Question

The sine of the angle between the straight line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ and the plane $2x-2y+z=5$ is

• A. $\frac{10}{6\sqrt{5}}$
• B. $\frac{4}{5\sqrt{2}}$
• C. $\frac{\sqrt{2}}{10}$
• D. $\frac{2\sqrt{3}}{5}$

Answer 1

The correct option is C $\frac{\sqrt{2}}{10}$

From eq. of line we find the direction vector

$\overrightarrow{s}=(3,4,5)(l,m,n)$

From eq. of plane we find the normal vector

$\overrightarrow{q}=(2,-2,1)(A,B,C)$

Using formula,

$\sin \theta =\frac{|Al+Bm+Cn|}{\sqrt{A^2+B^2+C^2}.\sqrt{l^2+m^2+n^2}}$

$\sin \theta =\frac{|6-8+5|}{\sqrt{4+4+1}.\sqrt{9+16+25}}$

$\sin \theta =\frac{3}{3.5\sqrt{2}}$

$\sin \theta =\frac{\sqrt{2}}{10}$