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

Answer 1

The correct option is D $336$

$\overrightarrow{u}={\lambda }_1\overrightarrow{a}+{\lambda }_2\overrightarrow{b}$
$\Rightarrow \overrightarrow{u}\cdot \overrightarrow{a}={\lambda }_1(\overrightarrow{a}\cdot \overrightarrow{a})+{\lambda }_2(\overrightarrow{a}\cdot \overrightarrow{b})$
$\Rightarrow 0={\lambda }_1(4+9+1)+{\lambda }_2(3-1)$
$\Rightarrow 0=14{\lambda }_1+2{\lambda }_2$
$\Rightarrow {\lambda }_2=-7{\lambda }_1$
$\therefore \overrightarrow{u}={\lambda }_1\overrightarrow{a}-7{\lambda }_1\overrightarrow{b}$
$\Rightarrow \overrightarrow{u}\cdot \overrightarrow{b}={\lambda }_1(\overrightarrow{a}\cdot \overrightarrow{b})-7{\lambda }_1(\overrightarrow{b}\cdot \overrightarrow{b})$
$\Rightarrow 24=2{\lambda }_1-14{\lambda }_1$
$\Rightarrow {\lambda }_1=-2$ and ${\lambda }_2=14$
Now,
$\overrightarrow{u}=-2(2\hat{i}+3\hat{j}-\hat{k})+14(\hat{j}+\hat{k})$
$\Rightarrow \overrightarrow{u}=-4\hat{i}+8\hat{j}+16\hat{k}$
$\therefore |\overrightarrow{u}{|}^2=336$