Question

Let $\overrightarrow{a}=\hat{i}+2\hat{j}+4\hat{k},\overrightarrow{b}=\hat{i}+\lambda \hat{j}+4\hat{k}$ and $\overrightarrow{c}=2\hat{i}+4\hat{j}+({\lambda }^2-1)\hat{k}$ be coplanar vectors. Then the non-zero vector $\overrightarrow{a}\times \overrightarrow{c}$ is :

• A. $-10\hat{i}-5\hat{j}$
• B. $-14\hat{i}-5\hat{j}$
• C. $-10\hat{i}+5\hat{j}$
• D. $-14\hat{i}+5\hat{j}$

Answer 1

The correct option is C $-10\hat{i}+5\hat{j}$

Since the vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are coplanar
$\therefore \left|\begin{array}{ccc}1& 2& 4\\ 1& \lambda & 4\\ 2& 4& ({\lambda }^2-1)\end{array}\right|=0$
$\Rightarrow \left(\lambda \right({\lambda }^2-1)-16)-2({\lambda }^2-1-8)+4(4-2\lambda )=0$
$\Rightarrow {\lambda }^3-2{\lambda }^2-9\lambda +18=0$
$\Rightarrow {\lambda }^2(\lambda -2)-9(\lambda -2)=0$
$\Rightarrow (\lambda -3)(\lambda +3)(\lambda -2)=0$
$\Rightarrow \lambda =2,3,-3$
So, $\lambda =2$ (as $\overrightarrow{a}$ is || to $\overrightarrow{c}$ for $\lambda =\pm 3$)
$\therefore \overrightarrow{a}\times \overrightarrow{c}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 4\\ 2& 4& 3\end{array}\right|=-10\hat{i}+5\hat{j}$