Please help. A ray of light coming from the point (-1, 3, 2) is travelling in the directionof the vector 4i + j -2k and meets the plane x + 3y + 2z - 24= 0. Find the angle that the ray of light makes with the plane.
Tks!
Hello, pantera!
The angle between two vectors $\displaystyle \vec{u}\text{ and }\vec{v}$ is given by: .$\displaystyle \cos\theta \:=\:\frac{|\vec{u}\cdot \vec{v}|}{|\vec{u}||\vec{v}|} $A ray of light coming from the point (-1, 3, 2) is travelling in direction $\displaystyle \langle 4, 1,-2\rangle$
and meets the plane $\displaystyle x + 3y + 2z - 24\:=\: 0$
Find the angle that the ray of light makes with the plane.
The ray has direction: .$\displaystyle \vec{u} \:=\:\langle4,1,-2\rangle$
The plane has normal direction $\displaystyle \vec{n} \:=\:\langle 1, 3, 2\rangle$
The angle $\displaystyle \theta$ between the ray and the normal is:
. . $\displaystyle \cos\theta \:=\:\frac{\langle 4,1,-2\rangle\cdot\langle1,3,2\rangle}{\sqrt{4^2+1^2+(-2)^2}\sqrt{1^2+3^2+2^2}} \;=\;\frac{3}{\sqrt{294}} \:=\:0.174963553 $
Hemce, $\displaystyle \theta \;=\;79.92346289 \;\approx\;80^o$
Therefore, the angle between the ray and the plane is about: .$\displaystyle 90^o - 80^o \;=\;\boxed{10^o}$