(14) Finding a normal vector

Here's the plane again with s and t being known vectors which lie in it.
finding the position vector of the point P
Working out the cross product of s and t will automatically give us a vector perpendicular to the plane in which s and t lie. This is the speediest method for finding a normal vector to this plane.
As an example, suppose s = 2i + j + 3k and t = 3i + 2j + 4k.
Then s x t = - 2i + j + k = N, a normal vector to the plane. (The working out of s x t is described at the end of the cross product section where I used the same vectors in an example there.)

It's also possible to find a normal vector by using the dot product. If we call the normal vector N, then N must be perpendicular to both s and t.

Therefore N.s = 0 and N.t = 0.
Now, using algebra it is possible to find an N which fits these two equations.
The next section uses these sections on lines, planes and normal vectors.

How to find angles between lines and planes

or back to the previous section.

back to the vectors homepage.