Question

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

• A. $0$
• B. $1$
• C. $-4$
• D. $-2$

Answer 1

Answer: (D)

$-2$

Given $\overline{a}=\hat{i}+\hat{j}+\hat{k}$

$\overline{b}=\hat{i}-\hat{j}+2\hat{k}$

$\overline{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}$

Also $\overline{c}$ lies in the plane of $\overline{a}$ and $\overline{b}$
$\Rightarrow \overline{a},\overline{b}$ and $\overline{c}$ are coplanar.

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

$\Rightarrow 2(1+x)+2=0$
$\therefore x=-2$
Hence, option $D$