1
JEE Main 2026 (Online) 5th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $\vec{a}=\sqrt{7} \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{j}+2 \hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a}+\vec{a} \times \vec{b}=\overrightarrow{0}$ and $\vec{r} \cdot \vec{a}=0$, then $|3 \vec{r}|^2$ is equal to:

A

44

B

54

C

86

D

132

2
JEE Main 2026 (Online) 4th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $\hat{u}$ and $\hat{v}$ be unit vectors inclined at an acute angle such that $|\hat{u} \times \hat{v}|=\frac{\sqrt{3}}{2}$. If $\overrightarrow{\mathrm{A}}=\lambda \hat{u}+\hat{v}+(\hat{u} \times \hat{v})$, then $\lambda$ is equal to:

A

$$ \frac{4}{3}(\overrightarrow{\mathrm{~A}} \cdot \hat{u})-\frac{2}{3}(\overrightarrow{\mathrm{~A}} \cdot \hat{v}) $$

B

$$ \frac{2}{3}(\overrightarrow{\mathrm{~A}} \cdot \hat{u})-\frac{1}{3}(\overrightarrow{\mathrm{~A}} \cdot \hat{v}) $$

C

$$ \frac{4}{3}(\overrightarrow{\mathrm{~A}} \cdot \hat{u})+\frac{2}{3}(\overrightarrow{\mathrm{~A}} \cdot \hat{v}) $$

D

$$ (\overrightarrow{\mathrm{A}} \cdot \hat{u})-\frac{1}{2}(\overrightarrow{\mathrm{~A}} \cdot \hat{v}) $$

3
JEE Main 2026 (Online) 2nd April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let the vectors $\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$. For some $\lambda, \mu \in \mathbb{R}$, let $\vec{c} = \lambda \vec{a} + \mu \vec{b}$.

If $\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10$ and $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2$, then $|\vec{c}|^2$ is equal to :

A

8

B

12

C

14

D

15

4
JEE Main 2026 (Online) 2nd April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Two adjacent sides of a parallelogram PQRS are given by $\overrightarrow{PQ} = \hat{j} + \hat{k}$ and $\overrightarrow{PS} = \hat{i} - \hat{j}$. If the side PS is rotated about the point P by an acute angle $\alpha$ in the plane of the parallelogram so that it becomes perpendicular to the side PQ, then $\sin^2\left(\frac{5\alpha}{2}\right) - \sin^2\left(\frac{\alpha}{2}\right)$ is equal to:

A

$ \frac{1}{2} $

B

$ \frac{\sqrt{3}}{2} $

C

$ \frac{\sqrt{3}}{4} $

D

$ \frac{2\sqrt{3}}{5} $

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