Question

The number of distinct real values of p, for which the vectors $-\left(p^2\right)i+j+k,i-\left(p^2\right)j+k$ and $i+j-\left(p^2\right)k$ are coplanar, is

• A. zero
• B. one
• C. two
• D. three

Answer 1

The correct option is C two

Consider the problem

For three vectors are co-planar,

Then,

$\left|\begin{array}{ccc}-p^2& 1& 1\\ 1& -p^2& 1\\ 1& 1& -p^2\end{array}\right|=0$

On expanding along $R_1$

$-p^2(p^4-1)-(-p^2-1)+(1+p^2)=0$

Therefore, $(p^2+1{)}^2(2-p^2)=0$

Hence, $p=\pm \sqrt{2}$

So, there are two real values of $p$.