Find an equation of the plane. The plane through the points (2, 1, 2), (3, -8, 6), and (-2, -3, 1)?

Solution:

Given three points (2,1, 2), (3, -8, 6) and (-2, -3, 1)

Vector equation of a plane passing through three points is

(\(\overrightarrow r - \overrightarrow a).{(\overrightarrow b - \overrightarrow a) × (\overrightarrow c - \overrightarrow a)} =0\)

 A(2, 1, 2) = 2i + j + 2k = a

 B(3, -8, 6) = 3i - 8j + 6k = b

 C(-2, -3, 1) = -2i - 3j + k = c

 (b - a) = (3i - 8j + 6k) - (2i + j + 2k)

 = i - 9j + 4k

 (c - a) = (-2i - 3j + k) - (2i + j + 2k)

 = -4i - 4j- k

Finding the cross product of the vectors, we get

 {(b - a) × (c - a)} = 25i -15j + 40k

 (r - a) = (xi + yj+ zk) - (2i + j + 2k)

= (x - 2)i + (y - 1)j + (z - 2)k

Finding the dot product of the vectors, we get

 (r - a). {(b - a) × (c - a)} = {(x - 2)i + (y - 1)j + (z - 2)k}.{25i -15j + 40k}

= 25(x - 2) - 15(y - 1) + 40(z - 2)

= 25x - 50 - 15y + 15 + 40z - 80

= 25x - 15y + 40z - 115

Required equation 25x - 15y + 40z - 115 = 0

Find an equation of the plane. The plane through the points (2, 1, 2), (3, -8, 6), and (-2, -3, 1)?

Summary:

The equation of the plane. The plane through the points (2, 1, 2), (3, -8, 6), and (-2, -3, 1) is 25x - 15y + 40z - 115 = 0.