Question

If the number of distinct real values of $\lambda$ for which the vectors $-{\lambda }^2\hat{i}+\hat{j}+\hat{k}$, $\hat{i}-{\lambda }^2\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}-{\lambda }^2\hat{k}$ are coplanar is '$n$' then $n$

• A. $0$
• B. $2$
• C. $1$
• D. $3$

Answer 1

The correct option is B $2$

$\left|\begin{array}{ccc}{\lambda }^2& 1& 1\\ 1& -{\lambda }^2& 1\\ 1& 1& -{\lambda }^2\end{array}\right|=0$ ( property of coplanarity )
$-{\lambda }^2({\lambda }^4-1)-1(-{\lambda }^2-1)+1(1+{\lambda }^2)$
$-{\lambda }^6+{\lambda }^2+{\lambda }^2+{\lambda }^2+1=0$
${\lambda }^6-3{\lambda }^2-2=0$
$({\lambda }^2+1)({\lambda }^4-{\lambda }^2-2)=0$
$({\lambda }^2+1)\left(\right({\lambda }^2-2\left)\right({\lambda }^2+1\left)\right)=0$
$\lambda =\underset{–}{+}\sqrt{2}$
So $n=2$