9. The scalar product





If one (or both) vectors are zero vectors then the above formula is obvious. Hence assume that u and v are nonzero. 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 = uv 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 = u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3} as claimed. Appendix: The law of cosinesThe law of cosines asserts that in an arbitrary triangle a^{2} = b^{2} + c^{2}  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 b^{2} cos ^{2} and rearranging we get a^{2} = b^{2} + c^{2}  2bc cos as required.


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