Question

Find the coordinates of the points which trisect the line segment $PQ$ formed by joining the points $P(2,0,-2)$ and $Q(6,-12,1)$

Answer 1

Let $A$ and $B$ be the points of trisection of the segment $PQ$, then

$PA=AB=BQ\Rightarrow 2PA=AQ$

$\Rightarrow \frac{PA}{AQ}=\frac{1}{2}$

$\Rightarrow A$ divides the line segment $PQ$ in the ratio $1:2$ internally.

$\therefore A=(\frac{1\times 6+2\times 2}{1+2},\frac{1\times -12+2\times 0}{1+2},\frac{1\times 1+2\times -2}{1+2})$

$\therefore A=(\frac{6+4}{3},\frac{-12}{3},\frac{1-4}{3})$

$\therefore A=(\frac{10}{3},\frac{-12}{3},\frac{-3}{3})$

$\therefore A=(\frac{10}{3},-4,-1)$

Also,$PA=AB=BQ\Rightarrow PB=2BQ$

$\Rightarrow \frac{PB}{BQ}=\frac{2}{1}$

$\Rightarrow B$ divides the line segment $PQ$ in the ratio $2:1$ internally.

$\therefore B=(\frac{2\times 6+1\times 2}{2+1},\frac{2\times -12+1\times 0}{2+1},\frac{2\times 1+1\times -2}{2+1})$

$=(\frac{12+2}{3},\frac{-24}{3},\frac{2-2}{2+1})$

$=(\frac{14}{3},\frac{-24}{3},\frac{0}{3})$

$\therefore B=(\frac{14}{3},-8,0)$