Let $\vec{a}=\hat{j}-\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$. Then vector $\vec{b}$ satisfying $\vec{a} \times \vec{b}+\vec{c}=\overrightarrow{0}$ and $\vec{a} \cdot \vec{b}=3$ is

(D) $-\hat{i}+\hat{j}-2 \hat{k}$

(A) $2 \hat{i}-\hat{j}+2 \hat{k}$

(C) $\hat{i}+\hat{j}-2 \hat{k}$

JEE Main · 2010 · 11 Apr 2010 · Offline · Vector Algebra

Let

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

and

c = \hat{i} - \hat{j} - \hat{k}

. Then vector

b

satisfying

a \times b + c = 0

and

a \cdot b = 3

Answer: (D)

We need

b

with

a \times b = - c

and

a \cdot b = 3

, where

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

and

c = \hat{i} - \hat{j} - \hat{k}

. Writing

b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}

, the dot condition

a \cdot b = b_{2} - b_{3} = 3

, and the cross-product condition worked out component by component forces

b_{1} = - 1

together with

b_{1} = b_{2} + b_{3}

, so

b_{2} + b_{3} = - 1

and with

b_{2} - b_{3} = 3

we get

b_{2} = 1, b_{3} = - 2

. Thus

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

; checking,

a \cdot b = 1 + 2 = 3

and

a \times b = - \hat{i} + \hat{j} + \hat{k} = - c

, as required.

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option