Question

Let O be the origin. Let

\(\overrightarrow{OP}=x\hat{i}+y\hat{j}-\hat{k}\) and \(\overrightarrow{OQ}=-\hat{i}+2\hat{j}+3x\hat{k}\), \(x,y\in\mathbb{R}\), \(x\gt 0\), be such that \(\left|\overrightarrow{PQ}\right|=\sqrt{20}\) and the vector \(\overrightarrow{OP}\) is perpendicular to \(\overrightarrow{OQ}\). If \(\overrightarrow{OR}=3\hat{i}+z\hat{j}-7\hat{k}\), \(z\in\mathbb{R}\), is coplanar with \(\overrightarrow{OP}\) and \(\overrightarrow{OQ}\), then the value of \(x^2+y^2+z^2\) is equal to :

• A. 2
• B. 9
• C. 7
• D. 1

Hint:

Use \(\overrightarrow{OP}\cdot\overrightarrow{OQ}=0\) and \(|\overrightarrow{PQ}|^2=20\) to find \(x,y\); then use the zero scalar triple product for coplanarity to get \(z\).

Answer 1

The Correct Option is B

\(\overrightarrow{OP}\cdot\overrightarrow{OQ}=-x+2y-3x=0\Rightarrow y=2x\).
\(\overrightarrow{PQ}=(-1-x,\,2-y,\,3x+1)\), so \(|\overrightarrow{PQ}|^2=(x+1)^2+(2-2x)^2+(3x+1)^2=14x^2+6=20\Rightarrow x^2=1\).
Since \(x\gt 0\): \(x=1,\ y=2\). So \(\overrightarrow{OP}=(1,2,-1)\), \(\overrightarrow{OQ}=(-1,2,3)\).
Coplanarity: \(\begin{vmatrix}1&2&-1\\-1&2&3\\3&z&-7\end{vmatrix}=0\).
Expanding: \(1(-14-3z)-2(7-9)+(-1)(-z-6)=-14-3z+4+z+6=-4-2z=0\Rightarrow z=-2\).
\(x^2+y^2+z^2=1+4+4=9\).
Answer: (2) 9