Vectors and planes

**pantera** · May 26th 2008, 08:57 AM

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!

**Soroban** · May 26th 2008, 09:20 AM

Hello, pantera!

A ray of light coming from the point (-1, 3, 2) is travelling in direction $\langle 4, 1,-2\rangle$
and meets the plane $x + 3y + 2z - 24\:=\: 0$
Find the angle that the ray of light makes with the plane.

The angle between two vectors $\vec{u}\text{ and }\vec{v}$ is given by: . $\cos\theta \:=\:\frac{|\vec{u}\cdot \vec{v}|}{|\vec{u}||\vec{v}|}$

The ray has direction: . $\vec{u} \:=\:\langle4,1,-2\rangle$
The plane has normal direction $\vec{n} \:=\:\langle 1, 3, 2\rangle$

The angle $\theta$ between the ray and the normal is:
. . $\cos\theta \:=\:\frac{\langle 4,1,-2\rangle\cdot\langle1,3,2\rangle}{\sqrt{4^2+1^2+(-2)^2}\sqrt{1^2+3^2+2^2}} \;=\;\frac{3}{\sqrt{294}} \:=\:0.174963553$

Hemce, $\theta \;=\;79.92346289 \;\approx\;80^o$

Therefore, the angle between the ray and the plane is about: . $90^o - 80^o \;=\;\boxed{10^o}$

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