1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $a$.
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.