Generalized Pythagoras through Vectors

Here’s a proof of the Pythagorean Theorem by way of vectors.  Of course, if your students already know vectors, they’re already way past the Pythagorean Theorem, but I thought Richard Pennington‘s statement of this on LinkedIn gave a pretty and stunningly brief (after all the definitions) proof of one of mathematics’ greatest equations.

Let O be the origin, and let \overrightarrow A and \overrightarrow B be two position vectors starting at O. The vector from \overrightarrow A to \overrightarrow B is simply \overrightarrow {B-A}, which I will call \overrightarrow C. Using properties of dot products,

\overrightarrow C\cdot\overrightarrow C = \overrightarrow {B-A}\cdot\overrightarrow {B-A} = \overrightarrow B\cdot\overrightarrow B-2\overrightarrow A\cdot\overrightarrow B+\overrightarrow A\cdot\overrightarrow A

The dot product of a vector with itself is the square of its magnitude, so

|\overrightarrow C|^2=|\overrightarrow B|^2-2|\overrightarrow A||\overrightarrow B|cos\theta+|\overrightarrow A|^2

where \theta is the angle between \overrightarrow A and \overrightarrow B.

This is the Law of Cosines–in my classes, I call it the generalized Pythagorean Theorem for all triangles.  If \theta=\frac{\pi}{2}, then \overrightarrow A and \overrightarrow B are the legs of a right triangle with hypotenuse \overrightarrow C which makes cos(\theta)=cos(\frac{\pi}{2})=0 and

|\overrightarrow C|^2=|\overrightarrow A|^2+|\overrightarrow B|^2

Pretty.

 

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