A unit vector perpendicular to both the vectors 4i-j+3k and -2i+j-2k is: Option 1: 1/3(i+2j+2k) Option 2: (-1)/3(i+2j+2k) Option 3: 1/3(-i+2j+2k) Option 4: 1/3(-i-2j+2k)

Answer: 1/3(-i+2j+2k). Solution: The cross product is (4,-1,3)x(-2,1,-2)=(-1,2,2). Its magnitude is √(1+4+4)=3. Dividing gives the unit vector 13(- i+2 j+2 k).

Algebra

Vector Algebra

Easy

core

A unit vector perpendicular to both the vectors $4\hat{i}-\hat{j}+3\hat{k}$ and $-2\hat{i}+\hat{j}-2\hat{k}$ is:

$\frac{1}{3}(\hat{i}+2\hat{j}+2\hat{k})$

$\frac{-1}{3}(\hat{i}+2\hat{j}+2\hat{k})$

$\frac{1}{3}(-\hat{i}+2\hat{j}+2\hat{k})$

$\frac{1}{3}(-\hat{i}-2\hat{j}+2\hat{k})$

✅ Correct Option: 3

The cross product is $(4,-1,3)\times(-2,1,-2)=(-1,2,2)$ .

Its magnitude is $\sqrt{1+4+4}=3$ .

Dividing gives the unit vector $\frac13(-\hat i+2\hat j+2\hat k)$ .

2025: 14 May Shift 2

Match List-I with List-II

Consider two vectors $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = -3\hat{i} - 6\hat{j} + 3\hat{k}$ , then

List-I	List-II
(A) Angle between $\vec{a}$ and $\vec{b}$ is	(I) $\cos^{-1}\left(\frac{1}{\sqrt{6}}\right)$
(B) Angle between $\vec{a}$ and $x$ -axis is	(II) $\cos^{-1}\left(\frac{2}{\sqrt{6}}\right)$
(C) Angle between $\vec{b}$ and $x$ -axis is	(III) $\pi$
(D) Angle between $\vec{a}$ and $y$ -axis is	(IV) $\cos^{-1}\left(-\frac{1}{\sqrt{6}}\right)$

Choose the correct answer from the options given below:

2026: 31 May Shift 2

If $|\vec{a}| = \sqrt{26}, |\vec{b}| = 7, |\vec{a}\times\vec{b}| = 35$ and $\theta$ be the angle between $\vec{a}$ and $\vec{b}$ then

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	$\sin\theta$	I.	$61$
B.	$\cos\theta$	II.	$5/\sqrt{26}$
C.	$\vec{a} \cdot \vec{b}$	III.	$1/\sqrt{26}$
D.	$\vert \vec{a} - \vec{b} \vert^2$	IV.	$7$

Choose the correct answer from the options given below:

2026: 18 May Shift 2

If $P(\vec{r})$ is a position vector of a point with scalar components $(x,y,z)$ , then the value of $(\vec{r}.\hat{i})(-\hat{i})-(\vec{r}.\hat{j})\hat{j}+(\vec{r}.(-\hat{k}))\hat{k}$ , is:

2022: 25 May Shift 1

Match List I with List II

LIST I	LIST II
A. The area of parallelogram determined by vectors $2\hat{i}$ and $3\hat{j}$	I. 2
B. The value of $(\hat{i} \times \hat{j}) \cdot \hat{k} + (\hat{j} \times \hat{k}) \cdot \hat{i}$	II. 4
C. The value of a for which the vectors $2\hat{i} - 3\hat{j} + 4\hat{k}$ and $a\hat{i} - 6\hat{j} + 8\hat{k}$ are collinear.	III. 0
D. The value of $\lambda$ for which the vectors $2\hat{i} + \hat{j} + \hat{k}$ and $2\hat{i} - 4\hat{j} + \lambda\hat{k}$ are perpendicular	IV. 6

Choose the correct answer from the options given below:

2022: 16 July Shift 2

If $|\vec{a}| = |\vec{b}| = |\vec{a} + \vec{b}| = 1$ , then $|\vec{a} - \vec{b}|$ is equal to:

2026: 22 May Shift 1

If $|\vec a|=2, |\vec b|=7$ and $\vec a\times\vec b=3\hat i+2\hat j+6\hat k$ , then angle between $\vec a$ and $\vec b$ is:

A unit vector perpendicular to both the vectors 4i^−j^+3k^4\hat{i}-\hat{j}+3\hat{k}4i^−j^​+3k^ and −2i^+j^−2k^-2\hat{i}+\hat{j}-2\hat{k}−2i^+j^​−2k^ is:

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

A unit vector perpendicular to both the vectors $4\hat{i}-\hat{j}+3\hat{k}$ and $-2\hat{i}+\hat{j}-2\hat{k}$ is: