Example 7

Find the plane through the three points P(-1, 3, 1), Q(1, 2, 3), S(-1, -1, 0).

The plane has position vector

P = -i + 3j + k

and the two direction vectors

C = Q - P = 2i - j + 2k,

D = S - P = -4j - k.

A normal vector of the plane is

N = C × D = ((-1)(-1) - 2(-4))i + (2 · 0 - 2(-1))j + (2(-4) - (-1) · 0)k = 9i + 2j - 8k.

Then

N · P = 9(-1) + 2 · 3 + (-8) · l = -11.

The plane has the vector equation (9i + 2j - 8k) · X = -11 and the scalar equation 9x + 2y - 8z = -11. The plane is shown in Figure 10.5.8.

Figure 10.5.8