Question

Three vectors $\overrightarrow{A},\overrightarrow{B}\&\overrightarrow{C}$ are given as $\overrightarrow{A}=2\hat{i}+3\hat{j}+4\hat{k},\overrightarrow{B}=\hat{i}+2\hat{j}+2\hat{k}$ and $\overrightarrow{C}=m\hat{i}+2\hat{j}+\hat{k}$. Find the value of $m$, for which these vectors are coplanar.

• A. $1$
• B. $2$
• C. $0.5$
• D. $0.25$

Answer 1

The correct option is C $0.5$

If these vectors are coplanar, then scalar triple product of these vector must equal to zero.
$\therefore \left[\overrightarrow{A}\overrightarrow{B}\overrightarrow{C}\right]=(\overrightarrow{A}\times \overrightarrow{B}).\overrightarrow{C}=\mathrm{0.....}\left(i\right)$
$\overrightarrow{A}\times \overrightarrow{B}=\left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 1& 2& 2\end{array}\right]$

$\overrightarrow{A}\times \overrightarrow{B}=\hat{i}[6-8]-\hat{j}[4-4]+\hat{k}[4-3]$
$\therefore \overrightarrow{A}\times \overrightarrow{B}=-2\hat{i}+\hat{k}$
Now taking dot product with $\overrightarrow{C}$:
$(\overrightarrow{A}\times \overrightarrow{B}).\overrightarrow{C}=(-2\hat{i}+\hat{k}).(m\hat{i}+2\hat{j}+\hat{k})$
$\Rightarrow (\overrightarrow{A}\times \overrightarrow{B}).\overrightarrow{C}=-2m+1$
From Eq. $i$:
$(\overrightarrow{A}\times \overrightarrow{B}).\overrightarrow{C}=0$
$\Rightarrow -2m+1=0$
$\therefore m=\frac{1}{2}$