Question

The number of distinct real values of $\lambda$ for which the vectors $-{\lambda }^2\overline{i}+\overline{j}+\overline{k}$, $\overline{i}-{\lambda }^2\overline{j}+\overline{k}$, $\overline{i}+\overline{j}-{\lambda }^2\overline{k}$ are coplanar is

• A. $2$
• B. $4$
• C. $5$
• D. $9$

Answer 1

Answer: (A)

$2$

$\begin{array}{c}\left|\begin{array}{ccc}-{\lambda }^2& 1& 1\\ 1& -{\lambda }^2& 1\\ 1& 1& -{\lambda }^2\end{array}\right|=0\\ \Rightarrow \left|\begin{array}{ccc}-{\lambda }^2& 1& 1\\ 1& -{\lambda }^2-1& 1\\ 1& 1+{\lambda }^2& -{\lambda }^2\end{array}\right|=0\\ \Rightarrow -{\lambda }^2\left[-{\lambda }^2\left(-{\lambda }^2-1\right)-1-{\lambda }^2\right]+1\left[1+{\lambda }^2+{\lambda }^2+1\right]=0\\ \Rightarrow -{\lambda }^2\left[{\lambda }^4+{\lambda }^2-1-{\lambda }^2\right]+2+2{\lambda }^2=0\\ \Rightarrow -{\lambda }^6-{\lambda }^4+{\lambda }^2+2+2{\lambda }^2=0\\ \Rightarrow -{\lambda }^6-3{\lambda }^2+2=0\\ \Rightarrow {\lambda }^6-3{\lambda }^2-2=0\\ y^3-3y-2=0\\ \Rightarrow \left(y+1\right)\left(y^2-y-2\right)=0\\ \Rightarrow \left(y+1\right)\left(y+1\right)\left(y-2\right)=0\\ y=-1,-1,2\\ \Rightarrow {\lambda }^2=-1,-1,2\\ \Rightarrow {\lambda }^2=2\\ \Rightarrow \lambda =\pm \sqrt{2}\\ Hencethereare2distinctrealvaluesof\lambda \end{array}$