Question

If the vector $\overrightarrow{OP}=\hat{i}+2\hat{j}+2\hat{k}$ rotates through a right angle about origin, passing through the positive $x-$axis on the way becomes $\overrightarrow{OQ}=x\hat{i}+y\hat{j}+z\hat{k}$, then the value of $x-y+z$ is

• A. $2\sqrt{2}$
• B. $\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{3}{\sqrt{2}}$

Answer 1

The correct option is A $2\sqrt{2}$

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


$\left|\overrightarrow{OP}\right|=\left|\overrightarrow{OQ}\right|\Rightarrow x^2+y^2+z^2=9\cdots \left(1\right)$

As $\overrightarrow{OP}$ is perpendicular to $\overrightarrow{OQ}$, so
$\left|\overrightarrow{OQ}\right|.\left|\overrightarrow{OP}\right|=0\Rightarrow x+2y+2z=0\cdots \left(2\right)$

$\cos \theta =\frac{1}{\left|\overrightarrow{OP}\right|}\Rightarrow \cos \theta =\frac{1}{\sqrt{1^2+2^2+2^2}}=\frac{1}{3}\cos (\frac{\pi }{2}-\theta )=\frac{x}{\left|\overrightarrow{OQ}\right|}\Rightarrow \sin \theta =\frac{x}{\left|\overrightarrow{OP}\right|}\Rightarrow \frac{2\sqrt{2}}{3}=\frac{x}{3}\Rightarrow x=2\sqrt{2}$

From equation $\left(1\right)$ and $\left(2\right)$, we get
$8+y^2+z^2=9\Rightarrow y^2+z^2=1\cdots \left(4\right)2\sqrt{2}+2y+2z=0\Rightarrow y+z+\sqrt{2}=0\cdots \left(5\right)$

From equation $\left(4\right)$ and $\left(5\right)$, we get
$y^2+(-\sqrt{2}-y{)}^2=1\Rightarrow 2y^2+2\sqrt{2}y+1=0\Rightarrow (\sqrt{2}y+1{)}^2=0\Rightarrow y=-\frac{1}{\sqrt{2}}\Rightarrow z=-\frac{1}{\sqrt{2}}\therefore x-y+z=2\sqrt{2}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=2\sqrt{2}$