Question

Let $\vec{a}=2 \hat{i}-\hat{j}+5 \hat{k}$ and $\vec{b}=\alpha \hat{i}+\beta \hat{j}+2 \hat{k}$. If $((\vec{a} \times \vec{b}) \times \hat{i}) \cdot \hat{k}=\frac{23}{2}$, then $|\vec{b} \times 2 \hat{j}|$ is equal to

• A. 4
• B. 5
• C. $\sqrt{21}$
• D. $\sqrt{17}$

Answer 1

Answer: (B)

$ \vec{a}=2 \hat{i}-\hat{j}+5 \hat{k}, \vec{b}=\alpha \hat{i}+\beta \hat{j}+2 \hat{k} $
$ ((\vec{a} \times \vec{b}) \times \hat{i}) \cdot \hat{k}=\frac{23}{2}, \text { then }|\vec{b} \times 2 \hat{j}| \text { is }$
$ ((\vec{a} \cdot \hat{i}) \vec{b}-(\vec{b} \cdot \hat{i}) \vec{a}) \cdot \hat{k}=\frac{23}{2}$
$ (\vec{a} \cdot \hat{i})(\vec{b} \cdot \hat{i})-(\vec{b} \cdot \hat{i})(\vec{a} \cdot \hat{k})=\frac{23}{2}$
$2 \times 2-\alpha \times 5=\frac{23}{2} \Rightarrow 5 \alpha=4-\frac{23}{2} \Rightarrow \alpha=\frac{-3}{2} $
$ \vec{b} \times 2 \hat{j}= \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 0 & \beta & 2 \\ 0 & 2 & 0\end{vmatrix}$
$=-4 \hat{i}+2 \alpha \hat{k}$
$ \therefore|\vec{b} \times 2 \hat{j}|=\sqrt{16+4 \alpha^2}=\sqrt{16+4 \times \frac{9}{4}}=5$