Question

Find the vector $\overline{r}$ perpendicular to $\overline{a}=\hat{i}-2\hat{j}+5\hat{k}$ and $\overline{b}=2\hat{i}+3\hat{j}-\hat{k}$ and $\overline{r}.(2\hat{i}+\hat{j}+\hat{k})+8=0$

• A. $-13\hat{i}+11\hat{j}+7\hat{k}$
• B. $13\hat{i}-11\hat{j}-7\hat{k}$
• C. $11\hat{i}+13\hat{j}+9\hat{k}$
• D. $7\hat{i}+13\hat{j}+11\hat{k}$

Answer 1

The correct option is A $-13\hat{i}+11\hat{j}+7\hat{k}$

Since $\overline{r}$ is $\perp$ to both $\overline{a}$ and $\overline{b}$
$\therefore$$\overline{r}=\lambda (\overline{a}\times \overline{b})$

$=\lambda \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 5\\ 2& 3& -1\end{array}\right|$
$=(-13\hat{i}+11\hat{j}+7\hat{k})\lambda$

Now, $\overline{r}.(2\hat{i}+\hat{j}+\hat{k})+8=0$

$\Rightarrow$ $-8\lambda +8=0$ $\Rightarrow \lambda =1$

$\therefore$ $\overline{r}=-13\hat{i}+11\hat{j}+7\hat{k}$