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The two lines $$x=ay+b,z=cy+d$$ and $$x = a'y + b',z = c'y + d'$$ will be perpendicular, if and only if :
Options:
[{"identifier": "A", "content": "$$aa' + cc' + 1 = 0$$ "}, {"identifier": "B", "content": "$$aa' + bb'cc' + 1 = 0$$ "}, {"identifier": "C", "content": "$$aa' + bb'cc' = 0$$ "}, {"identifier": "D", "conte... | ["A"]
Explanation:
$${{x - b} \over a} = {y \over 1} = {{z - d} \over c};$$
<br/><br/>$${{x - b'} \over {a'}}$$
$$ = {y \over 1} = {{z - d'} \over c'}$$
<br><br>For perpenedicularity of lines $$aa' + 1 + cc' = 0$$ |
A line makes the same angle $$\theta $$, with each of the $$x$$ and $$z$$ axis.
<br/><br/>If the angle $$\beta \,$$, which it makes with y-axis, is such that $$\,{\sin ^2}\beta = 3{\sin ^2}\theta ,$$ then $${\cos ^2}\theta $$ equals :
Options:
[{"identifier": "A", "content": "$${2 \\over 5}$$ "}, {"identifier": "B",... | ["C"]
Explanation:
<b>Concept :</b> If a line makes the angle $$\alpha ,\beta ,\gamma $$ with x, y, z axis respectively then
$$${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1$$$
<br><br>In this question given that the line makes angle θ with x and z-axis and β with y−axis.
<br><br>$$\therefore\: cos^2\theta... |
If a line makes an angle of $$\pi /4$$ with the positive directions of each of $$x$$-axis and $$y$$-axis, then the angle that the line makes with the positive direction of the $$z$$-axis is :
Options:
[{"identifier": "A", "content": "$${\\pi \\over 4}$$"}, {"identifier": "B", "content": "$${\\pi \\over 2}$$ "}, {"id... | ["B"]
Explanation:
Let the angle of line makes with the positive direction of $$z$$-axis is $$\alpha $$ direction cosines of line with the $$+ve$$ directions of $$x$$-axis, $$y$$-axis, and $$z$$-axis is $$l,$$ $$m,$$ $$n$$ respectively.
<br><br>$$\therefore$$ $$l = \cos {\pi \over 4},m = \cos {\pi \over 4},\,\,n = ... |
Let $$L$$ be the line of intersection of the planes $$2x+3y+z=1$$ and $$x+3y+2z=2.$$ If $$L$$ makes an angle $$\alpha $$ with the positive $$x$$-axis, then cos $$\alpha $$ equals
Options:
[{"identifier": "A", "content": "$$1$$ "}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 2 }}$$ "}, {"identifier": "C", "con... | ["C"]
Explanation:
Let the direction cosines of line $$L$$ be $$l,m,n,$$
<br><br>then $$2l+3m+n=0$$ $$\,\,\,\,\,\,\,....\left( i \right)$$
<br><br>and $$l + 3m + 2n = 0\,\,\,\,\,\,\,\,\,\,....\left( {ii} \right)$$
<br><br>on solving equation $$(i)$$ and $$(ii),$$ we get
<br><br>$${l \over {6 - 3}} = {m \over {1 - 4... |
The projections of a vector on the three coordinate axis are $$6,-3,2$$ respectively. The direction cosines of the vector are :
Options:
[{"identifier": "A", "content": "$${6 \\over 5},{{ - 3} \\over 5},{2 \\over 5}$$ "}, {"identifier": "B", "content": "$${6 \\over 7 },{{ - 3} \\over 7},{2 \\over 7}$$"}, {"identifier"... | ["B"]
Explanation:
Let $$P\left( {{x_1},{y_1},{z_1}} \right)$$ and $$Q\left( {{x_2},{y_2},{z_2}} \right)$$ be the initial and final points of the vector whose projections on the three coordinates axes are $${6, - 3,2}$$ then
<br><br>$${x_2} - {x_1}, = 6;\,\,{y_2} - {y_1} = - 3;\,\,{z_2} - {z_1} = 2$$
<br><br>So that... |
A line $$AB$$ in three-dimensional space makes angles $${45^ \circ }$$ and $${120^ \circ }$$ with the positive $$x$$-axis and the positive $$y$$-axis respectively. If $$AB$$ makes an acute angle $$\theta $$ with the positive $$z$$-axis, then $$\theta $$ equals :
Options:
[{"identifier": "A", "content": "$${45^ \\circ... | ["B"]
Explanation:
Direction cosines of the line :
<br><br>$$\ell = \cos {45^ \circ } = {1 \over {\sqrt 2 }},m = \cos {120^ \circ } = {{ - 1} \over 2},\pi = \cos \theta $$
<br><br>where $$\theta $$ is the angle, which line makes with positive $$z$$-axis.
<br><br>Now $${\ell ^2} + {m^2} + {n^2} = 1$$
<br><br>$$ \Righ... |
The angle between the lines whose direction cosines satisfy the equations $$l+m+n=0$$ and $${l^2} = {m^2} + {n^2}$$ is :
Options:
[{"identifier": "A", "content": "$${\\pi \\over 6}$$"}, {"identifier": "B", "content": "$${\\pi \\over 2}$$"}, {"identifier": "C", "content": "$${\\pi \\over 3}$$ "}, {"identifier": "D",... | ["C"]
Explanation:
Given
<br><br>$$l + m + n = 0$$ and $${l^2} = {m^2} + {n^2}$$
<br><br>Now, $${\left( { - m - n} \right)^2} = {m^2} + {n^2}$$
<br><br>$$ \Rightarrow mn = 0 \Rightarrow m = 0\,\,$$ or $$\,\,n = 0$$
<br><br>If $$m=0$$ then $$l=-n$$
<br><br>We know
<br><br>$${l^2} + {m^2} + {n^2} = 1 \Rightarrow n = ... |
ABC is a triangle in a plane with vertices
<br/><br> A(2, 3, 5), B(−1, 3, 2) and C($$\lambda $$, 5, $$\mu $$).
<br/><br/>If the median through A is equally inclined to the coordinate axes, then the value of ($$\lambda $$<sup>3</sup> + $$\mu $$<sup>3</sup> + 5) is : </br>
Options:
[{"identifier": "A", "content": "1130... | ["B"]
Explanation:
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267089/exam_images/rmtgi7eihcbphpv9sdfi.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 10th April Morning Slot Mathematics - 3D Geometry Question 270 English Expla... |
An angle between the lines whose direction cosines are gien by the equations,
<br/>$$l$$ + 3m + 5n = 0 and 5$$l$$m $$-$$ 2mn + 6n$$l$$ = 0, is :
Options:
[{"identifier": "A", "content": "$${\\cos ^{ - 1}}\\left( {{1 \\over 3}} \\right)$$"}, {"identifier": "B", "content": "$${\\cos ^{ - 1}}\\left( {{1 \\over 4}} \\ri... | ["C"]
Explanation:
Given
<br><br>l + 3m + 5n = 0
<br><br>and 5$$l$$m $$-$$ 2mn + 6n$$l$$ = 0
<br><br>From eq. (1) we have
<br><br>$$l$$ = $$-$$ 3m $$-$$ 5n
<br><br>Put the value of $$l$$ in eq. (2), we get ;
<br><br>5 ($$-$$3m $$-$$5n) m $$-$$ 2mn + 6n ($$-$$ 3m $$-$$ 5n) = 0
<br><br>$$ \Rightarrow $$ 15m<su... |
If a point R(4, y, z) lies on the line segment joining
the points P(2, –3, 4) and Q(8, 0, 10), then the
distance of R from the origin is :
Options:
[{"identifier": "A", "content": "$$2 \\sqrt {14}$$"}, {"identifier": "B", "content": "$$ \\sqrt {53}$$"}, {"identifier": "C", "content": "$$2 \\sqrt {21}$$"}, {"identifier... | ["A"]
Explanation:
Equation of PQ is
<br><br> $${{x - 2} \over {8 - 2}} = {{y + 3} \over {0 - \left( { - 3} \right)}} = {{z - 4} \over {10 - 4}}$$
<br><br>$$ \Rightarrow $$ $${{x - 2} \over 6} = {{y + 3} \over 3} = {{z - 4} \over 6}$$
<br><br>Point R (4, y, z) lies on this
<br><br>$$ \therefore $$ $${{4 - 2} \over 6} ... |
The projection of the line segment joining the
points (1, –1, 3) and (2, –4, 11) on the line
joining the points (–1, 2, 3) and (3, –2, 10)
is ____________.
Options:
[] | 8
Explanation:
Let A (1, – 1, 3), B(2, – 4, 11), C (–1, 2, 3) & D (3, –2, 10)
<br><br>$$ \therefore $$ $$\overrightarrow {AB} = \widehat i - 3\widehat j + 8\widehat k$$
<br><br>$$ \Rightarrow $$ $$\overrightarrow {CD} = 4\widehat i - 4\widehat j + 7\widehat k$$
<br><br>Projection of $$\overrightarrow {AB} $$ on ... |
Let $$\alpha$$ be the angle between the lines whose direction cosines satisfy the equations l + m $$-$$ n = 0 and l<sup>2</sup> + m<sup>2</sup> $$-$$ n<sup>2</sup> = 0. Then the value of sin<sup>4</sup>$$\alpha$$ + cos<sup>4</sup>$$\alpha$$ is :
Options:
[{"identifier": "A", "content": "$${{3 \\over 8}}$$"}, {"identif... | ["D"]
Explanation:
$${l^2} + {m^2} + {n^2} = 1$$<br><br>$$ \therefore $$ $$2{n^2} = 1 $$ ($$ \because $$ l<sup>2</sup> + m<sup>2</sup> $$-$$ n<sup>2</sup> = 0)
<br><br>$$\Rightarrow n = \pm {1 \over {\sqrt 2 }}$$<br><br>$$ \therefore $$ $${l^2} + {m^2} = {1 \over 2}$$ & $$l + m = {1 \over {\sqrt 2 }}$$<br><br>$$ ... |
The angle between the straight lines, whose direction cosines are given by the equations 2l + 2m $$-$$ n = 0 and mn + nl + lm = 0, is :
Options:
[{"identifier": "A", "content": "$${\\pi \\over 2}$$"}, {"identifier": "B", "content": "$$\\pi - {\\cos ^{ - 1}}\\left( {{4 \\over 9}} \\right)$$"}, {"identifier": "C", "co... | ["A"]
Explanation:
n = 2 (l + m)<br><br>lm + n(l + m) = 0<br><br>lm + 2(l + m)<sup>2</sup> = 0<br><br>2l<sup>2</sup> + 2m<sup>2</sup> + 5ml = 0<br><br>$$2{\left( {{l \over m}} \right)^2} + 2 + 5\left( {{l \over m}} \right) = 0$$<br><br>2t<sup>2</sup> + 5t + 2 = 0<br><br>(t + 2)(2t + 1) = 0<br><br>$$ \Rightarrow t = -... |
<p>If two straight lines whose direction cosines are given by the relations $$l + m - n = 0$$, $$3{l^2} + {m^2} + cnl = 0$$ are parallel, then the positive value of c is :</p>
Options:
[{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "c... | ["A"]
Explanation:
<p>Given that the direction cosines satisfy $l + m - n = 0$, we find that $n = l + m$.</p>
<p>The other equation is $3l^2 + m^2 + cnl = 0$, and substituting $n = l + m$ gives $3l^2 + m^2 + cl(l + m) = 0$.</p>
<p>This simplifies to $(3 + c)l^2 + clm + m^2 = 0$.</p>
<p>As the lines are parallel, they ... |
<p>Let $$\mathrm{P}(-2,-1,1)$$ and $$\mathrm{Q}\left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\right)$$ be the vertices of the rhombus PRQS. If the direction ratios of the diagonal RS are $$\alpha,-1, \beta$$, where both $$\alpha$$ and $$\beta$$ are integers of minimum absolute values, then $$\alpha^{2}+\beta^{2}$$ ... | 450
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7rb45xm/29b2975e-d397-4ba9-941a-4cc0bd7b15f1/c362a6a0-2e7f-11ed-8702-156c00ced081/file-1l7rb45xn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7rb45xm/29b2975e-d397-4ba9-941a-4cc0bd7b15f1/c362a6a0-2e7f-11ed... |
Consider a $\triangle A B C$ where $A(1,3,2), B(-2,8,0)$ and $C(3,6,7)$. If the angle bisector of $\angle B A C$ meets
the line $B C$ at $D$, then the length of the projection of the vector $\overrightarrow{A D}$ on the vector $\overrightarrow{A C}$ is :
Options:
[{"identifier": "A", "content": "$\\frac{37}{2 \\sqrt{3... | ["A"]
Explanation:
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqeg6dq/195e875b-1d41-47ba-b7a7-6f56da027c9a/8f22c6e0-cdc0-11ee-9f50-677e7e372eae/file-6y3zli1lsqeg6dr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqeg6dq/195e875b-1d41-47ba-b7a7-6f56da027c9a/8... |
<p>Let $$O$$ be the origin and the position vectors of $$A$$ and $$B$$ be $$2 \hat{i}+2 \hat{j}+\hat{k}$$ and $$2 \hat{i}+4 \hat{j}+4 \hat{k}$$ respectively. If the internal bisector of $$\angle \mathrm{AOB}$$ meets the line $$\mathrm{AB}$$ at $$\mathrm{C}$$, then the length of $$O C$$ is</p>
Options:
[{"identifier": ... | ["C"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt2yowkb/31073c70-5b0c-4f0f-80c0-62ea5ac61070/2035eab0-d4a9-11ee-bdd1-01c80c3e2d9a/file-1lt2yowkc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt2yowkb/31073c70-5b0c-4f0f-80c0-62ea5ac61070/2035eab0-d4a9-11... |
<p>Let $$\mathrm{P}(3,2,3), \mathrm{Q}(4,6,2)$$ and $$\mathrm{R}(7,3,2)$$ be the vertices of $$\triangle \mathrm{PQR}$$. Then, the angle $$\angle \mathrm{QPR}$$ is</p>
Options:
[{"identifier": "A", "content": "$$\\cos ^{-1}\\left(\\frac{7}{18}\\right)$$\n"}, {"identifier": "B", "content": "$$\\frac{\\pi}{6}$$\n"}, {"i... | ["D"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr8wgd3/6b24122f-fe39-4086-be4a-889687b965f8/a5aaee70-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8wgd4.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsr8wgd3/6b24122f-fe39-4086-be4a-889687b965f... |
<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 ... | ["A"]
Explanation:
<p>$$\begin{aligned}
& \overrightarrow{O P}=x \hat{i}+y \hat{j}+z \hat{k} \\
& \overrightarrow{O Q}=x \hat{i}+y \hat{j} \\
& |O P|=\gamma=\sqrt{x^2+y^2+z^2} \\
& \cos \theta=\frac{x}{\sqrt{x^2+y^2}} \Rightarrow \cos ^2 \theta=\frac{x^2}{\gamma^2-z^2}=\frac{x^2}{\gamma^2-\gamma^2 \cos ^2 \phi} \\
& \... |
<p>If the line $$\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$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "13"}, {"identifier": "C", "content": "5"}, {"id... | ["D"]
Explanation:
<p>$$\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 \m... |
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>
Options:
[{"identifier": "A", "content": "$$x-y+z=1$$"}, {"identifier": "B", "content": "$$x+y+z=5$$ "}, {"identifier": "C", "content": "$$x+2y-z=1$$ "}, {"identifier": "D",... | ["A"]
Explanation:
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... |
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 :
Options:
[{"identifier": "A", "content": "$${5 \\over 3}$$"}, {"iden... | ["A"]
Explanation:
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>$$\cos \left( {{\pi \over 2} - 0} \right) = {{\left( {\widehat i + 2\widehat j + 2\widehat k} \right).\left( {2\widehat i - \wideh... |
If the plane $$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$$ then a equals :</br></br>
Options:
[{"identifier": "A", "content": "$$-1$$"}, {"identifier": "B... | ["C"]
Explanation:
Centers of given spheres are $$\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$$ |
The distance between the line
<br/><br>$$\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$$ is </br></br>
Options:
[{"identifier": "A", "content": "$${{10} \\ove... | ["B"]
Explanation:
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>$$ = \left| {{{2 - 10 + 3 - 5} \over {\sqrt {1 + 25 + 1} }}} \right| = {{10} \over {3\sqrt 3 }}$$ |
The image of the point $$(-1, 3,4)$$ in the plane $$x-2y=0$$ is :
Options:
[{"identifier": "A", "content": "$$\\left( { - {{17} \\over 3}, - {{19} \\over 3},4} \\right)$$ "}, {"identifier": "B", "content": "$$(15,11,4)$$ "}, {"identifier": "C", "content": "$$\\left( { - {{17} \\over 3}, - {{19} \\over 3},1} \\right)$$... | ["D"]
Explanation:
<img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265452/exam_images/ibx2kf4tgkt5ewlrdikt.webp" loading="lazy" alt="AIEEE 2006 Mathematics - 3D Geometry Question 309 English Explanation">
<br><br>$$E{q^n}\,\,\,\,$$ of $$\,\,\,\,PN: $$-
<br><br>$${{x + 1} \over ... |
Let the line $$\,\,\,\,\,$$ $${{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)$$ equals
Options:
[{"identifier": "A", "content": "$$(-6,7)$$ "}, {"identifier": "B", "content": "$$(5,-15)$$ "}, {... | ["A"]
Explanation:
As the line $${{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 +... |
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>
Options:
[{"identifier": "A", "content": "$${3 \\over 2}$$"}, {"identifier": "B", "content": "$${2 \\ov... | ["D"]
Explanation:
If $$\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}} }}$$
... |
The image of the line $${{x - 1} \over 3} = {{y - 3} \over 1} = {{z - 4} \over { - 5}}\,$$ in the plane $$2x-y+z+3=0$$ is the line :
Options:
[{"identifier": "A", "content": "$${{x - 3} \\over 3} = {{y + 5} \\over 1} = {{z - 2} \\over { - 5}}$$ "}, {"identifier": "B", "content": "$${{x - 3} \\over { - 3}} = {{y + 5} \... | ["C"]
Explanation:
<img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266411/exam_images/lc7tycmcmkjqgjziprzg.webp" loading="lazy" alt="JEE Main 2014 (Offline) Mathematics - 3D Geometry Question 295 English Explanation">
<br><br>$${{a - 1} \over 2} = {{b - 3} \over { - 1}} = {{c - ... |
The equation of the plane containing the line $$2x-5y+z=3; x+y+4z=5,$$ and parallel to the plane, $$x+3y+6z=1,$$ is :
Options:
[{"identifier": "A", "content": "$$x+3y+6z=7$$"}, {"identifier": "B", "content": "$$2x+6y+12z=-13$$"}, {"identifier": "C", "content": "$$2x+6y+12z=13$$ "}, {"identifier": "D", "content": "$$x... | ["A"]
Explanation:
Equation of the plane containing the lines
<br><br>$$2x - 5y + z = 3$$ and $$x + y + 4z = 5$$ is
<br><br>$$2x - 5y + z - 3 + \lambda \left( {x + y + 4z - 5} \right) = 0$$
<br><br>$$ \Rightarrow \left( {2 + \lambda } \right)x + \left( { - 5 + \lambda } \right)y + \left( {1 + 4\lambda } \right)z + $... |
The distance of the point $$(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,$$ is :
Options:
[{"identifier": "A", "content": "$$3\\sqrt {21} $$ "}, {"identifier": "B", "content": "$$13$$"}, {"identifier": "C", "conte... | ["B"]
Explanation:
General point on given line $$ \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 + ... |
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 :
Options:
[{"identifier": "A", "content": "$$5$$ "}, {"identifier": "B", "content": "$$2$$ "}, {"identifier": "C", "content": "$$26$$"}, {"identifier": "D", "content... | ["B"]
Explanation:
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 ... |
The distance of the point $$(1,-5,9)$$ from the plane $$x-y+z=5$$ measured along the line $$x=y=z$$ is :
Options:
[{"identifier": "A", "content": "$${{10} \\over {\\sqrt 3 }}$$ "}, {"identifier": "B", "content": "$${20 \\over 3}$$"}, {"identifier": "C", "content": "$$3\\sqrt {10} $$"}, {"identifier": "D", "content": "... | ["D"]
Explanation:
<img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264547/exam_images/fmk5f20pf8f2dnokx8xv.webp" loading="lazy" alt="JEE Main 2016 (Offline) Mathematics - 3D Geometry Question 288 English Explanation">
<br><br>$$e{q^n}\,\,$$ of $$\,\,PO:\,\,$$ $${{x - 1} \over 1... |
The number of distinct real values of $$\lambda $$ for which the lines
<br/><br/>$${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over {{\lambda ^2}}}$$ and $${{x - 3} \over 1} = {{y - 2} \over {{\lambda ^2}}} = {{z - 1} \over 2}$$ are coplanar is :
Options:
[{"identifier": "A", "content": "4"}, {"identifier": ... | ["D"]
Explanation:
As lines are coplanar, so
<br><br>$$\left| {\matrix{
{3 - 1} & {2 - 2} & {1 - \left( { - 3} \right)} \cr
1 & 2 & {{\lambda ^2}} \cr
1 & {{\lambda ^2}} & 2 \cr
} } \right| $$ = 0
<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
2 & 0 & ... |
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:
Options:
[{"identifier": "A", "content": "$$2\\sqrt {42} $$"}, {"identifier": "B", "content": "$$\\sqrt {42} $$"}, {"identifier": "... | ["A"]
Explanation:
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: ... |
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>... | ["A"]
Explanation:
Let the plane be
<br><br>a(x $$-$$ 1) + b(y + 1) + c (z + 1) = 0
<br><br>Normal vector
<br><br>$$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & { - 2} & 3 \cr
2 & { - 1} & { - 1} \cr
} } \right| = 5\widehat i + 7\widehat j + 3\widehat k$... |
The coordinates of the foot of the perpendicular from the point (1, $$-$$2, 1) on the plane containing the lines, $${{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},$$ is :
Options:
[{"identifier": "A", "content": "(2, $$-$$4, 2)"}, {"identifi... | ["C"]
Explanation:
$$\overrightarrow n $$ = $$\overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $$ = $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
6 & 7 & 8 \cr
3 & 5 & 7 \cr
} } \right|$$
<br><br>= (9, $$-$$ 18, 9)
<br><br>= (1, $$-$$2, 1)
<br><br>$$ ... |
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 :
Options:
[{"identifier": "A", "content": "$${{x - {4 \\over 7}} \\over { - 2}} = {y \\over 7} =... | ["C"]
Explanation:
$$\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(... |
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 :
Options:
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content"... | ["C"]
Explanation:
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>$${{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + 4} \over { - 2}} = ... |
The length of the projection of the line segment joining the points (5, -1, 4) and (4, -1, 3) on the plane,
x + y + z = 7 is :
Options:
[{"identifier": "A", "content": "$$\\sqrt {{2 \\over 3}} $$"}, {"identifier": "B", "content": "$${2 \\over {\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"iden... | ["A"]
Explanation:
<img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266605/exam_images/sruywneq0jjxnahol5cq.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - 3D Geometry Question 286 English Explanation">
<br><br>PQ is the projection of line segment AB on the plan... |
An angle between the plane, x + y + z = 5 and the line of intersection of the planes, 3x + 4y + z $$-$$ 1 = 0 and 5x + 8y + 2z + 14 =0, is :
Options:
[{"identifier": "A", "content": "$${\\sin ^{ - 1}}\\left( {\\sqrt {{\\raise0.5ex\\hbox{$\\scriptstyle 3$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyl... | ["A"]
Explanation:
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 interse... |
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>
Options:
[{"identifier": "A", "content": "$${{x + 4} \\over 3} = {{y - 3} \\over {-1}} = {{z ... | ["A"]
Explanation:
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264981/exam_images/ehep4l3pw7lbyi0lsu7d.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Morning Slot Mathematics - 3D Geometry Question 268 English Expl... |
The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines
<br/>$$\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( { - \... | ["D"]
Explanation:
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)... |
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 :
Options:
[{"identifier": "A", "content": "$$2\\sqrt 7 $$"}, {"identifier": "B", "content": "14"}, {"identifier": "C"... | ["C"]
Explanation:
$${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 1} \over { - 1}} = \lambda $$<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>
$$ \Rightarrow 13\lambda + 13 = 0 \R... |
A perpendicular is drawn from a point on the line $${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {z \over 1}$$ to the plane x + y + z = 3 such that the
foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are :
Options:
[{"identifier": "A", "content": "(4, 0, \u2013 1)"}, {"id... | ["B"]
Explanation:
$${{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 - \lam... |
If the line, $${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 2} \over 4}$$ meets the plane,
x + 2y + 3z = 15 at a point P, then the distance of P from the origin is :
Options:
[{"identifier": "A", "content": "$${{\\sqrt 5 } \\over 2}$$"}, {"identifier": "B", "content": "2$$\\sqrt 5$$"}, {"identifier": "C", "content": ... | ["C"]
Explanation:
Let $${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 2} \over 4}$$ = $$\lambda $$
<br><br>Any arbitary point on the line is P( 2$$\lambda $$ + 1, 3$$\lambda $$ - 1, 4$$\lambda $$ + 2).
<br><br>This point also lies on the plane x + 2y + 3z = 15.
<br><br>$$ \therefore $$ (2$$\lambda $$ + 1) + 2(3$$\lam... |
The equation of a plane containing the line of
intersection of the planes 2x – y – 4 = 0 and
y + 2z – 4 = 0 and passing through the point
(1, 1, 0) is :
Options:
[{"identifier": "A", "content": "x \u2013 3y \u2013 2z = \u20132"}, {"identifier": "B", "content": "2x \u2013 z = 2"}, {"identifier": "C", "content": "x \u20... | ["C"]
Explanation:
The equation of any plane passing through the intersection of the planes 2x – y – 4 = 0 and
y + 2z – 4 = 0 is :
<br><br>(2x – y – 4) + $$\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 $$ (2 – 1 – 4... |
The vector equation of the plane through the line
of intersection of the planes x + y + z = 1 and 2x
+ 3y+ 4z = 5 which is perpendicular to the plane
x – y + z = 0 is :
Options:
[{"identifier": "A", "content": "$$\\mathop r\\limits^ \\to \\times \\left( {\\mathop i\\limits^ \\wedge - \\mathop k\\limits^ \\wedge }... | ["C"]
Explanation:
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 ... |
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 :
Options:
[{"identifier": "A", "content": "$$\\sqrt {{3 \\over 5}} $$"}, {"identifier": "B", "content":... | ["D"]
Explanation:
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$$\alpha $$ = $${{\left( {2\widehat i + \widehat j - 2\widehat k} \right).\left( {\widehat i - 2\widehat j - k\widehat k} \right)} \over {3\sqrt {1 + 4 + {k^2}} }}$$
... |
Two lines $${{x - 3} \over 1} = {{y + 1} \over 3} = {{z - 6} \over { - 1}}$$ and $${{x + 5} \over 7} = {{y - 2} \over { - 6}} = {{z - 3} \over 4}$$ intersect at the point R. The reflection of R in the xy-plane has coordinates :
Options:
[{"identifier": "A", "content": "(2, 4, 7)"}, {"identifier": "B", "content": "(2... | ["B"]
Explanation:
Point on L<sub>1</sub> ($$\lambda $$ + 3, 3$$\lambda $$ $$-$$ 1, $$-$$$$\lambda $$ + 6)
<br><br>Point on L<sub>2</sub> (7$$\mu $$ $$-$$ 5, $$-$$6$$\mu $$ + 2, 4$$\mu $$ + 3
<br><br>$$ \Rightarrow $$ $$\lambda $$ + 3 = 7$$\mu $$ $$-$$ 5 . . . . (i)
<br><b... |
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 ?
Options:
[{"identifier": "A", "content": "($$-$$ 2, 2, 2)"}, {"identifier": "B", "content": "(2, 2, 0)"}, ... | ["C"]
Explanation:
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 $$\left( { - 1,1,1} \right)$$
<... |
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 ?
Options:
[{"identifier": "A", "content": "$${{x - 4} \\over 1} = {{y - 5} \\over 1} = {{z - 5} \\over { - 1}}$$"}, {"identifier": "B", "content": "... | ["C"]
Explanation:
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>$$\la... |
The plane through the intersection of the planes x + y + z = 1 and 2x + 3y – z + 4 = 0 and parallel to y-axis
also passes through the point :
Options:
[{"identifier": "A", "content": "(\u20133, 0, -1) "}, {"identifier": "B", "content": "(3, 2, 1) "}, {"identifier": "C", "content": "(3, 3, -1)"}, {"identifier": "D", "... | ["B"]
Explanation:
The equation of plane
<br><br>(x + y + z $$-$$ 1) + $$\lambda $$ (2x + 3y $$-$$ z + 4) = 0
<br><br>$$ \Rightarrow $$ (1 + 2$$\lambda $$)x + (1 + 3$$\lambda $$)y + (1 $$-$$ $$\lambda $$)z + 4$$\lambda $$ $$-$$ 1 = 0
<br><br>As plane is parallel to y axis so the normal vector of plane and ... |
If the distance between the plane,
23x – 10y – 2z + 48 = 0 and the plane<br/><br/>
containing the lines
$${{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 ... | 3
Explanation:
Required distance = perpendicular distance of plane 23x – 10y – 2z + 48 = 0 either from (–1, 3, –1) or (–3, –2, 1)
<br><br>$$ \Rightarrow $$ $$\left| {{{ - 23 - 30 + 2 + 48} \over {\sqrt {{{\left( {23} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 2 \right)}^2}} }}} \right|$$ = $${k \over {\sqrt {6... |
A plane P meets the coordinate axes at A, B
and C respectively. The centroid of $$\Delta $$ABC is
given to be (1, 1, 2). Then the equation of the
line through this centroid and perpendicular to
the plane P is :
Options:
[{"identifier": "A", "content": "$${{x - 1} \\over 1} = {{y - 1} \\over 1} = {{z - 2} \\over 2}$$"}... | ["C"]
Explanation:
Let, Equation of plane is
<br><br>$${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$$ = 3, b = 3, c = 6
<br><br>... |
The shortest distance between the lines
<br/><br>$${{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}$$ <br/><br>and x + y + z + 1 = 0, 2x – y + z
+ 3 = 0 is :</br></br>
Options:
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over ... | ["D"]
Explanation:
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 $$ \Ri... |
If for some $$\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}$$ are coplanar,
<br/><br/>then the line L<sub>2</sub>
passes through the... | ["B"]
Explanation:
L<sub>1</sub> : $${{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} & ... |
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>
Options:
[{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$${1 \\over 7}$$"}, {"ide... | ["B"]
Explanation:
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,... |
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 :
Options:
[{"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"}, {"... | ["D"]
Explanation:
<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 Expl... |
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:
Options:
[{"identifier": "A", "content": "-10"}, {"identifier": "B", "content": "10"}, {"identifi... | ["C"]
Explanation:
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>
$$\overrightarrow n = \left| {\matrix{
{\widehat i} & ... |
The plane passing through the points (1, 2, 1),
<br/>(2, 1, 2) and parallel to the line, 2x = 3y, z = 1
<br/>also passes through the point :
Options:
[{"identifier": "A", "content": "(0, 6, \u20132)"}, {"identifier": "B", "content": "(\u20132, 0, 1)"}, {"identifier": "C", "content": "(0, \u20136, 2)"}, {"identifier": ... | ["B"]
Explanation:
Equation of plane passing through (2, 1, 2)<br><br>a(x $$-$$ 2) + b(y $$-$$ 1) + c(z $$-$$ 2) = 0 ......(1)<br><br>As point (1, 2, 1) also passes through the plane, so it satisfy the equation, <br><br>a(1 $$-$$ 2) + b(2 $$-$$ 1) + c(1 $$-$$ 2) = 0<br><br>$$ \Rightarrow $$ $$-$$a + b $$-$$ c = 0 ....... |
Let a plane P contain two lines
<br/>$$\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 $$) is the foot of th... | 5
Explanation:
Given lines,<br><br>$$\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)$$ parallel to $$(\wid... |
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 :
Options:
[{"identifier": "A", "content": "$$19\\sqrt 2 $$"}, {"identifier": "B", "content": "$$2\\sqrt {19} $$"}, {"identifier": "C", "content":... | ["D"]
Explanation:
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 = $$... |
The vector equation of the plane passing through the intersection<br/><br/> of the planes $$\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1$$ and $$\overrightarrow r .\left( {\widehat i - 2\widehat j} \right) = - 2$$, and the point (1, 0, 2) is :
Options:
[{"identifier": "A", "content": "... | ["B"]
Explanation:
Given, point (1, 0, 2)<br/><br/>Equation of plane = <br/><br/>$$\overrightarrow r\,.\,(\widehat i + \widehat j + \widehat k) = 1$$ and $$\overrightarrow r\,.\,(\widehat i - 2\widehat j) = - 2$$<br/><br/>Equation of plane passing through the intersection of given planes is<br/><br/>$$[\overrightarro... |
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 __________.
Options... | 8
Explanation:
<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 Explana... |
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 :
Options:
[{"identifier": "A", "content": "102"}, {"identifi... | ["A"]
Explanation:
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 li... |
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 :
Options:
[{"identifier": "A", "conten... | ["A"]
Explanation:
<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 Explan... |
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 _________.
Options:
[] | 2
Explanation:
<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 Explanatio... |
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 ____________.
Options:
[] | 4
Explanation:
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 ... |
If the equation of plane passing through the mirror image of a point (2, 3, 1) with respect to line $${{x + 1} \over 2} = {{y - 3} \over 1} = {{z + 2} \over { - 1}}$$ and containing the line $${{x - 2} \over 3} = {{1 - y} \over 2} = {{z + 1} \over 1}$$ is $$\alpha$$x + $$\beta$$y + $$\gamma$$z = 24, then $$\alpha$$ + $... | ["B"]
Explanation:
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267417/exam_images/rfcxq7voccwbfmoextkg.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267008/exam_images/neowpsgzybeeqozyfefn.webp"><im... |
Let P be a plane containing the line $${{x - 1} \over 3} = {{y + 6} \over 4} = {{z + 5} \over 2}$$ and parallel to the line $${{x - 1} \over 4} = {{y - 2} \over { - 3}} = {{z + 5} \over 7}$$. If the point (1, $$-$$1, $$\alpha$$) lies on the plane P, then the value of |5$$\alpha$$| is equal to ____________.
Options:
[] | 38
Explanation:
<p>Equation of required plane is $$\left| {\matrix{
{x - 1} & {y + 6} & {z + 5} \cr
3 & 4 & 2 \cr
4 & { - 3} & 7 \cr
} } \right| = 0$$</p>
<p>Since, (1, $$-$$1, $$\alpha$$) lies on it,</p>
<p>So, replace x by 1, y by ($$-$$1) and z and $$\alpha$$.</p>
<p>$$\left| {\matrix{
0 & 5 & {... |
Let P be a plane passing through the points (1, 0, 1), (1, $$-$$2, 1) and (0, 1, $$-$$2). Let a vector $$\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... | 81
Explanation:
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>$$\overrightarrow a = \alpha \widehat i + \beta \widehat j + ... |
Consider the line L given by the equation <br/><br/>$${{x - 3} \over 2} = {{y - 1} \over 1} = {{z - 2} \over 1}$$. <br/><br/>Let Q be the mirror image of the point (2, 3, $$-$$1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points i... | ["D"]
Explanation:
Plane p is $${ \bot ^r}$$ to line $${{x - 3} \over 2} = {{y - 1} \over 1} = {{z - 2} \over 1}$$ & passes through pt. (2, 3) equation of plane p <br><br>2(x $$-$$ 2) + 1(y $$-$$ 3) + 1 (z + 1) = 0<br><br>2x + y + z $$-$$ 6 = 0<br><br>Point (1, 2, 2) satisfies above equation |
Let L be the line of intersection of planes $$\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... | ["B"]
Explanation:
$${P_1}:x - y + 2z = 2$$<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 Even... |
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 ... | ["C"]
Explanation:
<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 Explana... |
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 $$\alpha$$ $$-$$ $$\beta... | ["D"]
Explanation:
First line is ($$\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$$ + 1 = 3q + 7 ...... (iii)<br><br>fo... |
The distance of the point P(3, 4, 4) from the point of intersection of the line joining the points. Q(3, $$-$$4, $$-$$5) and R(2, $$-$$3, 1) and the plane 2x + y + z = 7, is equal to ______________.
Options:
[] | 7
Explanation:
$$\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)$$<br><br>Now, satisfying it in the given plane.<br><br>We get r = $$-$$2<br><br>so, required point of intersection is T(1, $$-$$2, 7).<br><... |
If the lines $${{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}$$ are co-planar, then the value of k is _____________.
Options:
[] | 1
Explanation:
$$\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$$ |
Let a plane P pass through the point (3, 7, $$-$$7) and contain the line, $${{x - 2} \over { - 3}} = {{y - 3} \over 2} = {{z + 2} \over 1}$$. If distance of the plane P from the origin is d, then d<sup>2</sup> is equal to ______________.
Options:
[] | 3
Explanation:
$$\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|$... |
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?
Options:
[{"identifier": "A", "content": "($$-$$1, 1, 2)"}, {"identifier": "B", "content": "(0, 1, 1)"}, {"identifier": "C", "content": "(1, 0... | ["B"]
Explanation:
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>$$ \Rightarrow {\overrightarrow n _1} = (1 + \lambda )\wide... |
Let the line L be the projection of the line $${{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 4} \over 2}$$ in the plane x $$-$$ 2y $$-$$ z = 3. If d is the distance of the point (0, 0, 6) from L, then d<sup>2</sup> is equal to _______________.
Options:
[] | 26
Explanation:
To find the projection let's find the foot of perpendicular from $(1,3$,
4) to plane $x-2 y-z=3$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lfg2ca35/20453aa3-5bf8-445c-837d-c8b6d848626f/8ecab100-c6b2-11ed-b4b3-b306e87ca523/file-1lfg2ca36.png?format=png" data-orsrc="https:/... |
Let Q be the foot of the perpendicular from the point P(7, $$-$$2, 13) on the plane containing the lines $${{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}$$. Then (PQ)<sup>2</sup>, is equal to ___________.
Options:
[] | 96
Explanation:
Containing the line $$\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 ... |
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 :
Options:
[{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["D"]
Explanation:
<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 Expla... |
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 :
Options:
[{"identifier": "A", "content": "$$3x - y - 5z + 2 = 0$$"}, {"identifier": "B", "content": "$$3x - 4z + 3 = 0$$"}... | ["D"]
Explanation:
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, $${{|4\lambda - 1|} \over {\sqrt {{{(2\lambda + 1)}^2} + {{(\lambda - 1)}^2} + {{( - 3\lam... |
The equation of the plane passing through the line of intersection of the planes $$\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$$ and parallel to the x-axis is :
Options:
[{"identifier": "A", "con... | ["A"]
Explanation:
Equation of planes are<br><br>$$\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$$<br><br>equation of planes... |
Let S be the mirror image of the point Q(1, 3, 4) with respect to the plane 2x $$-$$ y + z + 3 = 0 and let R(3, 5, $$\gamma$$) be a point of this plane. Then the square of the length of the line segment SR is ___________.
Options:
[] | 72
Explanation:
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266720/exam_images/h71pebht6whdybpiqe1o.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Evening Shift Mathematics - 3D Geometry Question 173 English Explanat... |
Let the equation of the plane, that passes through the point (1, 4, $$-$$3) and contains the line of intersection of the <br/>planes 3x $$-$$ 2y + 4z $$-$$ 7 = 0 <br/>and x + 5y $$-$$ 2z + 9 = 0, be <br/>$$\alpha$$x + $$\beta$$y + $$\gamma$$z + 3 = 0, then $$\alpha$$ + $$\beta$$ + $$\gamma$$ is equal to :
Options:
[{"... | ["A"]
Explanation:
3x $$-$$ 2y + 4z $$-$$ 7 + $$\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$$ + 9$$\... |
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 __________.
Options:
[] | 61
Explanation:
$${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6} = \lambda $$<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>$$7\lambda = 7 \Rightarrow \lambda ... |
The distance of the point ($$-$$1, 2, $$-$$2) from the line of intersection of the planes 2x + 3y + 2z = 0 and x $$-$$ 2y + z = 0 is :
Options:
[{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${5 \\over 2}$$"}, {"identifier": "C", "content": "$${{\\sqrt {42} } \\over 2}$... | ["D"]
Explanation:
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 ... |
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 _______.
Options:
[] | 7
Explanation:
<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... |
<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... | 28
Explanation:
<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. $$ ... |
<p>Let $${{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$$ R. The shortest distance of the plane from the origin is :</p>
Options:
[{"identifier": "A", "content": "$$\\sqrt {{3 \\over {109}}} $$"}, {"identifier": "B", "content": "$$\\sqrt ... | ["B"]
Explanation:
$\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1}=\lambda$
<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/>
$2 p+q-8=0\} q=-22$
<b... |
<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... | ["B"]
Explanation:
$P(1,2,1)$ image in plane $x+2 y+2 z=16$
<br><br>
$$
\begin{aligned}
& \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=\frac{-2(1+2 \times 2+2 \times 1-16)}{1^{2}+2^{2}+2^{2}} \\\\
& \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=2 \\\\
& Q(3,6,5) \\\\
& \vec{r}=-\hat{k}+\lambda(\hat{i}+\hat{j}... |
<p>If two distinct point Q, R lie on the line of intersection of the planes $$ - x + 2y - z = 0$$ and $$3x - 5y + 2z = 0$$ and $$PQ = PR = \sqrt {18} $$ where the point P is (1, $$-$$2, 3), then the area of the triangle PQR is equal to :</p>
Options:
[{"identifier": "A", "content": "$${2 \\over 3}\\sqrt {38} $$"}, {"i... | ["B"]
Explanation:
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5obyhm8/67807093-6d4d-453b-a541-6801368b15df/45fa9b00-0544-11ed-987f-3938cfc0f7f1/file-1l5obyhm9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5obyhm8/67807093-6d4d-453b-a541-6801368b15df/45fa9b00-0544-1... |
<p>Let the plane $$P:\overrightarrow r \,.\,\overrightarrow a = d$$ contain the line of intersection of two planes $$\overrightarrow r \,.\,\left( {\widehat i + 3\widehat j - \widehat k} \right) = 6$$ and $$\overrightarrow r \,.\,\left( { - 6\widehat i + 5\widehat j - \widehat k} \right) = 7$$. If the plane P passes t... | ["B"]
Explanation:
<p>$${P_1}:x + 3y - z = 6$$</p>
<p>$${P_2}: - 6x + 5y - z = 7$$</p>
<p>Family of planes passing through line of intersection of P<sub>1</sub> and P<sub>2</sub> is given by $$x(1 - 6\lambda ) + y(3 + 5\lambda ) + z( - 1 - \lambda ) - (6 + 7\lambda ) = 0$$</p>
<p>It passes through $$\left( {2,3,{1 \ov... |
<p>Let the foot of the perpendicular from the point (1, 2, 4) on the line $${{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3}$$ be P. Then the distance of P from the plane $$3x + 4y + 12z + 23 = 0$$ is :</p>
Options:
[{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "$${{50} \\over {13}}$$"}... | ["A"]
Explanation:
<p>$$L:{{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3} = t$$</p>
<p>Let P = (4t $$-$$ 2, 2t + 1, 3t $$-$$ 1)</p>
<p>$$\because$$ P is the foot of perpendicular of (1, 2, 4)</p>
<p>$$\therefore$$ $$4(4t - 3) + 2(2t - 1) + 3(3t - 5) = 0$$</p>
<p>$$ \Rightarrow 29t = 29 \Rightarrow t = 1$$</p... |
<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>
Options:
[{"identifier": "A", "content": "$${a \\over 8} = {b... | ["A"]
Explanation:
<p>Consider the equation of plane,</p>
<p>$$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\l... |
<p>If the lines $$\overrightarrow r = \left( {\widehat i - \widehat j + \widehat k} \right) + \lambda \left( {3\widehat j - \widehat k} \right)$$ and $$\overrightarrow r = \left( {\alpha \widehat i - \widehat j} \right) + \mu \left( {2\widehat i - 3\widehat k} \right)$$ are co-planar, then the distance of the plane c... | ["B"]
Explanation:
<p>$$\because$$ Both lines are coplanar, so</p>
<p>$$\left| {\matrix{
{\alpha - 1} & 0 & { - 1} \cr
0 & 3 & { - 1} \cr
2 & 0 & { - 3} \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow \alpha = {5 \over 3}$$</p>
<p>Equation of plane containing both lines</p>
<p>$$\left| {\matrix{
{x -... |
<p>Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a line L passing through (1, $$-$$1, $$-$$1), parallel to the line PQ meets the plane S at R, then QR<sup>2</sup> is equal to :</p>
Options:
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "5"}, {"... | ["B"]
Explanation:
<p>As L is parallel to PQ d.r.s of S is <1, 1, 1></p>
<p>$$\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 \la... |
<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... | 5
Explanation:
<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>$$\left( {from\,\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 2 & 3 \cr
1 & 1 & 5 \cr
} } \right| = 7\widehat i - \widehat j(2) + \widehat k( - 1)} \right)$... |
<p>The distance of the point (3, 2, $$-$$1) from the plane $$3x - y + 4z + 1 = 0$$ along the line $${{2 - x} \over 2} = {{y - 3} \over 2} = {{z + 1} \over 1}$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "3"}, {"identifier": "D", ... | ["C"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l65uoj8c/397fdd39-a2f5-49e7-9863-c285adbe3668/d1a0d9c0-0ee6-11ed-a7de-eff776fdb55c/file-1l65uoj8d.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l65uoj8c/397fdd39-a2f5-49e7-9863-c285adbe3668/d1a0d9c0-0ee6-11... |
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