9. The scalar product
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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
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 cosinesThe 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.
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