Question

Let $\vec{u}$ be a vector coplanar with the vectors $\vec{a}=2\hat{i}+3\hat{j}-\hat{k}$ and $\vec{b}=\hat{j}+\hat{k}$. If $\vec{u}$ is perpendicular to $\vec{a}$ and $\vec{u}.\vec{b}=24,$ then $|\vec{u}{|}^2$ is equal to:

• A. 336
• B. 315
• C. 256
• D. 84

Answer 1

Answer: (A)

Clearly, $\vec{u}=\lambda (\vec{a}\times (\vec{a}\times \vec{b}))$
$\Rightarrow \vec{u}=\lambda \left((\vec{a}\cdot \vec{b})\vec{a}-|\vec{a}{|}^2\vec{b}\right)$
$\Rightarrow \vec{u}=\lambda (2\vec{a}-14\vec{b})=2\lambda \{(2\hat{i}+3\hat{j}-\hat{k})-7(\hat{j}+\hat{k})\}$
$\Rightarrow \vec{u}=2\lambda (2\hat{i}-4\hat{j}-8\hat{k})$
as, $\vec{u}\cdot \vec{b}=24$
$\Rightarrow 4\lambda (\hat{i}-2\hat{j}-4\hat{k})\cdot (\hat{j}+\hat{k})=24$
$\Rightarrow \lambda =-1$
So, $\vec{u}=-4(\hat{i}-2\hat{j}-4\hat{k})$
$\Rightarrow |\vec{u}{|}^2=336$