this post was submitted on 25 May 2024
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  • Show that cosθ=(u⋅v)/(|u||v|) for 2D vectors u and v.

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[–] [email protected] 4 points 5 months ago* (last edited 5 months ago) (1 children)

A good exercise to dust off trig. Hopefully no typos involving u, v, 1, 2, α, and β.

Trigonometric solution

By definition of |u| and Pythagoras, we can see that u1 and u2 can represent the lengths the legs of a right triangle with |u| being the length of the hypotenuse. Similarly for v. The triangles can be visualized in the usual way by plotting the vectors in the XY plane centered at the origin, with u1 and v1 constituting the X coordinates and u2 and v2 the Y coordinates of u and v, respectively.

Let α and β be the respective angles that u and v make with the X-axis. Then, θ = β - α, and

u1 = |u| cos α

u2 = |u| sin α

v1 = |v| cos β

v2 = |v| sin β

Now consider the expression:

|u| |v| cos θ

= |u| |v| cos (β - α)

= |u| |v| (cos β cos α + sin β sin α) (cosine angle difference formula)

= (|u| |v| cos β cos α) + (|u| |v| sin β sin α) (distribute)

= (|u| cos α)(|v| cos β) + (|u| sin α)(|v| sin β) (commutativity, associativity)

= (u1)(v1) + (u2)(v2) (substituting)

= uv (definition)

[–] [email protected] 2 points 5 months ago (1 children)
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[–] [email protected] 2 points 5 months ago