Question

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

• A. Zero
• B. One
• C. Two
• D. Three

Answer 1

The correct option is C Two

We know that three vectors are coplanar if their scalar triple product is zero.

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

$R_1\to R_1+R_2+R_3$
$\Rightarrow \left|\begin{array}{ccc}2-{\lambda }^2& 2-{\lambda }^2& 2-{\lambda }^2\\ 1& -{\lambda }^2& 1\\ 1& 1& -{\lambda }^2\end{array}\right|=0$
or $(2-{\lambda }^2)\left|\begin{array}{ccc}1& 1& 1\\ 1& -{\lambda }^2& 1\\ 1& 1& -{\lambda }^2\end{array}\right|=0$

$(R_2\to R_2-R_1,R_3\to R_3-R_1)$

or $(2-{\lambda }^2)\left|\begin{array}{ccc}1& 1& 1\\ 0& -(1+{\lambda }^2)& 0\\ 0& 0& -(1+{\lambda }^2)\end{array}\right|=0$

or $(2-{\lambda }^2)(1+{\lambda }^2{)}^2=0\Rightarrow \lambda =\pm \sqrt{2}$

Hence, two real solutions.