Where Does the Line Intersect the Plane in 3D Space?

[multicalcprob]

Let L be the line through the point Po(1,2,8) which is parallel to the vector R(3i-j-4k). Find the point at which l intersects the plane through the point p1(-4,0,3) having normal vector n(3i-2j+6k)

I did the following:
x=1+3t
y=2-t
z=8-4t

3(x+4)+6(z-3)=0
3x+6z-6=0

3(1+3t)+6(8-4t)-6=0
-15t+45=0
t=3

(10, -1, -4)

Did I do this right?
Thanks.

 

[weatherhead]

I got a different answer, I'm not quite sure how you're working out the equation of your plane. I ended up with the equation 3x-2y+6z-6=0, which doesn't give quite so nice an answer I'm afraid! Could you explain how your method for working out the equation of the plane works??

I did it like this:
equation of a plane is R.n=d
you are given n and you are given one specific R,
so equation is R.(3 -2 6) = d

now you substitute in the specific r to find the d, if that makes sense, so:

(-4 0 3).(3 -2 6) = -12 +0 +18 = 6

so you know the equation of the plane is
R.(3 -2 6) =6

or, in other words (taking R to be (x y z):
3x-2y+6z=6
3x-2y+6z-6=0

 

[dynamicsolo]

3(x+4)+6(z-3)=0
3x+6z-6=0

I believe you omitted the y-term in your plane equation.
Weatherhead's result for the plane looks right to me.