Question

A triangle is formed by joining the points $(1,0,0),(0,1,0)$ and $(0,0,1)$. Find the direction cosines of the medians.

Answer 1

$A\equiv (1,0,0)$

$B\equiv (0,1,0)$

$C\equiv (0,0,1)$

Middle point of side $AB\left(d\right)\equiv (\frac{1}{2},\frac{1}{2},0)$

Middle point of $BC\left(e\right)\equiv (0,\frac{1}{2},\frac{1}{2})$

Middle point of $AC\left(f\right)\equiv (\frac{1}{2},0,\frac{1}{2})$

$\therefore$ direction ratio of cd $=(\frac{1}{2},\frac{1}{2},-1)$

$\therefore$ direction ratio of bf $=(\frac{1}{2},-1,\frac{1}{2})$

$\therefore$ direction ratio of ae $=(-1,\frac{1}{2},\frac{11}{2})$

$\therefore$ direction cosine of cd =

$(\frac{1/2}{\sqrt{(}\frac{1}{2}{)}^2+(\frac{1}{2}{)}^2+1^2},\frac{1/2}{\sqrt{(\frac{1}{2}{)}^2+(\frac{1}{2}{)}^2}+1^2},\frac{-1}{\sqrt{(\frac{1}{2}{)}^2+(\frac{1}{2}{)}^2}+1^2})$

$=(\frac{1/2}{\sqrt{3/2}},\frac{+1/2}{\sqrt{3/2}},\frac{-1}{\sqrt{3/2}})$

$\therefore$ direction cosines of bf $:-$

$(\frac{1/2}{\sqrt{3/2}},\frac{-1}{\sqrt{3/2}},\frac{1/2}{\sqrt{3/2}})$

$\therefore$ direction cosines of ae

$=(\frac{-1}{\sqrt{3/2}},\frac{1}{2\sqrt{3/2}},\frac{1}{2\sqrt{3/2}})$