JEE Main 2026 — Mathematics Vectors & 3D Geometry

2026mcqhard

For a triangle ABC, let $\overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{CA}}$ and $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{BA}}$ . If $|\overrightarrow{\mathrm{p}}|=2 \sqrt{3},|\overrightarrow{\mathrm{q}}|=2$ and $\cos \theta=\frac{1}{\sqrt{3}}$ , where $\theta$ is the angle between $\vec{p}$ and $\vec{q}$ , then $|\vec{p} \times(\vec{q}-3 \vec{r})|^{2}+3|\vec{r}|^{2}$ is equal to :

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Options

A
410
B
340
C
200
D
220

Did you get this right?

Solution

Given $|\vec{p}| = 2\sqrt{3}$ , $|\vec{q}| = 2$ , and $\cos\theta = \frac{1}{\sqrt{3}}$ where $\theta$ is the angle between $\vec{p}$ and $\vec{q}$ .

Since $\vec{p} = \overrightarrow{BC}$ , $\vec{q} = \overrightarrow{CA}$ , and $\vec{r} = \overrightarrow{BA}$ , from triangle closure $\vec{r} = \vec{p} + \vec{q}$ .

$\vec{p} \cdot \vec{q} = |\vec{p}||\vec{q}|\cos\theta = 2\sqrt{3} \cdot 2 \cdot \frac{1}{\sqrt{3}} = 4$

$|\vec{r}|^2 = |\vec{p}|^2 + |\vec{q}|^2 + 2\vec{p} \cdot \vec{q} = 12 + 4 + 8 = 24$

$\vec{q} - 3\vec{r} = \vec{q} - 3(\vec{p} + \vec{q}) = -3\vec{p} - 2\vec{q}$

$\vec{p} \times (\vec{q} - 3\vec{r}) = -2(\vec{p} \times \vec{q})$

$\sin^2\theta = 1 - \frac{1}{3} = \frac{2}{3}$

$|\vec{p} \times \vec{q}|^2 = 12 \cdot 4 \cdot \frac{2}{3} = 32$

$|\vec{p} \times (\vec{q} - 3\vec{r})|^2 = 4 \cdot 32 = 128$

Result: $128 + 3(24) = 128 + 72 = 200$

Your note

JEE Main Vectors & 3D Geometry in other years

Vectors & 3D Geometry 2025 Vectors & 3D Geometry 2024 Vectors & 3D Geometry 2023 Vectors & 3D Geometry 2022 Vectors & 3D Geometry 2021 Vectors & 3D Geometry 2020 Vectors & 3D Geometry 2019 Vectors & 3D Geometry 2018 Vectors & 3D Geometry 2017 Vectors & 3D Geometry 2016 Vectors & 3D Geometry 2015 Vectors & 3D Geometry 2014

More JEE Main Vectors & 3D Geometry questions

Let the line $\mathrm{L}_{1}$ be parallel to the vector $-3 \hat{i}+2 \hat{j}+4 \hat{k}$ and pass through the point (2,6,7), and the line $\mathrm{L}_{2}$ be parallel to the vector $2 \hat{i}+\hat{j}+3 \hat{k}$ and pass through the point $(4,3,5)$. If the line $\mathrm{L}_{3}$ is parallel to the vector $-3 \hat{i}+5 \hat{j}+16 \hat{k}$ and intersects the lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ at the points C and D, respectively, then $|\overrightarrow{C D}|^{2}$ is equal to :

2026hard

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 __________.

2026hard

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

2026medium

Let $\overrightarrow{\mathrm{AB}}=2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\overrightarrow{\mathrm{AD}}=\hat{i}+2 \hat{j}+\lambda \hat{k}, \lambda \in \mathbb{R}$. Let the projection of the vector $\vec{v}=\hat{i}+\hat{j}+\hat{k}$ on the diagonal $\overrightarrow{\mathrm{AC}}$ of the parallelogram ABCD be of length one unit. If $\alpha, \beta$, where $\alpha>\beta$, be the roots of the equation $\lambda^{2} x^{2}-6 \lambda x+5=0$, then $2 \alpha-\beta$ is equal to

2026hard

Let $\vec{a}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}, \vec{b}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\vec{c}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$. Let $\vec{v}$ be the vector in the plane of the vectors $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\frac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to

2026medium

View all Vectors & 3D Geometry questions

Keep practicing Vectors & 3D Geometry

Work through every JEE Main Vectors & 3D Geometry PYQ, year by year.

Browse all Mathematics topics

Solution

Given $|\vec{p}| = 2\sqrt{3}$ , $|\vec{q}| = 2$ , and $\cos\theta = \frac{1}{\sqrt{3}}$ where $\theta$ is the angle between $\vec{p}$ and $\vec{q}$ .

Since $\vec{p} = \overrightarrow{BC}$ , $\vec{q} = \overrightarrow{CA}$ , and $\vec{r} = \overrightarrow{BA}$ , from triangle closure $\vec{r} = \vec{p} + \vec{q}$ .

$\vec{p} \cdot \vec{q} = |\vec{p}||\vec{q}|\cos\theta = 2\sqrt{3} \cdot 2 \cdot \frac{1}{\sqrt{3}} = 4$

$|\vec{r}|^2 = |\vec{p}|^2 + |\vec{q}|^2 + 2\vec{p} \cdot \vec{q} = 12 + 4 + 8 = 24$

$\vec{q} - 3\vec{r} = \vec{q} - 3(\vec{p} + \vec{q}) = -3\vec{p} - 2\vec{q}$

$\vec{p} \times (\vec{q} - 3\vec{r}) = -2(\vec{p} \times \vec{q})$

$\sin^2\theta = 1 - \frac{1}{3} = \frac{2}{3}$

$|\vec{p} \times \vec{q}|^2 = 12 \cdot 4 \cdot \frac{2}{3} = 32$

$|\vec{p} \times (\vec{q} - 3\vec{r})|^2 = 4 \cdot 32 = 128$

Result: $128 + 3(24) = 128 + 72 = 200$

JEE Main 2026 — Mathematics Vectors & 3D Geometry

Options

Did you get this right?

Solution

Your note

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

JEE Main 2026 — Mathematics Vectors & 3D Geometry

Options

Did you get this right?

Solution

Your note

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