Question

In which ratio the plane $YZ$ divides the lines joining the points $(2,1,2)$ and $(-6,3,4)$.

Answer 1

The coordinates of the point $C$ that divides the segment joining the points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ in the ratio $l:m$ internally is given by $C(\frac{l{x}_2+m{x}_1}{l+m},\frac{l{y}_2+m{y}_1}{l+m},\frac{l{z}_2+m{z}_1}{l+m})$

Let the given points be $A(2,1,2),B(-6,3,4)$

Let the point on the $YZ$ plane that divides $AB$ be $C$ and let the ratio in which it divides be $k:1$

Thus coordinates of $C$ is given by $C(\frac{-6k+2}{k+1},\frac{3k+1}{k+1},\frac{4k+2}{k+1})$

We know that $x$ coordinate of any point on $YZ$ plane is $0$.

$\Rightarrow \frac{-6k+2}{k+1}=0$

$\Rightarrow -6k+2=0$

$\Rightarrow -3k+1=0$

$\Rightarrow 3k=1$

$\Rightarrow k=\frac{1}{3}$

Therefore, the required ratio is $\frac{1}{3}:1=1:3$.