Question

Find the coordinates of the points which trisect the line segment $PQ$ formed by joining the points $P(1,-1,-3)$ and $Q(7,-13,2)$

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.

$\therefore A=(\frac{1\times 7+2\times 1}{1+2},\frac{1\times -13+2\times -1}{1+2},\frac{1\times 2+2\times -3}{1+2})$

$\therefore A=(\frac{7+2}{3},\frac{-13-2}{3},\frac{2-6}{3})$

$\therefore A=(\frac{9}{3},\frac{-15}{3},\frac{-4}{3})$

$\therefore A=(3,-5,\frac{-4}{3})$

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.

$\therefore B=(\frac{2\times 7+1\times 1}{2+1},\frac{2\times -13+1\times -1}{2+1},\frac{2\times 2+1\times -3}{2+1})$

$=(\frac{15}{3},\frac{-27}{3},\frac{1}{3})$

$\therefore B=(5,-9,\frac{1}{3})$