Question

If the vector $\overrightarrow{c}$ lies in the plane of $\overrightarrow{a}$ and $\overrightarrow{b}$ in which $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}$ then $x$ equals to:

• A. $0$
• B. $-2$
• C. $-1$
• D. $1$

Answer 1

The correct option is B $-2$

If $3$ vectors are co-planar, their scalar triple product is $0$.

Therefore, $\overrightarrow{a}.[\overrightarrow{b}\times \overrightarrow{c}]=0$

i.e

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

$\therefore 1[1-2(x-2\left)\right]-1[-1-2x]+1[x-2-(-x\left)\right]=0$

$\Rightarrow 1-2x+4+1+2x+x-2+x=0$

$\therefore x=-2$