Question

Let $\overrightarrow{u}$ be a vector co-planar with 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 $\overrightarrow{u}.\overrightarrow{b}=24$, then $|\overrightarrow{u}{|}^2$ is equal to.

Answer 1

$\overline{u}$ is coplanar with $\overline{a}$ and $\overline{b}$

$\therefore \overline{u}=k\overline{a}+\lambda \overline{b}$

$\overline{u}=2k\overline{i}+(3k+\lambda )\overline{j}+(\lambda -k)\overline{k}$

$\overline{u}$ is $\perp$ to $\overline{a}$

$\therefore \overline{u}.\overline{a}=0$

$\therefore 2k\times 2+(3k+\lambda )\times 3+(\lambda -k)\times -1=0$

$\therefore 4k+9k+3\lambda +k-\lambda =0$

$\therefore 14k+2\lambda =0$

$\therefore \lambda =-7k$

$\overline{u}.\overline{b}=24$

$\therefore (3k+\lambda )+(\lambda -k)=24$

but $\lambda =-7k$,

$\therefore 2k-14k=24$

$\therefore k=-2$

$\lambda =14$

$\overline{u}=-4\hat{i}+8\hat{j}+16\hat{k}$

$|\overline{u}{|}^2=4^2+8^2+{16}^2=336$