Question

The angle between the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and the plane $3x+2y-3z=4$, is

• A. ${45}^o$
• B. $0^o$
• C. ${\cos }^{-1}\left(\frac{24}{\sqrt{29\times 22}}\right)$
• D. ${90}^o$

Answer 1

Answer: (B)

$0^o$

Line$:\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ has directions $(2,3,4)$

Plane$:3x+2y-3z=4$ has normal with direction ratios $(3,2,-3)$

$\therefore$ angle between plane and line be $\theta$ then angle between line and its direction will be $90-\theta$.

$\cos (90-\theta )=\frac{2\times 3+3\times 2+4\times -3}{\sqrt{(2^2+3^2+4^2)(3^2+2^2+3^2)}}=0\therefore 90-\theta =90\Rightarrow \theta =0^{\circ }$