Question

If the planes $\overrightarrow{r}.(2\hat{i}-\hat{j}+2\hat{k})=4$ and $\overrightarrow{r}.(3\hat{i}+2\hat{j}+\lambda \hat{k})=3$ are perpendicular, then $\lambda =$

• A. $2$
• B. $-2$
• C. $3$
• D. $-3$

Answer 1

Answer: (B)

$-2$

Since, the planes $\overrightarrow{r}.(2\hat{i}-\hat{j}+2\hat{k})=4$ & $\overrightarrow{r}.(3\hat{i}+2\hat{j}+\lambda \hat{k})=3$ are perpendicular to each other
Therefore, $(2i-j+2k).(3i+2j+\lambda k)=0$
$\Rightarrow 6-2+2\lambda =0$
$\Rightarrow \lambda =-2$
Ans: B