Question

find the equation of the plane.

Answer 1

the plane that contains the line x=1+t ; y=2-t ; z=4-3t and parallel to the plane 5x+2y+z=1.

Answer 2

so far this is what i got.\[(1+t)+(2-t)+(4-3t)=0\]\[r=r_0+vt=(-i+2j+4k)+(-i+j+3k)t\]\[r_0=i+2j+4k\]\[v=-i+j+3k\]

Answer 3

so what's the next step... assuming i did this right. if not can you help me?

Answer 4

I am not sure about your directional vector v but your next step is to get a point P which is when t=0
then do the dot product of n.pr=0

Answer 5

if it is parallel to the other plane, you already know its normal. it is the normal of the parallel plane.

all you really needed to do was set t any anything, get a point on that line, ergo also on the plane and calculate the plane constant.