The University of Sydney - School of Mathematics and Statistics

Proof of the formula	Page 3 of 6

If one (or both) vectors are zero vectors then the above formula is obvious. Hence assume that u and v are non-zero.

The vectors u, v and u - v then form a triangle.

The edges of the above triangles have length |u|, |v| and |u - v|. By the law of cosines (see below)

By definition u · v = |u||v| cos , so

(1)

By the formula for the magnitude of a vector

Multiplying the right hand side out we get

If we substitute the above into (1 ) we get

so u · v = u₁v₁ + u₂v₂ + u₃v₃ as claimed.

Appendix: The law of cosines

The law of cosines asserts that in an arbitrary triangle a² = b² + c² - 2bc cos , where a, b, c and are as shown below.

The formula follows by applying the Theorem of Pythagoras to the triangles ADC and DBC. Note that the red line AD has length |b cos | and the green line DB has length c - b cos . Hence by the Theorem of Pythagoras

Multiplying the right hand side out we get

Adding b² cos ² and rearranging we get a² = b² + c² - 2bc cos as required.