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 vectors $\overrightarrow{c}$ lies in the plane of $\overline{a}$ and $\overline{b}$, then $x$ equals

• A. $-4$
• B. $-2$
• C. $0$
• D. $1$

Answer 1

The correct option is B $-2$

Given

$\overline{a}=\hat{i}+\hat{j}+\hat{k},\overline{b}=\hat{i}+\hat{j}+2\hat{k},$ and $\overline{c}=x\hat{i}+(x-2)\hat{j}-\hat{k},\left[\overline{a}\overline{b}\overline{c}\right]=0$

Since Vectors lies in a same plane

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

$\Rightarrow 1(1-2x+4)-1(-1-2x)+1(x-2+x)=0$
$\Rightarrow 5-2x+1+2x+2x-2=0$
$\Rightarrow x=-2$