##
(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** = 2**i** + **j** + 3**k** and
**t** = 3**i** + 2**j** + 4**k**.

Then **s** x **t** = - 2**i** + **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.