Question

The vectors $\lambda \hat{i}+\hat{j}+2\hat{k},\hat{i}+\lambda \hat{j}-\hat{k}\mathrm{and}2\hat{i}-\hat{j}+\lambda \hat{k}$ are coplanar, if λ =
(a) –2
(b) 0
(c) 1
(d) –1

Answer 1

$$\begin{gathered} \lambda \hat{i}+\hat{j}+2\hat{k},\hat{i}+\lambda \hat{j}-\hat{k}\mathrm{and}2\hat{i}-\hat{j}+\lambda \hat{k}\mathrm{are}\mathrm{coplanar}\mathrm{if} \\ \left|\begin{array}{ccc}\lambda & 1& 2\\ 1& \lambda & -1\\ 2& -1& \lambda \end{array}\right|=0 \\ \mathrm{i}.\mathrm{e}\lambda \left({\lambda }^2-1\right)-1\left(\lambda +2\right)+2\left(-1-2\lambda \right)=0 \\ \mathrm{i}.\mathrm{e}{\lambda }^3-\lambda -\lambda -2-2-4\lambda =0 \\ \mathrm{i}.\mathrm{e}{\lambda }^3-6\lambda -4=0 \\ \left\{\begin{array}{l}\mathrm{Since}\mathrm{for}\lambda =-2\\ {\left(-2\right)}^3-6\left(-2\right)-4\\ =-8+12-4=0\end{array}\right. \\ \mathrm{i}.\mathrm{e}\left(\lambda +2\right)\left({\lambda }^2-2\lambda -2\right)=0 \\ \mathrm{i}.\mathrm{e}\lambda =-2 \end{gathered}$$
Hence, the correct answer is option A.