MCQ
The angle between the line $\bar{r}=(2 \hat{i}+3 \hat{j}+\hat{k})+\lambda(\hat{i}+2 \hat{j}-\hat{k})$ and the plane $\bar{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=4$ is
  • A
    $\sin ^{-1}\left(\frac{1}{3}\right)$
  • $\sin ^{-1}\left(\frac{1}{6}\right)$
  • C
    $\frac{\pi}{3}$
  • D
    $\frac{\pi}{4}$

Answer

Correct option: B.
$\sin ^{-1}\left(\frac{1}{6}\right)$
(B)
Comparing the equations of line and plane with $\overline{ r }=\overline{ a }+\lambda \overline{ b }$ and $\overline{ r } . \overline{ n }= p$, we get $\bar{b}=\hat{i}+2 \hat{j}-\hat{k}$ and $\bar{n}=2 \hat{i}-\hat{j}+\hat{k}$
∴ The angle between the line and plane is
$\sin \theta=\left|\frac{\overline{ b } \cdot \overline{ n }}{|\overline{ b }| \cdot|\overline{ n }|}\right|=\left|\frac{1(2)+2(-1)-1(1)}{\sqrt{1+4+1} \sqrt{4+1+1}}\right|=\frac{1}{6}$
$\Rightarrow \theta=\sin ^{-1}\left(\frac{1}{6}\right)$

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