Question

If $\overrightarrow{a}=\hat{i}+2\hat{j}+4\hat{k},\overrightarrow{b}=\hat{i}+\lambda \hat{j}+4\hat{k},\overrightarrow{c}=2\hat{i}+4\hat{j}+({\lambda }^2-1)\hat{k},$ are coplanar vectors and $\overrightarrow{a},\overrightarrow{c}$ are not parallel, then $|\overrightarrow{a}\times \overrightarrow{c}|$ is equal to

• A. $3\sqrt{5}$
• B. $25$
• C. $2\sqrt{5}$
• D. $5\sqrt{5}$

Answer 1

The correct option is D $5\sqrt{5}$

Given: $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are coplanar.
$\Rightarrow \left|\begin{array}{ccc}1& 2& 4\\ 1& \lambda & 4\\ 2& 4& {\lambda }^2-1\end{array}\right|=0$
$\Rightarrow 1\left(\lambda \right({\lambda }^2-1)-16)-2({\lambda }^2-1-8)+4(4-2\lambda )=0$
$\Rightarrow {\lambda }^3-2{\lambda }^2-9\lambda +18=0$
$\Rightarrow (\lambda -2)({\lambda }^2-9)=0$
$\Rightarrow \lambda =2$ or $\lambda =\pm 3$
But for $\lambda =\pm 3,$ $\overrightarrow{a}$ is parallel to $\overrightarrow{c}.$
So, $\lambda =2$
$\therefore \overrightarrow{a}\times \overrightarrow{c}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 4\\ 2& 4& 3\end{array}\right|$
$\Rightarrow \overrightarrow{a}\times \overrightarrow{c}=\hat{i}(6-16)-\hat{j}(3-8)+\hat{k}(4-4)$
$\Rightarrow \overrightarrow{a}\times \overrightarrow{c}=-10\hat{i}+5\hat{j}$
Now, $|\overrightarrow{a}\times \overrightarrow{c}|=\sqrt{100+25}=5\sqrt{5}$