Mathematics Courses

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AMATH 301 — Beginning Scientific Computing — Winter 2017

https://www.youtube.com/watch?v=RYKyhUglVZA

Learn Differential Equations: Up Close with Gilbert Strang and Cleve Moler

A New Post

Dot Products, Norms, and Angles Between Vectors.

Recall that the Law of Cosines, a generalization on the Pythagorean Theorem, gives us the relationship between the side lengths of an arbitrary triangle. Specifically, if a triangle has side lengths aa, bb, and cc, then

$a^2 + b^2 - 2ab\cos \theta = c^2$

where θ is the angle between the sides of length aa and bb.

Consider the triangle that can be formed from the vectors xx, yy, and x−yx−y.

Applying the Law of Cosines to this triangle, we have $||x||^2 +||y||^2 - 2||x|| \, ||y||\cos \theta = ||x-y||^2$

But this implies, using our observations about the dot product made above, that

$\begin{array}{rcl} (x \cdot x) + (y \cdot y) - 2||x|| \, ||y||\cos \theta & = & (x-y) \cdot (x-y)\\ & = & x \cdot (x-y) - y \cdot (x-y)\\ & = & (x \cdot x) - (x \cdot y) - (y \cdot x) + (y \cdot y)\\ & = & (x \cdot x) - (x \cdot y) - (x \cdot y) + (y \cdot y)\\ & = & (x \cdot x) - 2(x \cdot y) + (y \cdot y)\\ \end{array}$

Subtracting the common (x⋅x)(x⋅x) and (y⋅y)(y⋅y) from both sides, we find −2||x||||y||cosθ=−2(x⋅y)

Which, solving for cosθcos⁡θ tells us

$\cos \theta = \frac{x \cdot y}{||x|| \, ||y||}$