Question

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

• A. 0
• B. 1
• C. -4
• D. -2

Answer 1

The correct option is A 0

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ all lie in same plane, then $(\overrightarrow{a}\times \overrightarrow{b}).\overrightarrow{c}=0$

Now. $\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}i& i& k\\ 1& 1& 1\\ 1& -1& 2\end{array}\right|=3j-j-2k$

$(3i-j-2k).(x\overrightarrow{i}+(x-2)\hat{i}+\hat{k})=3x-x+2-2=0$

$\Rightarrow x=0$