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)

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