Question

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

• A. $256$
• B. $84$
• C. $336$
• D. $315$

Answer 1

The correct option is C $336$

$\overrightarrow{a}=2\hat{i}+3\hat{i}-\hat{k}$

$\overrightarrow{b}=\hat{j}+\hat{k}$

Let $\overrightarrow{u}=x\hat{i}+y\hat{j}+z\hat{k}$

$\overrightarrow{u}\cdot \overrightarrow{a}=0$

$\Rightarrow 2x+3y-z=0$ ..(1)

$\overrightarrow{u}\cdot \overrightarrow{b}=24$

$y+z=24$ ..(2)

$\overrightarrow{u}$ is coplaner with $\overrightarrow{a}$ & $\overrightarrow{b}$, therefore $\overrightarrow{u}\cdot (\overrightarrow{a}\times \overrightarrow{b})=0$

$4x-2y+2z=0$ ..(3)

Solving eq(1) and (3), we get

$8x+4y=0$ or $y=-2x$ ..(4)

From (4) and (1), we get

$z=-4x$ ..(5)

From (5) and (2), we get $x=-4,y=8,z=16$

$|\overrightarrow{u}{|}^2=x^2+y^2+z^2=336$