Question

If $\theta$ denote the acute angle between the line
$\overrightarrow{r}=(\hat{i}+2\hat{j}-\hat{k})+\lambda (\hat{i}-\hat{j}+\hat{k})$ and the plane $\overrightarrow{r}.(2\hat{i}-\hat{j}+\hat{k})=4$, then $\sin \theta +\sqrt{2}\cos \theta =$

• A. $\frac{1}{\sqrt{2}}$
• B. 1
• C. $\sqrt{2}$
• D. $1+\sqrt{2}$

Answer 1

The correct option is C

$\sqrt{2}$

$\sin \theta =\frac{(\hat{i}-\hat{j}+\hat{k}).(2\hat{i}-\hat{j}+\hat{k})}{\sqrt{3}\sqrt{6}}$

$=\frac{4}{\sqrt{18}}=\frac{2\sqrt{2}}{3}$

$\cos \theta =\sqrt{1-\frac{8}{9}}=\frac{1}{3}$

$\Rightarrow \sin \theta +\sqrt{2}\cos \theta =\frac{2\sqrt{2}}{3}+\frac{\sqrt{2}}{3}=\sqrt{2}$