Question

If $\mathbf{a} = (1, - 1,1)$ and $\mathbf{c} = ( - 1, - 1,0),$ then the vector b satisfying $\mathbf{a} \times \mathbf{b} = \mathbf{c}$ and $\mathbf{a}.\mathbf{b} = 1$ is

• A. (1, 0, 0)
• B. (0, 0, 1)
• C. (0, –1, 0)
• D. None of these

Answer 1

Answer: (B)

(0, 0, 1)

Answer 2

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

But $(\mathbf{i} - \mathbf{j} + \mathbf{k}).(b_{1}\mathbf{i} + b_{2}\mathbf{j} + b_{3}\mathbf{k}) = 1 \Rightarrow b_{1} - b_{2} + b_{3} = 1$......(i)

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

$= - \mathbf{i}(b_{2} + b_{3}) + \mathbf{j}(b_{1} - b_{3}) + \mathbf{k}(b_{2} + b_{1})$

$\Rightarrow \mathbf{a} \times \mathbf{b} = \mathbf{c}$

Comparing the coefficients of $\mathbf{i},\mathbf{j}$ and $\mathbf{k}$ respectively,

we get $b_{2} + b_{3} = 1$ …..(ii)

$b_{1} - b_{3} = - 1$ …..(iii)

$b_{2} + b_{1} = 0$ …..(iv)

By solving the equations (i), (ii), (iii) and (iv), we get $b_{1} = 0,$

$b_{2} = 0$ and $b_{3} = 1$.