Continuing with some homework, we're going to use vector algebra to prove two geometric theorems.
[This is a similar diagram to the one from last time, but with labels switched around--sorry for any confusion].
We have vectors (
c) and the lengths of the corresponding sides (
|a| = a, etc.); also, the angle opposite side a is labeled A and so on.
The law of sines states that the ratio of the length of each side to the sine of the angle opposite is the same:
a/sin(A) = b/sin(B) = c/sin(C)
Recall that the area of the parallelogram formed by
bis given by the absolute value of the cross-product:
|a X b| = a b sin(C)
And the area of the triangle is one-half that. But we must obtain the same area no matter which two vectors we use to compute the cross-product, and no matter which orientation. Thus:
|c X -a| = a c sin(B)
|-b X c| = b c sin(A)
also have the same area.
a b sin(C) = a c sin(B) = b c sin(A)
This leads directly to the law of sines. Now let's relabel the triangle slightly.
b - alooks a little funny, just consider that
a + b - a = b
The law of cosines states that:
c2 = a2 + b2 - 2 a b cos(C)
c = |b - a|.
We can obtain this simply by expanding the dot product:
(b - a) • (b - a) =
= b • b - b • a - a • b + a • a
= b2 + a2 - 2 a b cos(C)
(b - a) • (b - a) = c2
c2 = b2 + a2 - 2 a b cos(C)
Using vectors makes it easy.