The line joining points $(3,5,-7)$ and $(-2,1,8)$ meets yz-plane at point
The line joining points $(3,5,-7)$ and $(-2,1,8)$ meets yz-plane at point
$\left(0, \frac{13}{5}, 2\right)$
$(0,13,2)$
$\left(0, \frac{13}{5},-3\right)$
$\left(0, \frac{-13}{5}, 2\right)$
Solution
Let $\mathrm{y}_{\mathrm{z}}$ plane divides the join in the ratio $\lambda: 1$ then
$\frac{\lambda \times(-2)+1 \times 3}{\lambda+1}=0 \Rightarrow \lambda=\frac{3}{2}$
So $\mathrm{y}_{\mathrm{z}}$ plane divides the join in the ratio $3: 2$.
Hence, the required point is
$\begin{aligned} & \left(\frac{3 \times(-2)+2 \times 3}{3+2}, \frac{3 \times 1+2 \times 5}{3+2}, \frac{3 \times 8+2 \times(-7)}{3+2}\right) \\ & =\left(0, \frac{13}{5}, 2\right)\end{aligned}$