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 09-25-2012, 02:59 AM #1 New Member   Join Date: Feb 2012 Posts: 20 how to find the angle Two vectors*A*and*B*have precisely equal magnitudes. In order for the magnitude of*A*+*B*to be*90*times larger than the magnitude ofA-*B, what must be the angle between them? Reply
 09-25-2012, 11:57 AM #2 Senior Member   Join Date: Mar 2012 Location: Cambridge, MA Posts: 1,913 Let θ be the angle between A and B. By the law of cosines, |A+B|^2 = |A|^2 + |B|^2 - 2|A||B| cos θ |A-B|^2 = |A|^2 + |B|^2 - 2|A||B| cos (π-θ) (I replaced |-B| with |B|). The expressions in the first equation are 90^2 = 8100 times larger than the expressions in the second equation. Also note that cos θ = -cos (π-θ). |A|^2 + |B|^2 - 2|A||B| cos θ = 8100(|A|^2 + |B|^2 + 2|A||B| cos θ) Replace |B| with |A| and factor: 2|A|^2 (1 - cos θ) = 8100 (2|A|^2)(1 + cos θ) Cancel out 2|A|^2 leaving 1 - cos θ = 8100(1 + cos θ) --> 8101 cos θ = -8099 --> cos θ = -8099/8101. To find θ just take arccos of both sides. --------- The problem can also be solved by drawing a right triangle with sides 1, 90, sqrt(8101), drawing the median to the hypotenuse, and using area formulas, but it's a little tricky explaining without a diagram. Reply

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