Question

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

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

Answer 1

The correct option is C

Two

Explanation for the correct option:

Find the value of $\lambda$.

Three vectors $-{\lambda }^{2}i+j+k$, $i-{\lambda }^{2}j+k$ and $i+j-{\lambda }^{2}k$ are given.

Since, the given three vectors are coplanar.

Therefore,

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

Therefore, the number of distinct real values of $\lambda$ for which the given vectors are coplanar is $2$.

Hence, option $C$ is the correct answer.