Question

Find the direction cosines of the sides of the triangle whose vertices are $(3,5,-2),(-1,1,2)$ and $(-5,-5,-2)$. What type of triangle is it?

Answer 1

Direction ratios of :

$AB=(4,4,-4)BC=(8,10,0)CA=(4,6,4)$

Directions of

$AB=(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}})BC=(\frac{4}{\sqrt{41}},\frac{5}{\sqrt{41}},0)CA=(\frac{2}{\sqrt{17}},\frac{3}{\sqrt{17}},\frac{2}{\sqrt{17}})$

angle between,

$AB\&BC={\cos }^{-1}(\frac{1}{\sqrt{3}}.\frac{4}{\sqrt{41}}+\frac{1}{\sqrt{3}}.\frac{5}{\sqrt{41}})={\cos }^{-1}\left(\frac{9}{\sqrt{123}}\right)BC\&CA={\cos }^{-1}(\frac{4}{\sqrt{41}}.\frac{2}{\sqrt{17}}+\frac{5}{\sqrt{41}}.\frac{3}{\sqrt{17}})={\cos }^{-1}\left(\frac{23}{\sqrt{697}}\right)AB\&CA={\cos }^{-1}(\frac{2}{\sqrt{17}}.\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}.\frac{3}{\sqrt{17}}-\frac{2}{\sqrt{17}}.\frac{1}{\sqrt{3}})={\cos }^{-1}\left(\frac{3}{\sqrt{51}}\right)$

All angles different therefore triangle is scalene.