Consider the points $u = (1,1,−1)$ , $v = (a,2,−1)$ and $w = (1,2,−1)$ in $R^3$, where $a \in R$. There are three possible values of $a$ for which $u, v$ and $w$ will form an isosceles triangle.

How do we find these values of a and hence how can we find the angle between the equal sides of the triangle?

  • $\begingroup$ Once you have a value of $a$ and know which two sides are equal, the usual approach is to bisect their included angle, which reduces the calculation to trigonometry on either of the resulting right triangles. $\endgroup$ – hardmath Mar 25 at 18:29
  • $\begingroup$ @hardmath How does one compute the value for a? $\endgroup$ – bigfocalchord Mar 25 at 18:51
  • $\begingroup$ I'll post an Answer if no one beats me to it, but note all three points $u,v,w$ have the same $z$-coordinate, so this is about a plane triangle (and only the $x,y$ coordinates matter). $\endgroup$ – hardmath Mar 25 at 18:55
  • $\begingroup$ Have you tried drawing a picture (in the plane $z=-1$)? $\endgroup$ – Andreas Blass Mar 25 at 19:28

We have two cases:

  1. $$|u-v|=|u-w|,$$ which gives $$(a-1)^2+1=1$$ or $a=1$, which is impossible, otherwise $v=w$;
  2. $$|v-w|=|u-w|,$$ which gives $$(a-1)^2=1,$$ which gives two values of $a$.
  • $\begingroup$ And from there, how does one find the angle between the equal sides of the triangle? $\endgroup$ – bigfocalchord Mar 25 at 19:11
  • $\begingroup$ @bigfocalchord You can use $\cos\measuredangle(\vec{a},\vec{b})=\frac{\vec{a}\vec{b}}{|\vec{a}||\vec{b}|}.$ $\endgroup$ – Michael Rozenberg Mar 25 at 19:13

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