Question

The points $P(3,2,-4),Q(5,4,-6)$ and $R(9,8,-10)$ are collinear. If the ratio in which $Q$ divides $PR$ is $1:a$, then $a$ is equal to____ .

Answer 1

Given that, points $P(3,2,-4),Q(5,4,-6)$ and $R(9,8,-10)$ are collinear.

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

We know that,

Co-ordinates of point $P(x,y,z)$ that divides line segment joining $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ in ratio $m:n$ is

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

Let point $Q(5,4,-6)$ divide line segment $P(3,2-4),$ $R(9,8,-10)$ in ratio $k:1$

Here,

$x_1=3,y_1=2,z_1=-4$

$x_2=9,y_2=8,z_2=-10$

and $m=k,n=1$

Putting values

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

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

Comparing $x-$ co-ordinate of $Q$

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

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

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

$\Rightarrow$ $4k=2$

$\therefore$ $k=\frac{1}{2}$

So, $k:1=1:2$

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

The value of $a$ is equal to $2$.