Adrian McMenamin

Vector dot product

This is another one of my “reminder” posts, so look away now if you are familiar with vector dot products.

If we had a two dimensional vector X and vector Y then the dot product would be (X.x)(Y.y) + (X.y)(Y.y) .

But this can also be expressed as \mid X \mid \mid Y \mid cos \theta (where \theta is the angle between the two vectors).

Let Q be the angle X makes to the x-axis and P the angle made by Y , then X.x = \mid X \mid cos Q, X.y = \mid X \mid sin Q, Y.x = \mid Y \mid cos P and Y.y = \mid Y \mid sin P .

Then the dot product can be rewritten as:

\mid X \mid cos Q \mid Y \mid cos P + \mid X \mid sin Q \mid Y \mid sin P

= \mid X \mid \mid Y \mid (cos P cos Q + sin P sin Q)

As P = \theta + Q we can rewrite this as:

\mid X \mid \mid Y \mid (cos(\theta + Q)cos Q + sin(\theta + Q) sin Q)

Now sin(a + b) = sin(a) cos( b )+ cos (a) sin( b) and cos (a + b) = cos( a) cos (b )- sin (a )sin (b)

so we can rewrite the dot product as \mid X \mid \mid Y \mid (cos^2 Q cos \theta - cos Q sin \theta sin Q + sin^2 Q cos \theta + sin Q cos Q sin \theta)

As sin^2 \alpha + cos ^2 \alpha = 1 we have \mid X \mid \mid Y \mid cos \theta .

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