Let $\vec{a}=-\hat{i}-\hat{j}+\hat{k}, \vec{a} \cdot \vec{b}=1$ and $\vec{a} \times \vec{b}=\hat{i}-\hat{j}$. Then $\overrightarrow{\mathrm{a}}-6 \overrightarrow{\mathrm{~b}}$ is equal to

(A) $3(\hat{i}-\hat{j}-\hat{k})$

(C) $3(\hat{i}-\hat{j}+\hat{k})$

(D) $3(\hat{i}+\hat{j}-\hat{k})$

JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Vector Algebra

Let

a = - \hat{i} - \hat{j} + \hat{k}, a \cdot b = 1

and

a \times b = \hat{i} - \hat{j}

. Then

a - 6 b

is equal to

Answer: (B)

With

a = - \hat{i} - \hat{j} + \hat{k}

, the identity

a \times (a \times b) = (a \cdot b) a - (a \cdot a) b

is the quickest route. Here

a \cdot a = 3

and

a \cdot b = 1

, while

a \times (a \times b) = a \times (\hat{i} - \hat{j}) = \hat{i} + \hat{j} + 2 \hat{k}

. Hence

a - 3 b = \hat{i} + \hat{j} + 2 \hat{k}

. Writing

a - 6 b = 2 (a - 3 b) - a

gives

2 (\hat{i} + \hat{j} + 2 \hat{k}) - (- \hat{i} - \hat{j} + \hat{k}) = 3 \hat{i} + 3 \hat{j} + 3 \hat{k}

, so

a - 6 b = 3 (\hat{i} + \hat{j} + \hat{k})

JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Single correct option