JEE Main 2026 — Mathematics Vectors & 3D Geometry

2026integerhard

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

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Held on 6 Apr 2026 · Verified 6 Jul 2026.

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Solution

Let $M_1$ be the foot of the perpendicular from $P(0, -5, 0)$ to the first line $L_1: \dfrac{x-1}{2} = \dfrac{y}{1} = \dfrac{z+1}{-2}$ .

A general point on $L_1$ is $M_1(2\lambda + 1, \lambda, -2\lambda - 1)$ .

The direction ratios of $PM_1$ are $(2\lambda + 1, \lambda + 5, -2\lambda - 1)$ .

Since $PM_1$ is perpendicular to $L_1$ , their dot product is zero:

$2(2\lambda + 1) + 1(\lambda + 5) - 2(-2\lambda - 1) = 0$

$4\lambda + 2 + \lambda + 5 + 4\lambda + 2 = 0 \Rightarrow 9\lambda + 9 = 0 \Rightarrow \lambda = -1$

Thus, $M_1$ is $(-1, -1, 1)$ .

Let $M_2$ be the foot of the perpendicular from $Q\left(0, \dfrac{-1}{2}, 0\right)$ to the second line $L_2: \dfrac{x-1}{-1} = \dfrac{y+9}{4} = \dfrac{z+1}{1}$ .

A general point on $L_2$ is $M_2(-\mu + 1, 4\mu - 9, \mu - 1)$ .

The direction ratios of $QM_2$ are $\left(-\mu + 1, 4\mu - \dfrac{17}{2}, \mu - 1\right)$ .

Since $QM_2$ is perpendicular to $L_2$ , their dot product is zero:

$-1(-\mu + 1) + 4\left(4\mu - \dfrac{17}{2}\right) + 1(\mu - 1) = 0$

$\mu - 1 + 16\mu - 34 + \mu - 1 = 0 \Rightarrow 18\mu - 36 = 0 \Rightarrow \mu = 2$

Thus, $M_2$ is $(-1, -1, 1)$ .

Since $M_1$ and $M_2$ are the same point $M(-1, -1, 1)$ , the diagonals $PR$ and $QS$ of the quadrilateral $PQRS$ bisect each other at $M$ . Therefore, $PQRS$ is a parallelogram.

The area of the parallelogram is given by $\dfrac{1}{2} |\overrightarrow{PR} \times \overrightarrow{QS}|$ .

We have:

$\overrightarrow{PR} = 2\overrightarrow{PM} = 2(-1 - 0, -1 - (-5), 1 - 0) = (-2, 8, 2)$

$\overrightarrow{QS} = 2\overrightarrow{QM} = 2\left(-1 - 0, -1 - \left(\dfrac{-1}{2}\right), 1 - 0\right) = (-2, -1, 2)$

Now, compute the cross product $\overrightarrow{PR} \times \overrightarrow{QS}$ :

$\overrightarrow{PR} \times \overrightarrow{QS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 8 & 2 \\ -2 & -1 & 2 \end{vmatrix} = \hat{i}(16 - (-2)) - \hat{j}(-4 - (-4)) + \hat{k}(2 - (-16)) = 18\hat{i} + 18\hat{k}$

The area of the parallelogram is:

$\text{Area} = \dfrac{1}{2} |18\hat{i} + 18\hat{k}| = \dfrac{1}{2} \sqrt{18^2 + 18^2} = \dfrac{18\sqrt{2}}{2} = 9\sqrt{2}$

The square of the area is $(9\sqrt{2})^2 = 162$ .

Answer: $162$

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Solution

Let $M_1$ be the foot of the perpendicular from $P(0, -5, 0)$ to the first line $L_1: \dfrac{x-1}{2} = \dfrac{y}{1} = \dfrac{z+1}{-2}$ .

A general point on $L_1$ is $M_1(2\lambda + 1, \lambda, -2\lambda - 1)$ .

The direction ratios of $PM_1$ are $(2\lambda + 1, \lambda + 5, -2\lambda - 1)$ .

Since $PM_1$ is perpendicular to $L_1$ , their dot product is zero:

$2(2\lambda + 1) + 1(\lambda + 5) - 2(-2\lambda - 1) = 0$

$4\lambda + 2 + \lambda + 5 + 4\lambda + 2 = 0 \Rightarrow 9\lambda + 9 = 0 \Rightarrow \lambda = -1$

Thus, $M_1$ is $(-1, -1, 1)$ .

Let $M_2$ be the foot of the perpendicular from $Q\left(0, \dfrac{-1}{2}, 0\right)$ to the second line $L_2: \dfrac{x-1}{-1} = \dfrac{y+9}{4} = \dfrac{z+1}{1}$ .

A general point on $L_2$ is $M_2(-\mu + 1, 4\mu - 9, \mu - 1)$ .

The direction ratios of $QM_2$ are $\left(-\mu + 1, 4\mu - \dfrac{17}{2}, \mu - 1\right)$ .

Since $QM_2$ is perpendicular to $L_2$ , their dot product is zero:

$-1(-\mu + 1) + 4\left(4\mu - \dfrac{17}{2}\right) + 1(\mu - 1) = 0$

$\mu - 1 + 16\mu - 34 + \mu - 1 = 0 \Rightarrow 18\mu - 36 = 0 \Rightarrow \mu = 2$

Thus, $M_2$ is $(-1, -1, 1)$ .

Since $M_1$ and $M_2$ are the same point $M(-1, -1, 1)$ , the diagonals $PR$ and $QS$ of the quadrilateral $PQRS$ bisect each other at $M$ . Therefore, $PQRS$ is a parallelogram.

The area of the parallelogram is given by $\dfrac{1}{2} |\overrightarrow{PR} \times \overrightarrow{QS}|$ .

We have:

$\overrightarrow{PR} = 2\overrightarrow{PM} = 2(-1 - 0, -1 - (-5), 1 - 0) = (-2, 8, 2)$

$\overrightarrow{QS} = 2\overrightarrow{QM} = 2\left(-1 - 0, -1 - \left(\dfrac{-1}{2}\right), 1 - 0\right) = (-2, -1, 2)$

Now, compute the cross product $\overrightarrow{PR} \times \overrightarrow{QS}$ :

The area of the parallelogram is:

$\text{Area} = \dfrac{1}{2} |18\hat{i} + 18\hat{k}| = \dfrac{1}{2} \sqrt{18^2 + 18^2} = \dfrac{18\sqrt{2}}{2} = 9\sqrt{2}$

The square of the area is $(9\sqrt{2})^2 = 162$ .

Answer: $162$

JEE Main 2026 — Mathematics Vectors & 3D Geometry

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JEE Main 2026 — Mathematics Vectors & 3D Geometry

Did you get this right?

Solution

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JEE Main Vectors & 3D Geometry in other years

More JEE Main Vectors & 3D Geometry questions

Other Mathematics topics for JEE Main

Algebra

Calculus

Coordinate Geometry

Probability & Statistics

Trigonometry

Keep practicing Vectors & 3D Geometry