Question

Find the coordinates of the points which trisect the line segment $PQ$ formed by joining the points $P(3,1,-5)$ and $Q(9,-15,4)$

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.

$\therefore A=(\frac{1\times 9+2\times 3}{1+2},\frac{1\times -15+2\times 1}{1+2},\frac{1\times 4+2\times -5}{1+2})$

$=(\frac{9+6}{3},\frac{-15+2}{3},\frac{4-10}{3})$

$=(\frac{15}{3},\frac{-13}{3},\frac{-6}{3})$

$\therefore A=(5,\frac{-13}{3},-2)$

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.

$\therefore B=(\frac{2\times 9+1\times 3}{2+1},\frac{2\times -15+1\times 1}{2+1},\frac{2\times 4+1\times -5}{2+1})$

$=(\frac{18+3}{3},\frac{-30+1}{3},\frac{8-5}{3})$

$=(\frac{21}{3},\frac{-29}{3},\frac{3}{3})$

$\therefore B=(7,\frac{-29}{3},1)$