Question

If $\mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k},\mathbf{a}.\mathbf{b} = 1$ and$\mathbf{a} \times \mathbf{b} = \mathbf{j} - \mathbf{k}$, then $\mathbf{b} =$

• A. $\mathbf{i}$
• B. $\mathbf{i} - \mathbf{j} + \mathbf{k}$
• C. $2\mathbf{j} - \mathbf{k}$
• D. $2\mathbf{i}$

Answer 1

Answer: (A)

$\mathbf{i}$

Answer 2

Let $\mathbf{b} = b_{1}\mathbf{i} + b_{2}\mathbf{j} + b_{3}\mathbf{k}$

Now, $\mathbf{j} - \mathbf{k} = \mathbf{a} \times \mathbf{b} = \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ b_{1} & b_{2} & b_{3} \end{matrix} \right|$

⇒ $b_{3} - b_{2} = 0,b_{1} - b_{3} = 1,b_{2} - b_{1} = - 1$ ⇒ $b_{3} = b_{2},b_{1} = b_{2} + 1$

Now, $\mathbf{a}.\mathbf{b} = 1$ ⇒ $b_{1} + b_{2} + b_{3} = 1$ ⇒ $3b_{2} + 1 = 1$ ⇒ $b_{2} = 0$

⇒$b_{1} = 1,b_{3} = 0$. Thus $\mathbf{b} = \mathbf{i}$