Question

Calc III: At what point does the curve (2t^2, 1-t, 3+t^2) intersect the plane 3x - 14y + z - 10 = 0

Answer 1

Substitute 2t^2 for x, 1-t for y and 3+t^2 for z in the equation for the plane. solve for t

Answer 2

oh, easy

Answer 3

thanks!

Answer 4

I get two different points, t= 1 and t= -3

Answer 5

that's the answer the book gives.
out of curiosity, how did you solve the quadratic?

Answer 6

3x - 14y + z - 10 = 0
2t^2 for x, 1-t for y and 3+t^2 for z

we get
6t^2 -14(1-t) + 3+t^2 -10 = 0
7t^2 +14t -21 =0
or, dividing by 7,

t^2 +2t -3=0

Now factor into
(t+3)(t-1)=0
so one or the other must be zero: t+3= 0, t=-3 or t-1=0, t=1
The actual x,y,z points using 2t^2 for x, 1-t for y and 3+t^2 for z, are
for t=1, (2, 0, 4)
t=-3 (18, 4, 12)