Target Exam

CUET

Subject

Section B1

Chapter

Vectors

Question:

Let $\vec{a}$ be any vector such that $|\vec{a}| = a$. The value of $|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2$ is

Options:

$a^2$

$2a^2$

$3a^2$

0

Correct Answer:

$2a^2$

Explanation:

The correct answer is Option (2) → $2a^2$ ##

Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$.

Given $|\vec{a}| = a$.

Then $\vec{a} \times \hat{i} = -a_2\hat{k} + a_3\hat{j} \quad [∵\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0]$

$\vec{a} \times \hat{j} = a_1\hat{k} - a_3\hat{i}$

$\vec{a} \times \hat{k} = -a_1\hat{j} + a_2\hat{i}$

$\text{and } ∴|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = a_2^2 + a_3^2 + a_1^2 + a_3^2 + a_1^2 + a_2^2$

$[∵|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}]$

$= 2(a_1^2 + a_2^2 + a_3^2)$

$= 2|\vec{a}|^2 = 2a^2$