Example: Find the intersection point and the angle between the planes: 4x + z − 2 = 0 and the line
given in parametric form: x =− 1 − 2t y = 5 z = 1 + t
Solution: Because the intersection point is common to the line and plane we can substitute the line parametric points
into the plane equation to get:
4(− 1 − 2t) + (1 + t) − 2 = 0 
t = − 5/7 = 0.71 
Now we can substitute the value of t into the line parametric equation to get the intersection point.
x = − 1 − 2(− 5/7) = 3/7 = 0.43 
y = 5 
z = 1 − 5/7 = 2/7 = 0.29 
And the intersection point is: (0.43 , 5 , 0.29).
The angle between the line and the plane can be calculated by the cross product of the line vector with
the vector representation of the plane which is perpendicular to the plane: v = 4i + k
The line vector representation is the t portion of the parametric line equation: n = 2i + k
And the angle between the plane and the line is: θ = π/2 − α = π/2 − 40.6 = 49.4 degree
