Question

Given that $P(3,2,-4);Q(5,4-6);R(9,8,-10)$ are collinear, find the ratio in which $Q$ divides $\left[PR\right]$.

Answer 1

$P(3,2,-4),Q(5,4,-6),R(9,8,-10)$ are collinear

$Q$ must divide line segment $PR$ in some ratio externally and internally

$(x,y,z)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

$Q(5,4,-6)=(\frac{m\left(9\right)+n\left(3\right)}{m+n},\frac{m\left(8\right)+n\left(2\right)}{m+n},\frac{m(-10)+n(-4)}{m+n})$

$m=k,n=1$

$(5,4,-6)=(\frac{9k+3}{k+1},\frac{8k+2}{k+1},\frac{-10k-4}{k+1})$

$5=\frac{9k+3}{k+1}$

$5(k+1)=9k+3$

$\Rightarrow$ $9k-5k=5-3$

$\Rightarrow$ $4k=2\Rightarrow$ $k=\frac{1}{2}$

i.e., $m:n=k:1$

$=\frac{1}{2}:1$

$=1:2$

$\therefore$ Point $Q$ divides $PR$ in the ratio $1:2$