Intersection of a Line and Plane:
If equation of a plane is ax + by + cz + d = 0, then the direction cosines of normal to this plane are a, b, c. So the angle between normal to the plane and a straight line having direction cosines l, m ,n can be given by 
.
Then angle between plane and straight line is Π/2-θ.
- The plane and straight line will be parallel if al + bm + cn = 0
- The plane and straight line will be perpendicular if
.
Example: Find equation of plane passing through intersection of planes 2x - 4y + 3z + 5= 0, x + y + z = 6 and parallel to straight line with direction cosines (1, -1, -1).
Solution: The equation of required plane be
(2x - 4y + 3z + 5) + l(x + y - z - 6) = 0
i.e. (2 + λ)x + (-4 + λ)y + z(3 - λ) + (5 - 6λ) = 0
This plane is parallel to a straight line. So, aλ + bm + cn = 0
1(2 + λ) + (-1)(-4 + λ) + (-1)(3 - λ) = 0 i.e. λ = -3
∴ Equation of required plane is -x - 7y + 6z + 23 = 0
i.e. x + 7y - 6z - 23 = 0.
Bisector Planes of Angle between 2 Planes:
The equation of planes bisecting angles between 2 given planes a1x +b1y +c1z +d1 = 0 and a2x + b2y + c2z +d2 = 0 is
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