Question

If $\overline{A}=2\hat{i}+2\hat{j}+4\hat{k},\overline{B}=-\hat{i}+2\hat{j}+\hat{k}$ and $\overline{C}=3\hat{i}+\hat{j}$ then $\overline{A}+t\overline{B}$ is perpendicular to $\overline{C}$ if $t$ is equal to

• A. $8$
• B. $4$
• C. $6$
• D. $2$

Answer 1

The correct option is A $8$

Given $\overline{A}=2\hat{i}+2\hat{j}+4\hat{k},\overline{B}=-\hat{i}+2\hat{j}+\hat{k}$ and $\overline{C}=3\hat{i}+\hat{j}$ and $\overline{A}+t\overline{B}$ is perpendicular to $\overline{C}$.
$\Rightarrow (\overline{A}+t\overline{B})\cdot \overline{C}=0$
$\Rightarrow \left(\right(2\hat{i}+2\hat{j}+4\hat{k})+t(-\hat{i}+2\hat{j}+\hat{k}\left)\right)\cdot (3\hat{i}+\hat{j})=0$
$\Rightarrow \left(\right(2-t)\hat{i}+(2+2t)\hat{j}+(4+t\left)\hat{k}\right)\cdot (3\hat{i}+\hat{j})=0$
$\Rightarrow (2-t)3+(2+2t)=0$
$\therefore t=8$
Hence, option A.