Question

Let $\overrightarrow{a}=\hat{i}-2\hat{j}+3\hat{k}$, $\overrightarrow{b}=2\hat{i}+3\hat{j}-\hat{k}$ and $\overrightarrow{c}=\lambda \hat{i}+\hat{j}+(2\lambda -1)\hat{k}$. If $\overrightarrow{c}$ is parallel to the plane of the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ then $\lambda$ is

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

Answer: (B)

$0$

In other words the three vectors are coplanar.

$\Rightarrow \left|\begin{array}{ccc}1& -2& 3\\ 2& 3& -1\\ \lambda & 1& 2\lambda -1\end{array}\right|=0$

$\Rightarrow 1(6\lambda -3+1)+2(4\lambda -2+\lambda )+3(2-3\lambda )=0$

$\Rightarrow 6\lambda -2+10\lambda -4+6-9\lambda =0$

$\Rightarrow \lambda =0$