Question

Find the acute angle the line $\frac{x}{1}=\frac{y}{3}=\frac{z}{0}$ and plane $2x+y=5$.

Answer 1

We know that, angle between the line $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ and the plane $(lx+my+nz+p=0$ is,

$\sin \theta =\frac{al+bm+cn}{\sqrt{l^2+m^2+n^2}\sqrt{a^2+b^2+c^2}}$

Here,

Equation of line: $\frac{x-0}{1}=\frac{y-0}{3}=\frac{z-0}{0}$

Equation of plane: $2x+y=5$

$\Rightarrow \sin \theta =\frac{1\left(2\right)+3\left(1\right)+0\left(0\right)}{\sqrt{1^2+3^2+0^2}\sqrt{2^2+1^2+0^2}}$

$\Rightarrow \sin \theta =\frac{2+3}{\sqrt{10\times 5}}$

$\Rightarrow \sin \theta =\frac{5}{5\sqrt{2}}$

$\Rightarrow \sin \theta =\frac{1}{\sqrt{2}}$

$\Rightarrow \sin \theta =\sin {45}^{\circ }$

$\Rightarrow \theta ={45}^o=\frac{\pi }{4}$

Hence, which is the required answer.