Browse Questions

# Find the ratio in which $YZ$ plane divides the segment formed by joining the points $(-2,4,7)$ and $(3,-5,8)$

$\begin{array}{1 1}3:2 \\ 2:3 \\ 3:4 \\ 4:3\end{array}$

Toolbox:
• Section formula: 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 $C\big(\large\frac{lx_2+mx_1}{l+m},\frac{ly_2+my_1}{l+m},\frac{lz_2+mz_1}{l+m}\big)$
• The $x$ coordinate of any point on $YZ$ plane is $0$
Let ther given points be $A(-2,4,7)$ and $B(3,-5,8)$
Let the point on $YZ$ plane that divides $AB$ be $C$ and
let the ratio in which it divides be $k:1$.
$\therefore$ From section formula the coordinates of $C$ is goven by
$C\big(\large\frac{3k-2}{k+1},\frac{-5k+4}{k+1},\frac{8k+7}{k+1}\big)$
But gives that $C$ lies on $YZ$ plane.
We know that the $x$ coordinate of any point on $YZ$ plane is $0$
$\Rightarrow\:\large\frac{3k-2}{k+1}$$=0 \Rightarrow\;3k-2=0 or k=\large\frac{2}{3} \therefore The required ratio is \large\frac{2}{3}$$:1=2:3$