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Determine the cosines of the angles of the triangle formed by (2,-1,1), (1,-3,-5), (3,-4,-4)

Consider the points (2,-1,1), \ (1,-3,-5), \ (3,-4,-4) in \mathbb{R}^3. Determine the cosines of the angles of the triangle whose vertices are on these points.


Let a denote the angle between (2,-1,1) and (1,-3,-5), let b denote the angle between (3,-4,-4) and (1,-3,-5) and let c denote the angle between (2,-1,1) and (3,-4,-4). Then we compute the cosines,

    \begin{align*}  \cos a &= \frac{(2,-1,1) \cdot (1,-3,-5)}{\sqrt{6} \sqrt{35}} = 0 \\[9pt]  \cos b &= \frac{(1,-3,-5)\cdot(3,-4,-4)}{\sqrt{35} \sqrt{41}} = \frac{35}{\sqrt{35}\sqrt{41}} = \sqrt{\frac{35}{41}} \\[9pt]  \cos c &= \frac{(2,-1,1) \cdot (3,-4,-4)}{\sqrt{6} \sqrt{41}} = \frac{6}{\sqrt{6} \sqrt{41}} = \sqrt{\frac{6}{41}}. \end{align*}

One comment

  1. Rafael Deiga says:

    That reasoning is wrong. Let A, B and C the given vectors. So, what we want is the angle between the vectors B-C and A-C, B-A and C-A, C-B and A-B.

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