# Prove the following

Question:

Let $\mathrm{O}$ be the origin. Let $\overrightarrow{\mathrm{OP}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and

$\overrightarrow{\mathrm{OQ}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \mathrm{x} \hat{\mathrm{k}}, \mathrm{x}, \mathrm{y} \in \mathrm{R}, \mathrm{x}>0$, be such that

$|\overrightarrow{\mathrm{PQ}}|=\sqrt{20}$ and the vector $\overrightarrow{\mathrm{OP}}$ is perpendicular to $\overrightarrow{\mathrm{OQ}}$. If

$\overrightarrow{\mathrm{OR}}=3 \hat{\mathrm{i}}+\mathrm{z} \hat{\mathrm{j}}-7 \hat{\mathrm{k}}, \mathrm{z} \in \mathrm{R}$, is coplanar with $\overrightarrow{\mathrm{OP}}$ and $\overrightarrow{\mathrm{OQ}}$, then the value

of $x^{2}+y^{2}+z^{2}$ is equal to

1. 7

2. 9

3. 2

4. 1

Correct Option: , 2

Solution:

$\overrightarrow{\mathrm{OP}} \perp \overrightarrow{\mathrm{OQ}}$

$\Rightarrow-x+2 y-3 x=0$

$\Rightarrow y=2 x$

$|\overrightarrow{\mathrm{PQ}}|^{2}=20$

$\Rightarrow(x+1)^{2}+(y-2)^{2}+(1+3 x)^{2}=20$

$\Rightarrow x=1$

$\overrightarrow{\mathrm{OP}}, \overrightarrow{\mathrm{OQ}}, \overrightarrow{\mathrm{OR}}$ are coplanar.

$\Rightarrow\left|\begin{array}{ccc}\mathrm{x} & \mathrm{y} & -1 \\ -1 & 2 & 3 \mathrm{x} \\ 3 & \mathrm{z} & -7\end{array}\right|=0$

$\Rightarrow\left|\begin{array}{ccc}1 & 2 & -1 \\ -1 & 2 & 3 \\ 3 & z & -7\end{array}\right|=0$

$\Rightarrow 1(-14-3 z)-2(7-9)-1(-z-6)=0$

$\Rightarrow \mathrm{z}=-2$

$\therefore x^{2}+y^{2}+z^{2}=1+4+4=9$