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Angles

# Line-Plane Angle

The angle between a line, r, and a plane, π, is the angle between r and its orthogonal projection onto π, r'.

That is, the angle between a line and a plane is equal to the complementary acute angle that forms between the direction vector of the line and the normal vector of the plane.

Let's name θ to the angle which line makes with plane, u=(u1,u2,u3) the direction vector of line and n=(A,B,C) the normal vector of the plane.

Determine the angle between the line:

$r\;\equiv\;\frac{x-1}{2}=\frac{y+1}{1}=\frac{z}{2}\;$

and the plane $\pi\;\equiv\;x+y-1=0\;$

Solution:

u=(2,1,2) and n=(1,1,0)

Therefore

θ=45º

Determine the angle between the line

 x+3y-z+3=0 2x-y-z-1=0

and the plane $\pi\;\equiv\;2x-y+3z+1=\;0\;$

Solution:

u=
 i j k 1 3 -1 2 -1 -1
=-4i-j-7k $\Rightarrow$ u=(-4,-1,-7)

n=(2,-1,3)

θ=22.91º