Practicing Success
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 |
-4 -2 0 1 |
-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 $3x-x+2+2=0$ $2x=-4$ $x=-2$ |