(14) Finding a normal vector
Here's the plane again with s and t being known vectors which
lie in it.
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.