Example 10: Plane Containing a Line Perpendicular to Another Plane

Find the plane p containing the line X = i + t(j + k) which is perpendicular to the plane x + 3y - 22 = 0.

The given plane q has the normal vector M = i + 3j - 2k, and the given line L has the direction vector D = j + k.

The required plane p must have a normal vector N which is perpendicular to both M and D, so we take

N = 5i - j + k.

The vector P = i is a position vector of L and therefore a position vector of p. So p has the vector equation

N · X = N · P,

(5i - j + k) · X = 5,

and the scalar equation

5x - y + z = 5

(see Figure 10.5.12).

Figure 10.5.12