Question

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

• A. 01
• B. 1.00
• C. 1
• D. 1.0

Answer 1

Answer: (A), (B), (C), (D)

Line $\overrightarrow{r}=\hat{i}+2\hat{j}+\lambda (4\hat{i}-3\hat{k})$
Vector along the direction of line $=4\hat{i}-3\hat{k}$
Normal unit vector of $xy-$plane = $\hat{k}$
Angle between line and plane is
$\sin \alpha =\left|\frac{(4\hat{i}-3\hat{k})\cdot \hat{k}}{\sqrt{4^2+3^2}\cdot 1}\right|=\frac{3}{5}$
$\Rightarrow \cos \alpha =\frac{4}{5}$

Angle between planes $x+2y=0$ and $2x+y=0$ is
$\cos \beta =\frac{(\hat{i}+2\hat{j})\cdot (2\hat{i}+\hat{j})}{\sqrt{5}\sqrt{5}}=\frac{2+2}{5}=\frac{4}{5}$
$\Rightarrow \sin \beta =\frac{3}{5}$
Now, ${\cos }^2\alpha +{\sin }^2\beta =\frac{4^2}{5^2}+\frac{3^2}{5^2}=\frac{25}{25}=1$