Example 6

Find the plane with position vector P = k and direction vectors C = -2i + j + k. D = -j. First we find a normal vector of the plane,

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

Then

N · P = 1 · 0 + 0 · 0 + 2 · 1 = 2.

The plane has the vector equation

(i + 2k) · X = 2

and the scalar equation

x + 2z = 2.

The plane is shown in Figure 10.5.7.

Figure 10.5.7