At what point the straight that passes by A(2,3,4) and B(1,0,-2) intercepts the xy plane?
Answer:
$\displaystyle (\frac{4}{3},1,0)$
The parametric line equations are:
$\displaystyle x=1+t, \;\ y=3t, \;\ z=-2+6t$
In order to intersect the xy plane, then z must = 0.
So, from z:
$\displaystyle -2+6t=0, \;\ t=1/3$
Insert this into x and y and we get:
$\displaystyle x=4/3, \;\ y=1, \;\ z=0$