I have to find intersection of two lines ($AH$ & $CD$)
$A(3.42,-1.84,8.56) $
$B(-3.42,3.84,-8.56) $
$C(0.00,16.25,0.00)$
$AH$ is the perpendicular; $CD$ is the median
I tried so:
Firstly I found coordinates of the point $H$:
$x_H=(3.42-3.42)/2=0$
$y_H=(-1.84+3.84)/2=1$
$z_H=(8.56-8.56)/2=0$
So $H=(0,1,0)$
After, I found the median by formula:
$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
$\frac{x-0}0=\frac{y-16.25}{1-16.25}=\frac{z-0}0$
-> it's mean median equation: $y=16.25$
All this is correct?
Please help me, I do not know what to do next ...