Question

Find the coordinates of the points which trisect the line segment $PQ$ formed by joining the points $P(4,2,-6)$ and $Q(10,-16,6)$

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.

$\therefore A=(\frac{1\times 10+2\times 4}{1+2},\frac{1\times -16+2\times 2}{1+2},\frac{1\times 6+2\times -6}{1+2})$

$=(\frac{10+8}{3},\frac{-16+4}{3},\frac{6-12}{3})$

$=(\frac{18}{3},\frac{-12}{3},\frac{-6}{3})$

$\therefore A=(6,-4,-2)$

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.

$\therefore B=(\frac{2\times 10+1\times 4}{2+1},\frac{2\times -16+1\times 2}{2+1},\frac{2\times 6+1\times -6}{2+1})$

$=(\frac{20+4}{3},\frac{-32+2}{3},\frac{12-6}{3})$

$=(\frac{24}{3},\frac{-30}{3},\frac{6}{3})$

$\therefore B=(8,-10,2)$