Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+2\hat{k},$ and $\vec{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}.$ If the vector $\vec{C}$ lies in the plane of $\vec{a}$ and $\vec{b}$, then
Answer & explanation
Correct answer: option 2
The correct answer is option (2) → -2
$\vec v=\vec a×\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\1&1&1\\1&-1&2\end{vmatrix}$
$\vec v=3\hat i-\hat j-2\hat k$
so $\vec v.\vec c=0$ as c is in plane of $\vec a,\vec b$
so $3x-x+2+2=0$
$2x=-4$
$x=-2$