If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{a} \cdot \vec{b} = 1$ and $\vec{a} \times \vec{b} = \hat{j} - \hat{k}$, then find $|\vec{b}|$.

1

The correct answer is Option (2) → 1 ##

Given, $\vec{a} = \hat{i} + \hat{j} + \hat{k}$

$\vec{a} \cdot \vec{b} = 1$ and $\vec{a} \times \vec{b} = \hat{j} - \hat{k}$

Let $\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$

Now, $\vec{a} \cdot \vec{b} = 1$

$\Rightarrow (\hat{i} + \hat{j} + \hat{k}) \cdot (b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}) = 1$

$\Rightarrow b_1 + b_2 + b_3 = 1 \quad ...(i)$

and $\vec{a} \times \vec{b} = \hat{j} - \hat{k}$

$\Rightarrow \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ b_1 & b_2 & b_3 \end{vmatrix} = \hat{j} - \hat{k}$

$\Rightarrow \hat{i}(b_3 - b_2) - \hat{j}(b_3 - b_1) + \hat{k}(b_2 - b_1) = \hat{j} - \hat{k}$

On comparing both sides, we get

$-(b_3 - b_1) = 1$ and $b_2 - b_1 = -1$

$\Rightarrow b_3 - b_1 = -1$ and $b_2 - b_1 = -1$

$\Rightarrow b_3 = -1 + b_1$ and $b_2 = -1 + b_1 \quad...(ii)$

Now from Eq. (i), we get

$b_1 + (-1 + b_1) + (-1 + b_1) = 1$

$\Rightarrow 3b_1 = 3$

$\Rightarrow b_1 = 1$

From Eq. (ii), we get

$b_2 = 0$ and $b_3 = 0$

$∴\vec{b} = \hat{i}$

Therefore, $|\vec{b}| = 1$