Plane P is given by its parametric equations xt2s y1t z32ts

Plane P is given by it\'s parametric equations: x=t+2s y=-1+t z=3+2t+s

ii.if it is parrallel, find the distance between L and P; if it is not parralell, find their point of intersection

I know how to find the distance between them, (or at least I think I do) Being the projection of vector OL onto vector OP? however the textbook does not give any examples similar nor bother to elaborate on the process when not given a predetermined point to work from.

Solution

Solution:given plan x=t+2s, y=-1+t & z=3+2t+s

t=1+y putting in eq. 3, s=z-2y-5

value of t & s put in eq. 1 x=1+y+2(z-2y-5)

after rearenging this variable we get plan x+3y-2z=-9

we see this is not parallela to given line eq.

now we find intersection point

eq. of line is x/-1=y/3=(z-2)/0 and we suppose this is equal to t, then x=-t, y=3t, z=2

this point put in plan eq. then we calculted value of t=-5/8

point is x=-5/8, y=-15/8, z=2