Three Dimensional Geometry — JEE Main PYQs

The line ${\mathrm{L}}_1$ is parallel to the vector $\overrightarrow{\mathrm{a}}=-3\hat{\mathrm{i}}+2\hat{\mathrm{j}}+4\hat{\mathrm{k}}$ and passes through the point $(7,6,2)$ and the line ${\mathrm{L}}_2$ is parallel to the vector …

Each of the angles $\beta$ and $\gamma$ that a given line makes with the positive y - and z -axes, respectively, is half of the angle that this line makes with the positive $x$-axes. Then the sum of all possible values of the angle $\beta$ is

The distance of the point $(7,10,11)$ from the line $\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}$ along the line $\frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6}$ is

Let the vertices $Q$ and $R$ of the triangle $PQR$ lie on the line $\frac{\mathrm{x}+3}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3},\mathrm{QR}=5$ and the coordinates of the point P be $(0,2,3)$. If the area of the triangle $PQR$ is $\frac{m}{n}$ then :

Let a line passing through the point $(4,1,0)$ intersect the line $L_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at the point $\mathrm{A}(\alpha ,\beta ,\gamma )$ and the line ${\mathrm{L}}_2:x-6=y=-z+4$ at the point $\mathrm{B}(\mathrm{a},\mathrm{b},\mathrm{c})$. …

Line ${\mathrm{L}}_1$ passes through the point $(1,2,3)$ and is parallel to z -axis. Line ${\mathrm{L}}_2$ passes through the point ( $\lambda$, 5, 6) and is parallel to y -axis. Let for $\lambda ={\lambda }_1,{\lambda }_2,{\lambda }_2<{\lambda }_1$, the shortest distance …

Let the values of p, for which the shortest distance between the lines $\frac{\mathrm{x}+1}{3}=\frac{\mathrm{y}}{4}=\frac{\mathrm{z}}{5}$ and $\vec{r}=(p\hat{i}+2\hat{j}+\hat{k})+\lambda (2\hat{i}+3\hat{j}+4\hat{k})$ is $\frac{1}{\sqrt{6}}$, be $a,b$, …

Let A be the point of intersection of the lines
${\mathrm{L}}_1:\frac{\mathrm{x}-7}{1}=\frac{\mathrm{y}-5}{0}=\frac{\mathrm{z}-3}{-1}$ and
${\mathrm{L}}_2:\frac{\mathrm{x}-1}{3}=\frac{\mathrm{y}+3}{4}=\frac{\mathrm{z}+7}{5}$. Let B and C be the point on the lines …

Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha }{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta }{4}$ be $3\sqrt{30}$. Then the positive value of $5\alpha +\beta$ is

Let A and B be two distinct points on the line $\mathrm{L}:\frac{\mathrm{x}-6}{3}=\frac{\mathrm{y}-7}{2}=\frac{\mathrm{z}-7}{-2}$. Both A and B are at a distance $2\sqrt{17}$ from the foot of perpendicular drawn from the point $(1,2,3)$ on the line L. If O is the origin, …

If the equation of the line passing through the point $\left(0,-\frac{1}{2},0\right)$ and perpendicular to the lines $\overrightarrow{\mathrm{r}}=\lambda (\hat{\mathrm{i}}+\mathrm{a}\hat{\mathrm{j}}+\mathrm{b}\hat{\mathrm{k}})$ and …

Consider the lines ${\mathrm{L}}_1:\mathrm{x}-1=\mathrm{y}-2=\mathrm{z}$ and ${\mathrm{L}}_2:\mathrm{x}-2=\mathrm{y}=\mathrm{z}-1$. Let the feet of the perpendiculars from the point $\mathrm{P}(5,1,-3)$ on the lines ${\mathrm{L}}_1$ and ${\mathrm{L}}_2$ be Q and R res …

If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x}{1}=\frac{y}{\alpha }=\frac{z-5}{1}$ is $\frac{5}{\sqrt{6}}$, then the sum of all possible values of $\alpha$ is

Let the line L pass through $(1,1,1)$ and intersect the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}$. Then, which of the following points lies on the line L?

Let the values of $\lambda$ for which the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-\lambda }{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is $\frac{1}{\sqrt{6}}$ be ${\lambda }_1$ and ${\lambda }_2$. Then the radius of the circle pa …

Let the area of the triangle formed by the lines $\mathrm{x}+2=\mathrm{y}-1=\mathrm{z}$, $\frac{\mathrm{x}-3}{5}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-1}{1}$ and $\frac{\mathrm{x}}{-3}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}-2}{1}$ be A . Then ${\mathrm{A}}^2$ is equal to ______

Let the point, on the line passing through $P(1,-2,3)$ and $Q(5,-4,7)$, farther from the origin and at a distance of 9 units from the point $P$, be $(\alpha ,\beta ,\gamma )$. Then ${\alpha }^2+{\beta }^2+{\gamma }^2$ is equal to

If the shortest distance between the lines $\frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$ and $\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$ is $\frac{38}{3\sqrt{5}}k$, and ${\int }_0^k[x^2]dx=\alpha -\sqrt{\alpha }$, where $[x]$ denotes the greatest integer function, then …

Let $P$ be the point of intersection of the lines $\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$ and $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$. Then, the shortest distance of $P$ from the line $4x=2y=z$ is

Consider a line $L$ passing through the points $P(1,2,1)$ and $Q(2,1,-1)$. If the mirror image of the point $A(2,2,2)$ in the line $L$ is $(\alpha ,\beta ,\gamma )$, then $\alpha +\beta +6\gamma$ is equal to ____.

If the line $\frac{2-x}{3}=\frac{3y-2}{4\lambda +1}=4-z$ makes a right angle with the line $\frac{x+3}{3\mu }=\frac{1-2y}{6}=\frac{5-z}{7}$, then $4\lambda +9\mu$ is equal to:

Let $d$ 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 $d^2+6$ is equal to

Let $(\alpha ,\beta ,\gamma )$ be the image of the point $(8,5,7)$ in the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{5}$. Then $\alpha +\beta +\gamma$ is equal to

The shortest distance between the line $\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$ and $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$ is

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 $PQ$ is equal to ____.

Let $\mathrm{P}(\alpha ,\beta ,\gamma )$ be the image of the point $\mathrm{Q}(3,-3,1)$ in the line $\frac{\mathrm{x}-0}{1}=\frac{\mathrm{y}-3}{1}=\frac{\mathrm{z}-1}{-1}$ and R be the point $(2,5,-1)$. If the area of the triangle ${\lambda }^2=74\mathrm{K}$ then K is equal to

If the shortest distance between the lines $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$ to $\frac{x-\lambda }{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and is $\frac{44}{\sqrt{30}}$ then the largest possible value of $|\lambda |$ is equal ____.

Let $\mathrm{P}(\mathrm{x},\mathrm{y},\mathrm{z})$ be a point in the first octant, whose projection in the xy -plane is the point Q . Let $\mathrm{OP}=\gamma$; the angle between OQ and the positive x -axis be $\theta$; and the angle between OP and the positive z -axis be …

If the shortest distance between the lines
${\mathrm{L}}_1:\overrightarrow{\mathrm{r}}=(2+\lambda )\hat{\mathrm{i}}+(1-3\lambda )\hat{\mathrm{j}}+(3+4\lambda )\hat{\mathrm{k}},\lambda \in \mathbb{R}$ …

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 :

Let $\mathrm{P}(\alpha ,\beta ,\gamma )$ be the image of the point $\mathrm{Q}(1,6,4)$ in the line $\frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}$. Then $2\alpha +\beta +\gamma$ is equal to

Let the line L intersect the lines $\mathrm{x}-2=-\mathrm{y}=\mathrm{z}-1,2(\mathrm{x}+1)=2(\mathrm{y}-1)=\mathrm{z}+1$ 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 L ?

The shortest distance between the line
$\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$ and $\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$ is :

Consider the line L passing through the points $(1,2,3)$ and $(2,3,5)$. The distance of the point $\left(\frac{11}{3},\frac{11}{3},\frac{19}{3}\right)$ from the line L along the line $\frac{3x-11}{2}=\frac{3y-11}{1}=\frac{3z-19}{2}$ is equal to :

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 ____.

Let O be the origin, and M and 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 MN is the shortest distance between the given lines. Then …

Let $(\alpha ,\beta ,\gamma )$ be 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

The shortest distance between lines ${\mathrm{L}}_1$ and ${\mathrm{L}}_2$, 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 $\mathrm{A}(-4,4,3)\cdot \mathrm{B}(-1,6,3)$ and perpendicular to the line …

A line passes through $\mathrm{A}(4,-6,-2)$ and $\mathrm{B}(16,-2,4)$. The point $\mathrm{P}(\mathrm{a},\mathrm{b},\mathrm{c})$ where $\mathrm{a},\mathrm{b},\mathrm{c}$ are non-negative integers, on the line AB lies at a distance of 21 units, from the point A. The dis …

Let $(\alpha ,\beta ,\gamma )$ be the foot of perpendicular from the point (1,2,3) on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$. Then $19(\alpha +\beta +\gamma )$ is equal to :

If $d_1$ is the shortest distance between the lines $x+1=2y=-12z$, $x=y+2=6z-6$ and $d_2$ 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 $\frac{32\sqrt{3}{d}_1}{d_2}$ i …

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 $\frac{2\pi }{3}$ a …

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 P from the line $\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$ is $l$, then $14l^2$ is equal to

The distance, of the point $(7,-2,11)$ from the line $\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is :

If the shortest distance between the lines $\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ 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 :

Let P and Q be the points on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ which are at a distance of 6 units from the point $\mathrm{R}(1,2,3)$. If the centroid of the triangle PQR is $(\alpha ,\beta ,\gamma )$, then ${\alpha }^2+{\beta }^2+{\gamma }^2$ is:

If the mirror image of the point $\mathrm{P}(3,4,9)$ in the line $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}$ is $(\alpha ,\beta ,\gamma )$, then $14(\alpha +\beta +\gamma )$ is :

The distance of the point $Q(0,2,-2)$ from the line passing through the point $P(5,-4,3)$ and perpendicular to the lines $\vec{r}=(-3\hat{i}+2\hat{k})+\lambda (2\hat{i}+3\hat{j}+5\hat{k}),\lambda \in \mathbb{R}$ and …

Let Q and R be the feet of perpendiculars from the point P(a, a, a) on the lines x=y, z=1 and x=-y, z=-1 respectively. If ∠QPR is a right angle, then $12a^2$ is equal to ____

If the shortest distance between the lines
$\frac{x-\lambda }{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ 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 :