trianglesvectors
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?
Let $ABC$ be a triangle with $\angle BCA=\angle ABC$. Then observe that $ABC\equiv ACB$ (the order of the vertices is important), because they satisfy $ASA$ criterion. Therefore, we have that $CA=AB$, i.e. $ABC$ is isosceles.
Construct equilateral triangle $ABX$ as in the picture.
Note that the concave angle $BXA$ equals $300^\circ = 2\cdot 150^\circ = 2\angle BPA$, hence $P$ lies on the circle with center $X$ and radius $XA=XB$. It follows that $\angle PXB = 2\angle PAB = 40^\circ$.
On the other hand, $AB=AX=AC$, so $A$ is the circumcenter of triangle $BXC$, hence $\angle CXB = \frac 12 \cdot \angle CAB = \frac 12 \cdot 80^\circ = 40^\circ$.
Since $\angle PXB = 40^\circ = \angle CXB$, points $X, P, C$ are collinear.
Since $AC=AX$, we have $\angle PCA = \angle AXP = 2\cdot \angle ABP = 20^\circ$, hence $$\angle APC = 180^\circ - \angle CAP - \angle PCA = 180^\circ - 60^\circ - 20^\circ = 100^\circ.$$
Best Answer
We have two cases: