<p>Two vectors<em>A</em>and<em>B</em>have precisely equal magnitudes. In order for the magnitude of<em>A</em>+<em>B</em>to be<em>90</em>times larger than the magnitude ofA-*B, what must be the angle between them?</p>
<p>Let θ be the angle between A and B. By the law of cosines,</p>
<p>|A+B|^2 = |A|^2 + |B|^2 - 2|A||B| cos θ</p>
<p>|A-B|^2 = |A|^2 + |B|^2 - 2|A||B| cos (π-θ) (I replaced |-B| with |B|).</p>
<p>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).</p>
<p>|A|^2 + |B|^2 - 2|A||B| cos θ = 8100(|A|^2 + |B|^2 + 2|A||B| cos θ)</p>
<p>Replace |B| with |A| and factor:</p>
<p>2|A|^2 (1 - cos θ) = 8100 (2|A|^2)(1 + cos θ)</p>
<p>Cancel out 2|A|^2 leaving</p>
<p>1 - cos θ = 8100(1 + cos θ) → 8101 cos θ = -8099 → cos θ = -8099/8101. To find θ just take arccos of both sides.</p>
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<p>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.</p>