Plane determined by point and normal vector
Given a point P, there are many planes that contain P. However, assuming that we are living in three-dimensional space (R3), a plane is uniquely determined if we also specify a normal vector n (i.e., a vector that is perpendicular to the plane).
Here’s a CVT that illustrates this fact. Given the point P (shown in red), you can change the normal vector n (shown in green) by dragging the ball at its end. See how the plane changes. You can also drag the point P around to change the plane. (I’ve drawn vector n in a nonstandard way with a ball rather than an arrowhead at its endpoint; see an earlier reading for a discussion on this.)
Note that the plane doesn’t care about the length of n, only its direction. In fact, if you change n to point in the opposite direction, you still get the same plane back. Of course, changing P to any other point on the plane, wouldn’t change the plane either. (However, in this CVT, dragging the red ball around changes n unless you move both green and red balls by the same amount.)
We can infer the equation for the plane from these properties. Let a be the vector representing the point P (i.e., the vector from the origin to P). How can we know if a point Q represented by the vector x = (x,y,z) is on the plane? The following CVT may help answer the question. Here, you can drag also drag a point Q (in yellow) that is constrained to be in the plane. The vector from P to Q is x - a, and is also drawn in yellow. (Sometimes the vector is partially obscured by the plane.)
If the point represented by x is in the plane (as it is in the above CVT), the vector x - a must be parallel to the plane, hence perpendicular to the normal vector n. Two vectors are perpendicular if their dot product is zero. We conclude that for any point represented by x that is in the plane, the following equation must be satisfied:
| n ⋅ (x - a) = 0. | (1) |
Plane determined by three points
We just determined that to write the equation for a plane, we want a point P in the plane and a normal vector n. But most of us know that three points determine a plane (as long as they aren’t collinear, i.e., lie in straight line). Here is a plane determined by the points represented by the vectors P, Q, and R, which are drawn by the three dots whose colors match the letters. You can drag the points around to change the plane.
Since a plane is given by a point (say P) and normal vector, somehow the addition of the points Q, and R must determine the normal vector, which is shown in cyan. The hint is that I’ve also drawn the vector from P to Q (in green) and the vector from P to R (in blue). Both of these vectors are parallel to the plane. Hence, the normal vector of the plane must be perpendicular to both vectors. One way to obtain a vector perpendicular to two vectors is take their cross product. The cyan normal vector is indeed the cross product of the vector from P to Q and the vector from P to R.
In summary, if you are given three points, you can take the cross product of the vectors between two pairs of points to determine a normal vector n. Pick one of the three points, and let a be the vector representing that point. Then equation (1) is the equation for the plane going through the three points.
In the exercises, you get to try other ways of finding the equation for the plane, all which boil down to finding a and n. It’s something worth becoming proficient at.
If you don’t want to wait for the exercise, but want to see some examples now, follow this link.
