Question

If angle between line $\overrightarrow{r}=\hat{i}+2\widehat{+}2\hat{k}+\lambda (4\hat{j}-3\hat{k})$ and $XY$ plane is $\alpha$, and angle between the planes $x+2y=0$ and $2x+y=0$ is $\beta$, then $\frac{{\cos }^2\alpha }{{\sin }^2\beta }=$

• A. $1$
• B. $\frac{9}{16}$
• C. $\frac{16}{9}$
• D. $\frac{9}{25}$

Answer 1

The correct option is C

$\frac{16}{9}$

The angle between the line $\overrightarrow{r}=\hat{i}+2\widehat{+}2\hat{k}+\lambda (4\hat{j}-3\hat{k})$ and the plane $XY$ can be found out as:

$\sin \alpha =\frac{(4\hat{j}-3\hat{k}).\hat{k}}{\sqrt{4^2+3^2}\times 1}=\frac{-3}{\sqrt{25}}=-\frac{3}{5}$

$\Rightarrow {\sin }^2\alpha =\frac{9}{25}$

$\Rightarrow 1-{\cos }^2\alpha =\frac{9}{25}$

$\Rightarrow {\cos }^2\alpha =\frac{16}{25}$

The angle between the planes $x+2y=0$ and $2x+y=0$ can be found out as:

$\cos \beta =\frac{1\times 2+2\times 1}{\sqrt{1^2+2^2}\times \sqrt{1^2+2^2}}=\frac{4}{\sqrt{5}\times \sqrt{5}}=\frac{4}{5}$

$\Rightarrow {\cos }^2\beta =\frac{16}{25}$

$\Rightarrow 1-{\sin }^2\beta =\frac{16}{25}$

$\Rightarrow {\sin }^2\beta =\frac{9}{25}$

$\Rightarrow \frac{{\cos }^2\alpha }{{\sin }^2\beta }=\frac{16/25}{9/25}=\frac{16}{25}\times \frac{25}{9}=\frac{16}{9}$