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 $