Question

Let $O$ be the origin. Let $OP=xi+yj–k$ and $OQ=–i+2j+3xk,x,y\in R,x>0$, be such that $\left|PQ\right|=\sqrt{20}$ and the vector $OP$ is perpendicular to $OQ$. If $OR=3i+zk–7k,z\in R$, is coplanar with OP and OQ, then the value of $x^2+y^2+z^2$ is equal to:

• A. $2$
• B. $9$
• C. $1$
• D. $7$

Answer 1

The correct option is B

$9$

Explanation for the correct option:

Step 1: Finding the dot product

Given, $OP=xi+yj-k$ , $OQ=-i+2j+3xk$

$\left|PQ\right|=\sqrt{20}$ and also $OP\perp OQ$

Now dot product of two perpendicular vectors is equal to zero, hence

$$\begin{gathered} \left(xi+yj-k\right)\left(-i+2j+3k\right)=0 \\ \Rightarrow -x+2y-3x=0 \\ \Rightarrow -4x+2y=0 \\ \Rightarrow 4x=2y \\ \Rightarrow y=2x\dots \dots \left(1\right) \end{gathered}$$

Step 2: Solving for $\mathrm{x}$ and $\mathrm{y}$

Now we know that the position vector can be calculated by using the method shown below,

$$\begin{gathered} PQ=OQ-OP \\ \Rightarrow PQ=\left(-1-x\right)i+\left(2-y\right)j+\left(3x+1\right)k \\ \Rightarrow \left|PQ\right|=\sqrt{{\left(-1-x\right)}^2+{\left(2-y\right)}^2+{\left(3x+1\right)}^2} \\ \Rightarrow \left|PQ\right|=\sqrt{{\left(x+1\right)}^2+{\left(y-2\right)}^2+{\left(3x+1\right)}^2} \\ \Rightarrow \sqrt{20}=\sqrt{{\left(x+1\right)}^2+{\left(y-2\right)}^2+{\left(3x+1\right)}^2} \end{gathered}$$

Squaring on both sides we get,

$$\begin{gathered} 20=x^2+2x+1+y^2+4-4y+9x^2+1+6x \\ \Rightarrow 20=x^2+2x+1+4x^2+4-8x+9x^2+1+6x\left(\because y=2x\right) \\ \Rightarrow 20=14x^2+6 \\ \Rightarrow 14x^2=14 \\ \Rightarrow x^2=1 \\ \Rightarrow x=1\left(\because x>0\right) \end{gathered}$$

Therefore from equation $\left(1\right)$ we have,

$$\begin{gathered} y=2x \\ \Rightarrow y=2\times 1 \\ \Rightarrow y=2 \end{gathered}$$

Step 3: Solving for $\mathrm{z}$

When three vectors are coplanar then their determinant is equal to zero, hence, putting the coefficients of $OP,OQ,OR$ respectively in between two vertical lines, we get,

$$\begin{gathered} \left|\begin{array}{ccc}1& 2& -1\\ -1& 2& 3\\ 3& z& -7\end{array}\right|=0 \\ \Rightarrow 1\left(-14-3z\right)-2\left(7-9\right)-1\left(-z-6\right)=0 \\ \Rightarrow -14-3z+4+z+6=0 \\ \Rightarrow 2z=-14+10 \\ \Rightarrow z=\frac{-4}{2} \\ \Rightarrow z=-2 \end{gathered}$$

Step 4: Calculating $x^2+y^2+z^2$

Therefore the value of an expression is equal to,

$$\begin{gathered} x^2+y^2+z^2=1^2+2^2+{\left(-2\right)}^2 \\ =1+4+4 \\ =9 \end{gathered}$$

Hence, option (B) is the correct answer.