Question

The vectors $\overrightarrow{a}=\lambda \hat{i}+\hat{j}+2\hat{k},\overrightarrow{b}=\hat{i}+\lambda \hat{j}-\hat{k}$ and $\overrightarrow{c}=2\hat{i}-\hat{j}+\lambda \hat{k},(\lambda \in Z)$ are coplanar, then the value of $\lambda$ is

• A. $0$
• B. $1$
• C. $-2$
• D. $-1$

Answer 1

The correct option is C $-2$

$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are coplanar.
$\Rightarrow \overrightarrow{a}\cdot (\overrightarrow{b}\times \overrightarrow{c})=0$

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

$\Rightarrow {\lambda }^3-6\lambda -4=0\Rightarrow (\lambda +2)({\lambda }^2-2\lambda -2)=0$
$\Rightarrow \lambda =-2,1\pm \sqrt{3}$