Three Dimensional Geometry — JEE Main PYQs

Let the point A be the foot of perpendicular drawn from the point $P(a, b, 0)$ on the line $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-\alpha}{3}$. If the midpoint of the line segment PA is $\left(0, \frac{3}{4}, \frac{-1}{4}\right)$, then the value of $a^2 + b^2 + \alpha^2$ is equal to :

A line with direction ratios $1, -1, 2$ intersects the lines $\frac{x}{2} = \frac{y}{3} = \frac{z+1}{3}$ and $\frac{x+1}{-1} = \frac{y-2}{1} = \frac{z}{4}$ at the points P and Q, respectively. If the length of the line segment PQ is $\alpha$, then $225\alpha^2$ is equal to:

The square of the distance of the point $(-2, -8, 6)$ from the line $\frac{x-1}{1} = \frac{y-1}{2} = \frac{z}{-1}$ along the line $\frac{x+5}{1} = \frac{y+5}{-1} = \frac{z}{2}$ is equal to:

The shortest distance between the lines

$$\vec{r} = \left(\frac{1}{3}\hat{i} + 2\hat{j} + \frac{8}{3}\hat{k}\right) + \lambda\left(2\hat{i} - 5\hat{j} + 6\hat{k}\right)$$

and $\vec{r} = \left(-\dfrac{2}{3}\hat{i} - \dfrac{1}{3}\hat{k}\right) + \mu\left(\hat{j} - \hat{k}\right)$, $\lambda, \mu \in \mathbb{R}$, is:

If $\left(2\alpha + 1,\ \alpha^2 - 3\alpha,\ \dfrac{\alpha - 1}{2}\right)$ is the image of $(\alpha, 2\alpha, 1)$ in the line $\dfrac{x-2}{3} = \dfrac{y-1}{2} = \dfrac{z}{1}$, then the possible value(s) of $\alpha$ is (are)

The square of the distance of the point $P(5, 6, 7)$ from the line $\frac{x-2}{2} = \frac{y-5}{3} = \frac{z-2}{4}$ is equal to:

The square of the distance of the point of intersection of the lines $\vec{r} = \left( \hat{i} + \hat{j} - \hat{k} \right) + \lambda\left( a\hat{i} - \hat{j} \right)$, $a \neq 0$ and $\vec{r} = \left( 4\hat{i} - \hat{k} \right) + \mu\left( 2\hat{i} + a\hat{k} \right)$ from the origin is:

Let a triangle $PQR$ be such that $P$ and $Q$ lie on the line $\dfrac{x+3}{8} = \dfrac{y-4}{2} = \dfrac{z+1}{2}$ and are at a distance of 6 units from $R\,(1, 2, 3)$. If $(\alpha, \beta, \gamma)$ is the centroid of $\Delta PQR$, then $\alpha + \beta + \gamma$ is equal to :

If the distance of the point $(a, 2, 5)$ from the image of the point $(1, 2, 7)$ in the line $\dfrac{x}{1} = \dfrac{y-1}{1} = \dfrac{z-2}{2}$ is 4, then the sum of all possible values of $a$ is equal to :

Let the image of the point $P(1, 6, a)$ in the line $L: \frac{x}{1}=\frac{y-1}{2}=\frac{z-a+1}{b}$, $b>0$, be $\left(\frac{a}{3}, 0, a+c\right)$. If $S(\alpha, \beta, \gamma)$, $\alpha>0$, is the point on L such that the distance of S from the foot of perpendicular from the point P on L is $2\sqrt{14}$, then $\alpha+\beta+\gamma$ is equal to:

Let a line L be perpendicular to both the lines

$$L_1: \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} \text{ and } L_2: \frac{x-2}{1}=\frac{y-4}{4}=\frac{z-6}{7}.$$

If $\theta$ is the acute angle between the lines L and

$$L_3: \frac{x-\frac{8}{7}}{2}=\frac{y-\frac{4}{7}}{1}=\frac{z}{2},$$

then $\tan\theta$ is equal to:

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

Let the image of the point $P(0, -5, 0)$ in the line $\frac{x - 1}{2} = \frac{y}{1} = \frac{z + 1}{-2}$ be the point R and the image of the point $Q\left(0, \frac{-1}{2}, 0\right)$ in the line $\frac{x - 1}{-1} = \frac{y + 9}{4} = \frac{z + 1}{1}$ be the point S. Then the square of the area of the parallelogram PQRS is ________.

Let the foot of perpendicular from the point $(\lambda, 2, 3)$ on the line $\frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1}$ be the point $(1, \mu, 2)$. Then the distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}$ and $\frac{x-\lambda}{2} = \frac{y-\mu}{3} = \frac{z+5}{6}$ is equal to :

Let a line $L_1$ pass through the origin and be perpendicular to the lines

$$L_2 : \vec{r} = (3 + t)\hat{i} + (2t - 1)\hat{j} + (2t + 4)\hat{k} \text{ and}$$

$$L_3 : \vec{r} = (3 + 2s)\hat{i} + (3 + 2s)\hat{j} + (2 + s)\hat{k},\; t, s \in \mathbb{R}.$$

If $(a, b, c)$, $a \in \mathbb{Z}$, is the point on $L_3$ at a distance of $\sqrt{17}$ from the point of intersection of $L_1$ and $L_2$, then $(a + b + c)^2$ is equal to __________.

Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the $\Delta$BCD is equal to :

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

If the image of the point $P(1, 0, 3)$ in the line joining the points $A(4, 7, 1)$ and $B(3, 5, 3)$ is $Q(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to :

The line $L_1$ is parallel to the vector $\vec{a}=-3\hat{i}+2\hat{j}+4\hat{k}$ and passes through the point $(7, 6, 2)$ and the line $L_2$ is parallel to the vector $\vec{b}=2\hat{i}+\hat{j}+3\hat{k}$ and passes through the point $(5, 3, 4)$. The shortest distance between the lines $L_1$ and $L_2$ is :

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 between the two lines be 3. Then the square of the distance of the point $(\lambda_1, \lambda_2, 7)$ from the line $\mathrm{L}_1$ is

Let a line passing through the point $(4, 1, 0)$ intersect the line $\mathrm{L}_1 : \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $\mathrm{L}_2 : x - 6 = y = -z + 4$ at the point $B(a, b, c)$. Then $\begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix}$ is equal to

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 $\dfrac{x - 4}{1} = \dfrac{y - 4}{0} = \dfrac{z - 2}{3}$ along the line $\dfrac{x - 9}{2} = \dfrac{y - 13}{3} = \dfrac{z - 17}{6}$ is

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 $L : \frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$. 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, then $\overrightarrow{OA} \cdot \overrightarrow{OB}$ is equal to

Let the values of $p$, for which the shortest distance between the lines $\dfrac{x+1}{3}=\dfrac{y}{4}=\dfrac{z}{5}$ and $\vec{r}=\left(p\hat{i}+2\hat{j}+\hat{k}\right)+\lambda\left(2\hat{i}+3\hat{j}+4\hat{k}\right)$ is $\dfrac{1}{\sqrt{6}}$, be $a, b$, $(a<b)$. Then the length of the latus rectum of the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ is :

Let A be the point of intersection of the lines $\mathrm{L}_1:\dfrac{x-7}{1}=\dfrac{y-5}{0}=\dfrac{z-3}{-1}$ and $\mathrm{L}_2:\dfrac{x-1}{3}=\dfrac{y+3}{4}=\dfrac{z+7}{5}$. Let B and C be the points on the lines $\mathrm{L}_1$ and $\mathrm{L}_2$ respectively such that $\mathrm{AB}=\mathrm{AC}=\sqrt{15}$. Then the square of the area of the triangle ABC 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?

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

If the equation of the line passing through the point $\left(0, -\dfrac{1}{2}, 0\right)$ and perpendicular to the lines $\vec{r} = \lambda\left(\hat{i} + a\hat{j} + b\hat{k}\right)$ and $\vec{r} = \left(\hat{i} - \hat{j} - 6\hat{k}\right) + \mu\left(-b\hat{i} + a\hat{j} + 5\hat{k}\right)$ is $\dfrac{x - 1}{-2} = \dfrac{y + 4}{d} = \dfrac{z - c}{-4}$, then $a + b + c + d$ is equal to :

Consider the lines $L_1 : x - 1 = y - 2 = z$ and $L_2 : x - 2 = y = z - 1$. Let the feet of the perpendiculars from the point $P(5, 1, -3)$ on the lines $L_1$ and $L_2$ be $Q$ and $R$ respectively. If the area of the triangle $PQR$ is $A$, then $4A^2$ is equal to :

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 passing through the points $(0, 0)$, $(\lambda_1, \lambda_2)$ and $(\lambda_2, \lambda_1)$ is

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

Let $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$ ?

Let $L_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $L_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}$, $\alpha \in \mathbb{R}$, be two lines, which intersect at the point B. If P is the foot of perpendicular from the point A$(1,1,-1)$ on $L_2$, then the value of $26\,\alpha(PB)^2$ is ________.

The perpendicular distance, of the line $\dfrac{x-1}{2} = \dfrac{y+2}{-1} = \dfrac{z+3}{2}$ from the point $P(2, -10, 1)$, is :

Let a line pass through two distinct points $P(-2, -1, 3)$ and $Q$, and be parallel to the vector $3\hat{i} + 2\hat{j} + 2\hat{k}$. If the distance of the point $Q$ from the point $R(1, 3, 3)$ is 5, then the square of the area of $\triangle PQR$ is equal to :

Let $P$ be the foot of the perpendicular from the point $Q(10, -3, -1)$ on the line $\frac{x-3}{7} = \frac{y-2}{-1} = \frac{z+1}{-2}$. Then the area of the right angled triangle $PQR$, where $R$ is the point $(3, -2, 1)$, is

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

If the square of the shortest distance between the lines $\dfrac{x-2}{1} = \dfrac{y-1}{2} = \dfrac{z+3}{-3}$ and $\dfrac{x+1}{2} = \dfrac{y+3}{4} = \dfrac{z+5}{-5}$ is $\dfrac{\mathrm{m}}{\mathrm{n}}$, where m, n are coprime numbers, then $\mathrm{m} + \mathrm{n}$ is equal to :

Let in a $\triangle ABC$, the length of the side AC be 6, the vertex B be $(1, 2, 3)$ and the vertices A, C lie on the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle ABC$ is:

Let the line passing through the points $(-1, 2, 1)$ and parallel to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ intersect the line $\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}$ at the point P. Then the distance of P from the point $Q(4, -5, 1)$ is

Let P be the image of the point $\mathrm{Q}(7, -2, 5)$ in the line $\mathrm{L} : \dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z}{4}$ and $\mathrm{R}(5, \mathrm{p}, \mathrm{q})$ be a point on L. Then the square of the area of $\Delta\mathrm{PQR}$ is __________.

Let A $(x, y, z)$ be a point in $xy$-plane, which is equidistant from three points $(0, 3, 2)$, $(2, 0, 3)$ and $(0, 0, 1)$. Let B $= (1, 4, -1)$ and C $= (2, 0, -2)$. Then among the statements (S1) : $\triangle$ABC is an isosceles right angled triangle, and (S2) : the area of $\triangle$ABC is $\frac{9\sqrt{2}}{2}$,

If the image of the point $(4, 4, 3)$ in the line $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3}$ is $(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma$ is equal to

The square of the distance of the point $\left(\frac{15}{7},\ \frac{32}{7},\ 7\right)$ from the line $\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}$ in the direction of the vector $\hat{i} + 4\hat{j} + 7\hat{k}$ is :

Let the point, on the line passing through the points $P(1, -2, 3)$ 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 $\int\limits_{0}^{k}[x^{2}]\, dx=\alpha-\sqrt{\alpha}$, where $[x]$ denotes the greatest integer function, then $6\alpha^{3}$ is equal to ________.

Let P be the point of intersection of the lines $\dfrac{x-2}{1} = \dfrac{y-4}{5} = \dfrac{z-2}{1}$ and $\dfrac{x-3}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{2}$. 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 $\dfrac{2-x}{3} = \dfrac{3y-2}{4\lambda+1} = 4-z$ makes a right angle with the line $\dfrac{x+3}{3\mu} = \dfrac{1-2y}{6} = \dfrac{5-z}{7}$, then $4\lambda + 9\mu$ is equal to :

Let d be the distance of the point of intersection of the lines $\dfrac{x+6}{3} = \dfrac{y}{2} = \dfrac{z+1}{1}$ and $\dfrac{x-7}{4} = \dfrac{y-9}{3} = \dfrac{z-4}{2}$ 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 $\dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-2}{5}$. Then $\alpha + \beta + \gamma$ is equal to :

Let the point $(-1, \alpha, \beta)$ lie on the line of the shortest distance between the lines $\dfrac{x + 2}{-3} = \dfrac{y - 2}{4} = \dfrac{z - 5}{2}$ and $\dfrac{x + 2}{-1} = \dfrac{y + 6}{2} = \dfrac{z - 1}{0}$. Then $(\alpha - \beta)^2$ is equal to ________.

The shortest distance between the lines $\dfrac{x-3}{2} = \dfrac{y+15}{-7} = \dfrac{z-9}{5}$ and $\dfrac{x+1}{2} = \dfrac{y-1}{1} = \dfrac{z-9}{-3}$ 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 $P(\alpha, \beta, \gamma)$ be the image of the point $Q(3, -3, 1)$ in the line $\dfrac{x-0}{1}=\dfrac{y-3}{1}=\dfrac{z-1}{-1}$ and $R$ be the point $(2, 5, -1)$. If the area of the triangle $PQR$ is $\lambda$ and $\lambda^2=14K$, then $K$ is equal to :

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

If the shortest distance between the lines $$L_1 : \vec{r}=(2+\lambda)\hat{i}+(1-3\lambda)\hat{j}+(3+4\lambda)\hat{k}, \quad \lambda\in\mathbb{R}$$ $$L_2 : \vec{r}=2(1+\mu)\hat{i}+3(1+\mu)\hat{j}+(5+\mu)\hat{k}, \quad \mu\in\mathbb{R}$$ is $\frac{m}{\sqrt{n}}$, where $\gcd(m,n)=1$, then the value of $m+n$ equals

Let $P(x, y, z)$ be a point in the first octant, whose projection in the $xy$-plane is the point $Q$. Let $OP=\gamma$; the angle between $OQ$ and the positive $x$-axis be $\theta$; and the angle between $OP$ and the positive $z$-axis be $\phi$, where $O$ is the origin. Then the distance of $P$ from the $x$-axis is

If the shortest distance between the lines $\dfrac{x - \lambda}{2} = \dfrac{y - 4}{3} = \dfrac{z - 3}{4}$ and $\dfrac{x - 2}{4} = \dfrac{y - 4}{6} = \dfrac{z - 7}{8}$ is $\dfrac{13}{\sqrt{29}}$, 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 $\dfrac{x}{1} = \dfrac{y-1}{2} = \dfrac{z-2}{3}$. Then $2\alpha + \beta + \gamma$ is equal to ________