2 \\
3 & 2 & 0
\end{array}\right|}{\left|\overrigh... | mcq | jee-main-2024-online-31st-january-evening-shift | 4,525 |
jaoe38c1lsd5a4ux | maths | 3d-geometry | lines-in-space | <p>A line passes through $$A(4,-6,-2)$$ and $$B(16,-2,4)$$. The point $$P(a, b, c)$$, where $$a, b, c$$ are non-negative integers, on the line $$A B$$ lies at a distance of 21 units, from the point $$A$$. The distance between the points $$P(a, b, c)$$ and $$Q(4,-12,3)$$ is equal to __________.</p> | [] | null | 22 | <p>$$\begin{aligned}
& \frac{x-4}{12}=\frac{x+6}{4}=\frac{z+2}{6} \\
& \frac{x-4}{\frac{6}{7}}=\frac{y+6}{\frac{2}{7}}=\frac{z+2}{\frac{3}{7}}=21 \\
& \left(21 \times \frac{6}{7}+4, \frac{2}{7} \times 21-6, \frac{3}{7} \times 21-2\right) \\
& =(22,0,7)=(a, b, c) \\
& \therefore \sqrt{324+144+16}=22
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-evening-shift | 4,526 |
jaoe38c1lse5wwlu | maths | 3d-geometry | lines-in-space | <p>Let $$\mathrm{Q}$$ and $$\mathrm{R}$$ be the feet of perpendiculars from the point $$\mathrm{P}(a, a, a)$$ on the lines $$x=y, z=1$$ and $$x=-y, z=-1$$ respectively. If $$\angle \mathrm{QPR}$$ is a right angle, then $$12 a^2$$ is equal to _________.</p> | [] | null | 12 | <p>$$\begin{aligned}
& \frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=r \rightarrow Q(r, r, 1) \\
& \frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=k \rightarrow R(k,-k,-1) \\
& \overline{P Q}=(a-r) \hat{i}+(a-r) \hat{j}+(a-1) \hat{k} \\
& a=r+a-r=0 \\
& 2 a=2 r \rightarrow a=r \\
& \overline{P R}=(a-k) i+(a+k) \hat{j}+(a+1) \hat{k} \\
... | integer | jee-main-2024-online-31st-january-morning-shift | 4,527 |
jaoe38c1lsfkztkt | maths | 3d-geometry | lines-in-space | <p>Let O be the origin, and M and $$\mathrm{N}$$ be the points on the lines $$\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$$ and $$\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$$ respectively such that $$\mathrm{MN}$$ is the shortest distance between the given lines. Then $$\overrightarrow{O M} \cdot \overrightarrow{O N}$$ ... | [] | null | 9 | <p>$$\mathrm{L}_1: \frac{\mathrm{x}-5}{4}=\frac{\mathrm{y}-4}{1}=\frac{\mathrm{z}-5}{3}=\lambda \quad \operatorname{drs}(4,1,3)=\mathrm{b}_1$$</p>
<p>$$\begin{aligned}
& \mathrm{M}(4 \lambda+5, \lambda+4,3 \lambda+5) \\
& \mathrm{L}_2: \frac{\mathrm{x}+8}{12}=\frac{\mathrm{y}+2}{5}=\frac{\mathrm{z}+11}{9}=\mu \\
& \mat... | integer | jee-main-2024-online-29th-january-evening-shift | 4,530 |
1lsg4acyi | maths | 3d-geometry | lines-in-space | <p>Let $$L_1: \vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k}), \lambda \in \mathbb{R}$$,</p>
<p>$$L_2: \vec{r}=(\hat{j}-\hat{k})+\mu(3 \hat{i}+\hat{j}+p \hat{k}), \mu \in \mathbb{R} \text {, and } L_3: \vec{r}=\delta(\ell \hat{i}+m \hat{j}+n \hat{k}), \delta \in \mathbb{R}$$</p>
<p>be three lines... | [{"identifier": "A", "content": "$$(1,7,-4)$$\n"}, {"identifier": "B", "content": "$$(1,-7,4)$$\n"}, {"identifier": "C", "content": "$$(-1,7,4)$$\n"}, {"identifier": "D", "content": "$$(-, 1-7,4)$$"}] | ["C"] | null | <p>$$\mathrm{L}_1 \perp \mathrm{L}_2 \quad \mathrm{~L}_3 \perp \mathrm{L}_1, \mathrm{~L}_2$$</p>
<p>$$\begin{aligned}
& 3-1+2 \mathrm{P}=0 \\
& \mathrm{P}=-1 \\
& \left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
1 & -1 & 2 \\
3 & 1 & -1
\end{array}\right|=-\hat{\mathrm{i}}+7 \hat{\math... | mcq | jee-main-2024-online-30th-january-evening-shift | 4,531 |
1lsgclxib | maths | 3d-geometry | lines-in-space | <p>If $$\mathrm{d}_1$$ is the shortest distance between the lines $$x+1=2 y=-12 z, x=y+2=6 z-6$$ and $$\mathrm{d}_2$$ is the shortest distance between the lines $$\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$$, then the value of $$\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}... | [] | null | 16 | <p>$$\mathrm{L}_1: \frac{\mathrm{x}+1}{1}=\frac{\mathrm{y}}{1 / 2}=\frac{\mathrm{z}}{-1 / 12}, \mathrm{~L}_2: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}+2}{1}=\frac{\mathrm{z}-1}{\frac{1}{6}}$$</p>
<p>$$\mathrm{d}_1=$$ shortest distance between $$\mathrm{L}_1 ~\& \mathrm{~L}_2$$</p>
<p>$$\begin{aligned}
& =\left|\frac{\left... | integer | jee-main-2024-online-30th-january-morning-shift | 4,534 |
luxwden8 | maths | 3d-geometry | lines-in-space | <p>The square of the distance of the image of the point $$(6,1,5)$$ in the line $$\frac{x-1}{3}=\frac{y}{2}=\frac{z-2}{4}$$, from the origin is __________.</p> | [] | null | 62 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw1my772/a7f8567c-57ab-46bd-a498-b733b7e930a6/12a4b0e0-0f53-11ef-91cd-f19f7dc20f18/file-1lw1my773.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw1my772/a7f8567c-57ab-46bd-a498-b733b7e930a6/12a4b0e0-0f53-11ef-91cd-f19f7dc20f18... | integer | jee-main-2024-online-9th-april-evening-shift | 4,536 |
luy6z5hh | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $$\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$$ and $$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$$ is:</p> | [{"identifier": "A", "content": "$$\\frac{185}{\\sqrt{563}}$$\n"}, {"identifier": "B", "content": "$$\\frac{187}{\\sqrt{563}}$$\n"}, {"identifier": "C", "content": "$$\\frac{178}{\\sqrt{563}}$$\n"}, {"identifier": "D", "content": "$$\\frac{179}{\\sqrt{563}}$$"}] | ["B"] | null | <p>Given lines are</p>
<p>$$\frac{x-3}{4}=\frac{y-(-7)}{-11}=\frac{z-1}{5}$$ and</p>
<p>$$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z-(-2)}{1}$$</p>
<p>Shortest distance between two lines,</p>
<p>$$\begin{aligned}
& d=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\left... | mcq | jee-main-2024-online-9th-april-morning-shift | 4,537 |
luy6z4qf | maths | 3d-geometry | lines-in-space | <p>Let the line $$\mathrm{L}$$ intersect the lines $$x-2=-y=z-1,2(x+1)=2(y-1)=z+1$$ and be parallel to the line $$\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$$. Then which of the following points lies on $$\mathrm{L}$$ ?</p> | [{"identifier": "A", "content": "$$\\left(-\\frac{1}{3}, 1,-1\\right)$$\n"}, {"identifier": "B", "content": "$$\\left(-\\frac{1}{3},-1,1\\right)$$\n"}, {"identifier": "C", "content": "$$\\left(-\\frac{1}{3},-1,-1\\right)$$\n"}, {"identifier": "D", "content": "$$\\left(-\\frac{1}{3}, 1,1\\right)$$"}] | ["A"] | null | <p>$$\begin{aligned}
& L_1: \frac{x-2}{1}=\frac{y}{-1}=\frac{z-1}{1}=\lambda \\
& L_2: \frac{x+1}{(1 / 2)}=\frac{y-1}{(1 / 2)}=\frac{z+1}{1}=\mu
\end{aligned}$$</p>
<p>Any point of $$L_1$$ and $$L_2$$ will be $$(\lambda+2,-\lambda, \lambda+1)$$ and
$$\left(\frac{\mu}{2}-1, \frac{\mu}{2}+1, \mu-1\right)$$</p>
<p>Now Dr ... | mcq | jee-main-2024-online-9th-april-morning-shift | 4,538 |
lv2er9tw | maths | 3d-geometry | lines-in-space | <p>Consider a line $$\mathrm{L}$$ passing through the points $$\mathrm{P}(1,2,1)$$ and $$\mathrm{Q}(2,1,-1)$$. If the mirror image of the point $$\mathrm{A}(2,2,2)$$ in the line $$\mathrm{L}$$ is $$(\alpha, \beta, \gamma)$$, then $$\alpha+\beta+6 \gamma$$ is equal to __________.</p> | [] | null | 6 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwhk80oi/a0341b0b-ecfa-449a-9d8f-e828635e1020/f2b47520-1814-11ef-a08c-4f00e8f9a4da/file-1lwhk80oj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwhk80oi/a0341b0b-ecfa-449a-9d8f-e828635e1020/f2b47520-1814-11ef-a08c-4f00e8f9a4da... | integer | jee-main-2024-online-4th-april-evening-shift | 4,541 |
lv3vef5l | maths | 3d-geometry | lines-in-space | <p>If the shortest distance between the lines $$\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$$ and $$\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$$ is $$\frac{13}{\sqrt{29}}$$, then a value of $$\lambda$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{13}{25}$$\n"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$-$$1\n"}, {"identifier": "D", "content": "$$-\\frac{13}{25}$$"}] | ["B"] | null | <p>$$\begin{aligned}
& \vec{l}_1=2 \hat{i}+3 \hat{j}+4 \hat{k} \\
& \vec{l}_2=4 \hat{i}+6 \hat{j}+8 \hat{k}
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4lkbz1/158cd7a8-1619-430b-bf76-8b2ddd399871/01d314d0-10f4-11ef-8553-fdfc6347789d/file-1lw4lkbz2.png?format=png" da... | mcq | jee-main-2024-online-8th-april-evening-shift | 4,542 |
lv3vefgs | maths | 3d-geometry | lines-in-space | <p>Let $$\mathrm{P}(\alpha, \beta, \gamma)$$ be the image of the point $$\mathrm{Q}(1,6,4)$$ in the line $$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$$. Then $$2 \alpha+\beta+\gamma$$ is equal to ________</p> | [] | null | 11 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4mrnop/6ea8698b-9387-47c8-8aca-29425e7a83aa/b6aa6990-10f8-11ef-b50f-395513bc7af2/file-1lw4mrnoq.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw4mrnop/6ea8698b-9387-47c8-8aca-29425e7a83aa/b6aa6990-10f8-11ef-b50f-395513bc7af2... | integer | jee-main-2024-online-8th-april-evening-shift | 4,543 |
lv5gsxy6 | maths | 3d-geometry | lines-in-space | <p>If the shortest distance between the lines</p>
<p>$$\begin{array}{ll}
L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}, & \lambda \in \mathbb{R} \\
L_2: \vec{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, & \mu \in \mathbb{R}
\end{array}$$</p>
<p>is $$\frac{m}{\sqrt{n}}$$, ... | [{"identifier": "A", "content": "384"}, {"identifier": "B", "content": "387"}, {"identifier": "C", "content": "390"}, {"identifier": "D", "content": "377"}] | ["B"] | null | <p>$$\begin{aligned}
& L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k} \\
& L_1=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(\hat{i}-3 \hat{j}+4 \hat{k}) \\
& L_2: \vec{r}=2 \hat{i}+3 \hat{j}+5 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k}) \\
& \vec{a}_1=2 \hat{i}+\hat{j}+3 \hat{k} \\
& \vec{a}_2=2 \hat{... | mcq | jee-main-2024-online-8th-april-morning-shift | 4,544 |
lv7v3jvb | maths | 3d-geometry | lines-in-space | <p>Let $$\mathrm{d}$$ be the distance of the point of intersection of the lines $$\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$$ and $$\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$$ from the point $$(7,8,9)$$. Then $$\mathrm{d}^2+6$$ is equal to :</p> | [{"identifier": "A", "content": "75"}, {"identifier": "B", "content": "78"}, {"identifier": "C", "content": "72"}, {"identifier": "D", "content": "69"}] | ["A"] | null | <p>$$\begin{aligned}
& P_1:(3 k-6,2 k, k-1) \\
& P_2(4 \alpha+7,3 \alpha+9,2 \alpha+4) \\
& P_1 \equiv P_2 \\
& 3 k-6=4 \alpha+7 \Rightarrow 3 k-4 \alpha=13 \\
& 2 k=3 \alpha+9 \Rightarrow 2 k-3 \alpha=9 \\
& \therefore k=3, \alpha=-1 \\
& \therefore P_1:(3,6,2)
\end{aligned}$$</p>
<p>Distance of $$(3,6,2)$$ and $$(7,8... | mcq | jee-main-2024-online-5th-april-morning-shift | 4,545 |
lv9s20kw | maths | 3d-geometry | lines-in-space | <p>Let the point $$(-1, \alpha, \beta)$$ lie on the line of the shortest distance between the lines $$\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$$ and $$\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$$. Then $$(\alpha-\beta)^2$$ is equal to _________.</p> | [] | null | 25 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lweuubn4/2752e8cb-54f7-4c91-8212-691518b0463e/20511700-1698-11ef-ad74-71e0a1829759/file-1lweuubn5.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lweuubn4/2752e8cb-54f7-4c91-8212-691518b0463e/20511700-1698-11ef-ad74-71e0a1829759... | integer | jee-main-2024-online-5th-april-evening-shift | 4,547 |
lvb294m1 | maths | 3d-geometry | lines-in-space | <p>Let $$\mathrm{P}(\alpha, \beta, \gamma)$$ be the image of the point $$\mathrm{Q}(3,-3,1)$$ in the line $$\frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}$$ and $$\mathrm{R}$$ be the point $$(2,5,-1)$$. If the area of the triangle $$\mathrm{PQR}$$ is $$\lambda$$ and $$\lambda^2=14 \mathrm{~K}$$, then $$\mathrm{K}$$ is equa... | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "81"}, {"identifier": "C", "content": "72"}, {"identifier": "D", "content": "36"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwad7urv/960c0402-e139-4dd9-89a4-53df9b260717/11b135b0-1420-11ef-90c1-6758c45d5635/file-1lwad7urw.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwad7urv/960c0402-e139-4dd9-89a4-53df9b260717/11b135b0-1420-11ef-90c1-6758c45d5635... | mcq | jee-main-2024-online-6th-april-evening-shift | 4,548 |
lvc58e18 | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $$\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$$ and $$\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$$ is</p> | [{"identifier": "A", "content": "$$8 \\sqrt{3}$$\n"}, {"identifier": "B", "content": "$$6 \\sqrt{3}$$\n"}, {"identifier": "C", "content": "$$5 \\sqrt{3}$$\n"}, {"identifier": "D", "content": "$$4 \\sqrt{3}$$"}] | ["D"] | null | <p>Given two lines are represented as:</p>
<p>$ \frac{x-3}{2} = \frac{y+15}{-7} = \frac{z-9}{5} $</p>
<p>and</p>
<p>$ \frac{x+1}{2} = \frac{y-1}{1} = \frac{z-9}{-3} $</p>
<p>The formula for the shortest distance between two lines is:</p>
<p>$ d = \frac{|\left(\vec{a}_2 - \vec{a}_1\right) \cdot \left(\vec{b}_1 \tim... | mcq | jee-main-2024-online-6th-april-morning-shift | 4,550 |
lvc58e9z | maths | 3d-geometry | lines-in-space | <p>Let $$P$$ be the point $$(10,-2,-1)$$ and $$Q$$ be the foot of the perpendicular drawn from the point $$R(1,7,6)$$ on the line passing through the points $$(2,-5,11)$$ and $$(-6,7,-5)$$. Then the length of the line segment $$P Q$$ is equal to _________.</p> | [] | null | 13 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwd5pdpk/204a0778-5148-47b3-a659-091fdb696421/09fd8280-15a9-11ef-99cc-63deb36dcc4a/file-1lwd5pdpl.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwd5pdpk/204a0778-5148-47b3-a659-091fdb696421/09fd8280-15a9-11ef-99cc-63deb36dcc4a... | integer | jee-main-2024-online-6th-april-morning-shift | 4,551 |
ZZ2mbSw3xnje5rne | maths | 3d-geometry | plane-in-space | The $$d.r.$$ of normal to the plane through $$(1, 0, 0), (0, 1, 0)$$ which makes an angle $$\pi /4$$ with plane $$x+y=3$$ are : | [{"identifier": "A", "content": "$$1,\\sqrt 2 ,1$$ "}, {"identifier": "B", "content": "$$1,1,\\sqrt 2 $$ "}, {"identifier": "C", "content": "$$1, 1, 2$$"}, {"identifier": "D", "content": "$$\\sqrt 2 ,1,1$$ "}] | ["B"] | null | Equation of plane through $$\left( {1,0,0} \right)$$ is
<br><br>$$a\left( {x - 1} \right) + by + cz = 0\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>$$(i)$$ passes through $$\left( {0,1,0} \right).$$
<br><br>$$ - a + b = 0 \Rightarrow b = a;$$
<br><br>Also, $$\cos {45^ \circ }$$
<br><br>$$ = {{a + a} \over {\sqrt ... | mcq | aieee-2002 | 4,552 |
CFmX8122sUhvE9Dl | maths | 3d-geometry | plane-in-space | Two systems of rectangular axes have the same origin. If a plane cuts then at distances $$a,b,c$$ and $$a', b', c'$$ from the origin then | [{"identifier": "A", "content": "$${1 \\over {{a^2}}} + {1 \\over {{b^2}}} + {1 \\over {{c^2}}} - {1 \\over {a{'^2}}} - {1 \\over {b{'^2}}} - {1 \\over {c{'^2}}} = 0$$ "}, {"identifier": "B", "content": "$$\\,{1 \\over {{a^2}}} + {1 \\over {{b^2}}} + {1 \\over {{c^2}}} + {1 \\over {a{'^2}}} + {1 \\over {b{'^2}}} + {1 \... | ["A"] | null | Equation of planes be $${x \over a} + {y \over b} + {z \over c} = 1\,\,\& \,\,{x \over {a'}} + {y \over {b'}} + {z \over {c'}} = 1$$
<br><br>So the distance from (0, 0, 0) to both the plane is same.
<br><br>$$ \therefore $$ $$\left| {{{ - 1} \over {\sqrt {{1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}}} }... | mcq | aieee-2003 | 4,554 |
HYn9nj1UWdOirDBu | maths | 3d-geometry | plane-in-space | The radius of the circle in which the sphere
<br/><br>$${x^2} + {y^2} + {z^2} + 2x - 2y - 4z - 19 = 0$$ is cut by the plane
<br/><br>$$x+2y+2z+7=0$$ is </br></br> | [{"identifier": "A", "content": "$$4$$"}, {"identifier": "B", "content": "$$1$$"}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$$3$$"}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265346/exam_images/c0dqhctziwbekfugrolb.webp" loading="lazy" alt="AIEEE 2003 Mathematics - 3D Geometry Question 321 English Explanation">
<br><br>Center of sphere $$ = \left( { - 1,1,2} \right)$$
<br><br>Radius of sphere $$\sqrt {1 ... | mcq | aieee-2003 | 4,555 |
wXNJAE9OixV0o1Ul | maths | 3d-geometry | plane-in-space | The intersection of the spheres
<br/>$${x^2} + {y^2} + {z^2} + 7x - 2y - z = 13$$ and
<br/>$${x^2} + {y^2} + {z^2} - 3x + 3y + 4z = 8$$
<br/>is the same as the intersection of one of the sphere and the plane | [{"identifier": "A", "content": "$$2x-y-z=1$$ "}, {"identifier": "B", "content": "$$x-2y-z=1$$"}, {"identifier": "C", "content": "$$x-y-2z=1$$ "}, {"identifier": "D", "content": "$$x-y-z=1$$ "}] | ["A"] | null | The equation of spheres are
<br><br>$${S_1}:{x^2} + {y^2} + {z^2} + 7x - 2y - z - 13 = 0$$ and
<br><br>$${S_2}:{x^2} + {y^2} + {z^2} - 3x + 3y + 4z - 8 = 0$$
<br><br>Their plane of intersection is $${S_1} - {S_2} = 0$$
<br><br>$$ \Rightarrow 10x - 5y - 5z - 5 = 0$$
<br><br>$$ \Rightarrow 2x - y - z = 1$$ | mcq | aieee-2004 | 4,556 |
wUkvqWxVsZ5qPXxv | maths | 3d-geometry | plane-in-space | Distance between two parallel planes
<br/><br>$$\,2x + y + 2z = 8$$ and $$4x + 2y + 4z + 5 = 0$$ is :</br> | [{"identifier": "A", "content": "$${9 \\over 2}$$ "}, {"identifier": "B", "content": "$${5 \\over 2}$$"}, {"identifier": "C", "content": "$${7 \\over 2}$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}] | ["C"] | null | The planes are $$2x + y + 2x - 8 = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>and $$4x + 2y + 4z + 5 = 0$$
<br><br>or $$2x + y + 2z + {5 \over 2} = 0\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>$$\therefore$$ Distance between $$\left( 1 \right)$$ and $$\,\left( 2 \right)$$
<br><br>$$ = \left| {{{{5 \over ... | mcq | aieee-2004 | 4,557 |
gsaP2wxbhrwI4MNS | maths | 3d-geometry | plane-in-space | The plane $$x+2y-z=4$$ cuts the sphere $${x^2} + {y^2} + {z^2} - x + z - 2 = 0$$ in a circle of radius | [{"identifier": "A", "content": "$$3$$ "}, {"identifier": "B", "content": "$$1$$ "}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$${\\sqrt 2 }$$"}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265266/exam_images/iljyyh25rrxltbqfibh1.webp" loading="lazy" alt="AIEEE 2005 Mathematics - 3D Geometry Question 311 English Explanation">
<br><br>Perpendicular distance of center $$\left( {{1 \over 2},0, - {1 \over 2}} \right)$$
<br... | mcq | aieee-2005 | 4,558 |
jSTkPlfBJ5I6fd65 | maths | 3d-geometry | plane-in-space | <b>Statement-1 :</b> The point $$A(3, 1, 6)$$ is the mirror image of the point $$B(1, 3, 4)$$ in the plane $$x-y+z=5.$$
<br/><br><b>Statement-2 :</b> The plane $$x-y+z=5$$ bisects the line segment joining $$A(3, 1, 6)$$ and $$B(1, 3, 4).$$ </br> | [{"identifier": "A", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is <b>not</b> a correct explanation for Statement - 1."}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is false."}, {"identifier": "C", "content": "Statement - 1 is false , Statement - 2 is true."}, {"id... | ["A"] | null | $$A\left( {3,1,6} \right);\,\,B = \left( {1,3,4} \right)$$
<br><br>Mid-point of $$AB = \left( {2,2,5} \right)$$ lies on the plane.
<br><br>and d.r's of $$AB=(2,-2,2)$$
<br><br>d.r's of normal to plane $$ = \left( {1, - 1,1} \right).$$
<br><br>Direction ratio of $$AB$$ and normal to the plane are proportional therefore... | mcq | aieee-2010 | 4,559 |
1rZ38hZcB3hCW5WP | maths | 3d-geometry | plane-in-space | Distance between two parallel planes $$2x+y+2z=8$$ and $$4x+2y+4z+5=0$$ is : | [{"identifier": "A", "content": "$${3 \\over 2}$$ "}, {"identifier": "B", "content": "$${5 \\over 2}$$"}, {"identifier": "C", "content": "$${7 \\over 2}$$"}, {"identifier": "D", "content": "$${9 \\over 2}$$"}] | ["C"] | null | $$2x + y + 2z - 8 = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( {Plane\,\,1} \right)$$
<br><br>$$2x + y + 2z + {5 \over 2} = 0\,\,\,\,\,\,\,\,\,\,\,...\left( {Plane\,\,2} \right)$$
<br><br>Distance between Plane $$1$$ and $$2$$
<br><br>$$ = \left| {{{ - 8 - {5 \over 2}} \over {\sqrt {{2^2} + {1^2} + {2^2}} }}} \right|$$
<br><... | mcq | jee-main-2013-offline | 4,561 |
HsclIbPpXGLLYwNuOmjqV | maths | 3d-geometry | plane-in-space | If x = a, y = b, z = c is a solution of the system of linear equations
<br/><br/>x + 8y + 7z = 0
<br/><br/>9x + 2y + 3z = 0
<br/><br/>x + y + z = 0
<br/><br/>such that the point (a, b, c) lies on the plane x + 2y + z = 6, then 2a + b + c equals : | [{"identifier": "A", "content": "$$-$$ 1"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["C"] | null | Given,
<br><br>x + 8y + 7z = 0
<br><br>9x + 2y + 3z = 0
<br><br>x + y + z = 0
<br><br>Solving those equations, we get
<br><br>x = $$\lambda $$, y = 6$$\lambda $$, z = -7$$\lambda $$
<br><br>This point lies on the plane x + 2y + z = 6
<br><br>$$ \therefore $$ $$\lambda $$ + 2(6$$\lambda $$) + (-7$$\lambda $$) = 0
<br>... | mcq | jee-main-2017-online-9th-april-morning-slot | 4,563 |
QOo7bAqiG06GOKThuvm05 | maths | 3d-geometry | plane-in-space | If a variable plane, at a distance of 3 units from the origin, intersects the coordinate
axes at A, B and C, then the locus of the centroid of $$\Delta $$ABC is :
| [{"identifier": "A", "content": "$${1 \\over {{x^2}}} + {1 \\over {{y^2}}} + {1 \\over {{z^2}}} = 1$$ "}, {"identifier": "B", "content": "$${1 \\over {{x^2}}} + {1 \\over {{y^2}}} + {1 \\over {{z^2}}} = 3$$ "}, {"identifier": "C", "content": "$${1 \\over {{x^2}}} + {1 \\over {{y^2}}} + {1 \\over {{z^2}}} = {1 \\over 9}... | ["A"] | null | Suppose centroid be (h, k, $$l$$)
<br><br>$$ \therefore $$ x $$-$$ intp $$=$$ 3h, y $$-$$ intp $$=$$ 3k, z $$-$$ intp $$=$$ 3$$l$$
<br><br>Equation $${x \over {3h}} + {y \over {3k}} + {z \over {3l}} = 1$$
<br><br>$$ \therefore $$ Distance from (0, 0, 0)
<br><br>$$\left| {{{ - 1} \over ... | mcq | jee-main-2017-online-9th-april-morning-slot | 4,564 |
By2W9q22rMEe3evacfhKc | maths | 3d-geometry | plane-in-space | The sum of the intercepts on the coordinate axes of the plane passing through the point ($$-$$2, $$-2,$$ 2) and containing the line joining the points (1, $$-$$1, 2) and (1, 1, 1) is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "$$-$$ 4"}, {"identifier": "C", "content": "$$-$$ 8"}, {"identifier": "D", "content": "12"}] | ["B"] | null | Equation of plane passing through three given points is :
<br><br>$$\left| {\matrix{
{x - {x_1}} & {y - {y_1}} & {z - {z_1}} \cr
{{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr
{{x_3} - {x_1}} & {{y_3} - {y_1}} & {{z_3} - {z_1}} \cr
} } \right| = 0$$
<br><br>$$ \Rightar... | mcq | jee-main-2018-online-16th-april-morning-slot | 4,565 |
EJOo70wLbGOQ1WD7CS7IU | maths | 3d-geometry | plane-in-space | A plane bisects the line segment joining the points (1, 2, 3) and ($$-$$ 3, 4, 5) at rigt angles. Then this plane also passes through the point : | [{"identifier": "A", "content": "($$-$$ 3, 2, 1)"}, {"identifier": "B", "content": "(3, 2, 1)"}, {"identifier": "C", "content": "($$-$$ 1, 2, 3)"}, {"identifier": "D", "content": "(1, 2, $$-$$ 3)"}] | ["A"] | null | Since the plane bisects the line joining the points (1, 2, 3) and ($$-$$3, 4, 5) then the plane passes through the midpoint of the line which is :
<br><br>$$\left( {{{1 - 3} \over 2},{{2 + 4} \over 2},{{5 + 3} \over 2}} \right)$$ $$ \equiv $$ $$\left( {{{ - 2} \over 2},{6 \over 2},{8 \over 2}} \right)$$ $$... | mcq | jee-main-2018-online-15th-april-evening-slot | 4,566 |
pFJKQ1n9aFLfryRPPa1Lz | maths | 3d-geometry | plane-in-space | A variable plane passes through a fixed point (3,2,1) and meets x, y and z axes at A, B and C respectively. A plane is drawn parallel to yz -plane through A, a second plane is drawn parallel zx-plane through B and a third plane is drawn parallel to xy-plane through C. Then the locus of the point of intersection of thes... | [{"identifier": "A", "content": "$${x \\over 3} + {y \\over 2} + {z \\over 1} = 1$$ "}, {"identifier": "B", "content": "x + y + z = 6"}, {"identifier": "C", "content": "$${1 \\over x} + {1 \\over y} + {1 \\over z} = {{11} \\over 6}$$"}, {"identifier": "D", "content": "$${3 \\over x} + {2 \\over y} + {1 \\over z} = 1$$"... | ["D"] | null | If a, b, c are the intercepts of the variable plane on the x,y,z axes respectively, then the equation of the plane is <br/><br/>$${x \over a} + {y \over b} + {z \over c} = 1$$
<br><br>And the point of intersection of the planes parallel to the xy, yz and zx planes is $$\left( {a,b,c} \right)$$.
<br><br>As the point (... | mcq | jee-main-2018-online-15th-april-morning-slot | 4,567 |
LfSBYkVJfNOMRk2S | maths | 3d-geometry | plane-in-space | If L<sub>1</sub> is the line of intersection of the planes 2x - 2y + 3z - 2 = 0, x - y + z + 1 = 0 and L<sub>2</sub> is the line of
intersection of the planes x + 2y - z - 3 = 0, 3x - y + 2z - 1 = 0, then the distance of the origin from the
plane, containing the lines L<sub>1</sub> and L<sub>2</sub>, is : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${1 \\over {4\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$${1 \\over {3\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${1 \\over {2\\sqrt 2 }}$$"}] | ["C"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265244/exam_images/mtsizoslixc1qq9ocqig.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - 3D Geometry Question 287 English Explanation">
<br><br>L<sub>1</sub> is the line of intersection of plane 1 and plane 2.
<br><br... | mcq | jee-main-2018-offline | 4,568 |
h2kYrw078v0a3uPgCCURP | maths | 3d-geometry | plane-in-space | The perpendicular distance from the origin to the plane containing the two lines, <br/><br/>$${{x + 2} \over 3} = {{y - 2} \over 5} = {{z + 5} \over 7}$$ and <br/><br/>$${{x - 1} \over 1} = {{y - 4} \over 4} = {{z + 4} \over 7},$$ is : | [{"identifier": "A", "content": "$$6\\sqrt {11} $$"}, {"identifier": "B", "content": "$${{11} \\over {\\sqrt 6 }}$$"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "11$$\\sqrt 6 $$"}] | ["B"] | null | $$\left| {\matrix{
i & j & k \cr
3 & 5 & 7 \cr
1 & 4 & 7 \cr
} } \right|$$
<br><br>= $$\widehat i$$(35 $$-$$ 28) $$-$$ $$\widehat j$$(21.7) + $$\widehat k$$(12 $$-$$ 5)
<br><br>= 7$$\widehat i$$ $$-$$ 14$$\widehat j$$ + 7$$\widehat k$$
<br><br>= $$\widehat i$$ $$-$$ 2$$\widehat ... | mcq | jee-main-2019-online-12th-january-morning-slot | 4,569 |
ZNjWs09KjUBWLXz3C53rsa0w2w9jxadan6d | maths | 3d-geometry | plane-in-space | A plane which bisects the angle between the two given planes 2x – y + 2z – 4 = 0 and x + 2y + 2z – 2 = 0,
passes through the point : | [{"identifier": "A", "content": "(1, \u20134, 1)"}, {"identifier": "B", "content": "(1, 4, \u20131)"}, {"identifier": "C", "content": "(2, 4, 1)"}, {"identifier": "D", "content": "(2, \u20134, 1)"}] | ["D"] | null | Planes bisecting the given planes are<br><br>
$${{2x - y + 2z - 4} \over 3} = \pm {{x + 2y + 2z - 2} \over 3}$$<br><br>
$$ \Rightarrow $$ x - 3y = 2 or 3x + y + 4z = 6<br><br>
Out of the four given points in question's options only (2, -4, 1) lies on the plane 3x + y + 4z = 6 | mcq | jee-main-2019-online-12th-april-evening-slot | 4,570 |
pOnUJRdMYw4ui1JSKs3rsa0w2w9jx1zs3oi | maths | 3d-geometry | plane-in-space | If the plane 2x – y + 2z + 3 = 0 has the distances
$${1 \over 3}$$
and
$${2 \over 3}$$
units from the planes 4x – 2y + 4z + $$\lambda $$ = 0 and
2x – y + 2z + $$\mu $$ = 0, respectively, then the maximum value of $$\lambda $$ + $$\mu $$ is equal to : | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "15"}] | ["A"] | null | Distance formula<br><br>
(i) $${{\left| {\lambda - 6} \right|} \over {\sqrt {16 + 4 + 16} }} = \left| {{{\lambda - 6} \over 6}} \right| = {1 \over 3}$$<br><br>
$$ \Rightarrow $$ $$\left| {\lambda - 6} \right| = 2$$<br><br>
$$ \Rightarrow $$ $$\lambda = 8,4$$<br><br>
(ii) $${{\left| {\mu - 3} \right|} \over {\sqrt ... | mcq | jee-main-2019-online-10th-april-evening-slot | 4,571 |
ACCrs5Q2N2AyKcnuNg3rsa0w2w9jwxv0uci | maths | 3d-geometry | plane-in-space | If Q(0, –1, –3) is the image of the point P in the plane 3x – y + 4z = 2 and R is the point (3, –1, –2), then the
area (in sq. units) of $$\Delta $$PQR is : | [{"identifier": "A", "content": "$${{\\sqrt {65} } \\over 2}$$"}, {"identifier": "B", "content": "$$2\\sqrt {13} $$"}, {"identifier": "C", "content": "$${{\\sqrt {91} } \\over 2}$$"}, {"identifier": "D", "content": "$${{\\sqrt {91} } \\over 4}$$"}] | ["C"] | null | Image of Q in plane<br><br>
$${{\left( {x - 0} \right)} \over 3} = {{\left( {y + 1} \right)} \over { - 1}} = {{z + 3} \over { + 4}} = {{ - 2(1 - 12 - 2)} \over {9 + 1 + 16}} = 1$$<br><br>
x = 3, y = –2, z = 1<br><br>
P(3, –2, 1), Q(0, –1, –3), R(3, –1, –2)<br><br>
Now area of $$\Delta $$PQR is<br><br>
$${1 \over 2}\lef... | mcq | jee-main-2019-online-10th-april-morning-slot | 4,572 |
yodXktCxgD0IJ4xePL18hoxe66ijvwvwdsy | maths | 3d-geometry | plane-in-space | Let P be the plane, which contains the line of
intersection of the planes, x + y + z – 6 = 0 and
2x + 3y + z + 5 = 0 and it is perpendicular to the
xy-plane. Then the distance of the point (0, 0, 256)
from P is equal to : | [{"identifier": "A", "content": "205$$\\sqrt5$$"}, {"identifier": "B", "content": "63$$\\sqrt5$$"}, {"identifier": "C", "content": "11/$$\\sqrt5$$"}, {"identifier": "D", "content": "17/$$\\sqrt5$$"}] | ["C"] | null | P<sub>1</sub> : x + y + z – 6 = 0
<br><br>P<sub>2</sub> : 2x + 3y + z + 5 = 0
<br><br>Equation of plane which passes through the line of intersection of P<sub>1</sub> and P<sub>2</sub> is
<br><br>P<sub>1</sub> + $$\lambda $$P<sub>2</sub> = 0
<br><br>$$ \Rightarrow $$ (x + y + z – 6) + $$\lambda $$(2x + 3y + z + 5) = 0
... | mcq | jee-main-2019-online-9th-april-evening-slot | 4,573 |
oW3MTlGq7uoYSy1HduycQ | maths | 3d-geometry | plane-in-space | A plane passing through the points (0, –1, 0)
and (0, 0, 1) and making an angle $${\pi \over 4}$$ with the
plane y – z + 5 = 0, also passes through the
point | [{"identifier": "A", "content": "$$\\left( {\\sqrt 2 ,1,4} \\right)$$"}, {"identifier": "B", "content": "$$\\left(- {\\sqrt 2 ,1,4} \\right)$$"}, {"identifier": "C", "content": "$$\\left( -{\\sqrt 2 ,-1,-4} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {\\sqrt 2 ,-1,4} \\right)$$"}] | ["A"] | null | Let ax + by + cz = 1 be the equation of the plane
<br><br>it passed through point (0, –1, 0).
<br><br>$$ \therefore $$ -b = 1
<br><br>$$ \Rightarrow $$ b = -1
<br><br>Also it passes through point (0, 0, 1)
<br><br>$$ \therefore $$ c = 1
<br><br>So the plane is ax - y + z = 1.
<br><br>This plane an angle $${\pi \over 4... | mcq | jee-main-2019-online-9th-april-morning-slot | 4,574 |
A3dJQnt8WpOkVMARcRG7E | maths | 3d-geometry | plane-in-space | Let S be the set of all real values of $$\lambda $$ such that a plane passing through the points (–$$\lambda $$<sup>2</sup>, 1, 1), (1, –$$\lambda $$<sup>2</sup>, 1) and (1, 1, – $$\lambda $$<sup>2</sup>) also passes through the point (–1, –1, 1). Then S is equal to : | [{"identifier": "A", "content": "{1, $$-$$1}"}, {"identifier": "B", "content": "{3, $$-$$ 3}"}, {"identifier": "C", "content": "$$\\left\\{ {\\sqrt 3 } \\right\\}$$"}, {"identifier": "D", "content": "$$\\left\\{ {\\sqrt 3 , - \\sqrt 3 } \\right\\}$$"}] | ["D"] | null | All four points are coplanar so
<br><br>$$\left| {\matrix{
{1 - {\lambda ^2}} & 2 & 0 \cr
2 & { - {\lambda ^2} + 1} & 0 \cr
2 & 2 & { - {\lambda ^2} - 1} \cr
} } \right| = 0$$
<br><br>($$\lambda $$<sup>2</sup> + 1)<sup>2</sup> (3 $$-$$ $$\lambda $$<sup>2</sup>) = 0
<br><br>$$\la... | mcq | jee-main-2019-online-12th-january-evening-slot | 4,576 |
vCpFMyMOEky4bJBVl2v9s | maths | 3d-geometry | plane-in-space | If the point (2, $$\alpha $$, $$\beta $$) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x – 5y = 15, then 2$$\alpha $$ – 3$$\beta $$ is equal to | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "17"}, {"identifier": "D", "content": "5"}] | ["B"] | null | Normal vector of plane
<br><br>$$ = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
2 & { - 5} & 0 \cr
4 & { - 4} & 5 \cr
} } \right|$$
<br><br>$$ = - 4\left( {5\widehat i + 2\widehat j - 3\widehat k} \right)$$
<br><br>equation of plane is
<br><br>5(x $$-$$ 7) ... | mcq | jee-main-2019-online-11th-january-evening-slot | 4,577 |
y1O9ASDQ4I3iqhyAwRaOV | maths | 3d-geometry | plane-in-space | The direction ratios of normal to the plane through the points (0, –1, 0) and (0, 0, 1) and making an angle $${\pi \over 4}$$ with the plane y $$-$$ z + 5 = 0 are : | [{"identifier": "A", "content": "2, $$-$$1, 1"}, {"identifier": "B", "content": "$$2\\sqrt 3 ,1, - 1$$"}, {"identifier": "C", "content": "$$\\sqrt 2 ,1, - 1$$"}, {"identifier": "D", "content": "$$\\sqrt 2 , - \\sqrt 2 $$"}] | ["C"] | null | Let the equation of plane be
<br><br>a(x $$-$$ 0) + b(y + 1) + c(z $$-$$ 0) = 0
<br><br>It passes through (0, 0, 1) then
<br><br>b + c = 0 . . . . (1)
<br><br>Now cos $${\pi \over 4}$$ = $${{a\left( 0 \right) + b\left( 1 \right) + c\left( { - 1} \right)} \over {\sqrt 2 \sqrt {{a^2} + {b^2} + {... | mcq | jee-main-2019-online-11th-january-morning-slot | 4,578 |
FZp6IAeQMHnGO3LU7gCrN | maths | 3d-geometry | plane-in-space | The plane which bisects the line segment joining the points (–3, –3, 4) and (3, 7, 6) at right angles, passes through which one of the following points ? | [{"identifier": "A", "content": "(2, 1, 3)"}, {"identifier": "B", "content": "(4, $$-$$ 1, 2)"}, {"identifier": "C", "content": "(4, 1, $$-$$ 2)"}, {"identifier": "D", "content": "($$-$$ 2, 3, 5)"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264937/exam_images/uqar8tycbulapjs4smas.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Evening Slot Mathematics - 3D Geometry Question 261 English Explanation">
<br>p : 3... | mcq | jee-main-2019-online-10th-january-evening-slot | 4,579 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.