Question

Parameterize the plane that contains the three points (−3,1,−2), (−6,−10,−4) and (15,5,20).

Answer 1

First find any two vectors in the plane, and take their cross product. The cross product will be perpendicular to both of them.

To find any vectors in the plane, take any two of the given points at a time, and treating them as vectors, subtract them. These vector differences are parallel to the plane we want to find.

(-3, 1, -2) - (-6, -10, -4) = (3, 11, 2)

(-3, 1, -2) - (15, 5, 20) = (-18, -4, -22)

Take the cross product to get the normal vector to the plane:

(3, 11, 2) x (-18, -4, -22) = (-234, 30, 186)

Let (x, y, z) be any point on the plane. The vector

(x, y, z) - (-3, 1, -2) = (x + 3, y - 1, z + 2)

runs parallel to the plane, so it's perpendicular to the plane's normal, which means their dot product is 0. This gives us the Cartesian equation for the plane,

(-234, 30, 186) • (x + 3, y - 1, z + 2) = 0

-234(x + 3) + 30(y - 1) + 186(z + 2) = 0

-234x + 30y + 186 = 360

Every coefficient has a GCD of 6, so the equation is equivalent to

-39x + 5y + 31z = 60

Finally - and this is the easiest step - write this in parametric form. Let x = s and y = t, then

z = (60 + 39s - 5t)/31

So the plane can be parameterized by

P(s, t) = (s, t, (60 + 39s - 5t)/31)

where s and t are any real numbers.