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

This becomes

or

Often, we prefer to write this as

(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 x c

$$\left\vert\vphantom{ \begin{array}{rrr} \mathbf{i} & \mathbf{j} & \mathbf{k} -3 & 0 & 2 0 & 6 & 1 \end{array} }\right. \begin{array}{rrr} \mathbf{i} & \mathbf{j} & \mathbf{k} -3 & 0 & 2 0 & 6 & 1 \end{array} \left.\vphantom{ \begin{array}{rrr} \mathbf{i} & \mathbf{j} & \mathbf{k} -3 & 0 & 2 0 & 6 & 1 \end{array} }\right\vert$$ = 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

which we rewrite as

or

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:

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 x c

$$\left\vert\vphantom{ \begin{array}{rrr} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & 2 & 3\\ 4 & 8 & 12 \end{array} }\right. \begin{array}{rrr} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & 2 & 3\\ 4 & 8 & 12 \end{array} \left.\vphantom{ \begin{array}{rrr} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & 2 & 3\\ 4 & 8 & 12 \end{array} }\right\vert$$ = 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.