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Q.
The coordinates of the points which trisect the line segments joining the points $P(4,2,-6)$ and $Q(10,-16,6)$, are
Introduction to Three Dimensional Geometry
Solution:
Let the point $R_1$ and $R_2$ trisects the line $P Q$ i.e., $R_1$ divides the line in the ratio $1: 2$.
$\Rightarrow R_1 =\left(\frac{1 \times 10+2 \times 4}{1+2}, \frac{1 \times(-16)+2 \times 2}{1+2}, \frac{1 \times 6+2 \times(-6)}{1+2}\right)$
$ =\left(\frac{10+8}{3}, \frac{-16+4}{3}, \frac{6-12}{3}\right)=\left(\frac{18}{3}, \frac{-12}{3}, \frac{-6}{3}\right)$
$ =(6,-4,-2)$
Again, let the point $R_2$ divides $P Q$ internally in the ratio $2: 1$. Then,
$\Rightarrow R_2 =\left(\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)}{1+2}\right) $
$ =\left(\frac{20+4}{3}, \frac{-32+2}{3}, \frac{12-6}{3}\right)=\left(\frac{24}{3}, \frac{-30}{3}, \frac{6}{3}\right) $
$ =(8,-10,2)$
Hence, required points are $(6,-4,-2)$ and $(8,-10,2)$.
