Forming plane examples

Example 1

Find the equation for the plane through the point (0, 1,-7) perpendicular to the vector (4,-1, 6).

Solution: Let a = (0, 1,-7). Let n = (4,-1, 6). Then, for x = (x,y,z), the equation for the plane is

n ⋅ (x - a) = 0.
This becomes
(4,-1, 6) ⋅ (x - 0,y - 1,z + 7) = 0
or
4x - (y - 1) + 6(z + 7) = 0.
Often, we prefer to write this as
4x - y + 6z + 43 = 0.
(Note that you can read the normal vector n = (4,-1, 6) right from the equation for the plane; the components of n are simply the coefficients of x, y, and z.)

Example 2

Find the equation for the plane through the points (0, 1,-7), (3, 1,-9), and (0,-5,-8).

Solution: Let b = (0, 1,-7)-(3, 1,-9) = (-3, 0, 2). Let c = (0, 1,-7)-(0-5,-8) = (0, 6, 1). Then, a normal vector is

n = b × c
= || i j k || || || | - 3 0 2 | | 0 6 1 |
= i(0 - 12) - j(-3 - 0) + k(-18 - 0)
= (-12, 3,-18).
We’ll pick the first point and let a = (0, 1,-7). The equation for the plane becomes
(-12, 3,-18) ⋅ (x - 0,y - 1,z + 7) = 0
which we rewrite as
- 12x + 3(y - 1) - 18(z + 7) = 0
or
- 12x + 3y - 18z - 129 = 0.

I hope you noticed that the planes from both examples when through the same point (0, 1,-7). Did you also notice that their normal vectors are parallel to each other? If you multiply the n from Example 1 by -3, you obtain the n from Example 2:

- 3(4,-1, 6) = (-12, 3,-18).

What does that mean about the relationship between the two planes? (This was discussed in the pre-lecture reading.) The two planes must be equal. In fact, if you divide both sides of the equation of the second plane by -3, you get the equation of the first plane.

Example 3

Find the equation for the plane through the points (1, 2, 3), (2, 4, 6), and (-3,-6,-9).

Solution: Let b = (2, 4, 6)-(1, 2, 3) = (1, 2, 3) and let c = (1, 2, 3)-(-3,-6,-9) = (4, 8, 12). Then, we attempt to find a normal vector as their cross product:

n = b × c
= || i j k || || || |1 2 3 | |4 8 12 |
= i(24 - 24) - j(12 - 12) + k(8 - 8)
= (0, 0, 0).

Hmm, something went wrong, as we weren’t able to find a normal direction. Do the three points actually determine a plane? If you are still puzzled, go back to the pre-lecture reading, and look for any conditions on the ability of three points to determine a plane.