Question: find points at which plane and curve intersect : 8x+15y+3z=20 and R(t): <t,t^2,3t^2>.
R(t) = <x(t), y(t), z(t)>, where x(t) = t, y(t) = t^2, z(t) = 3t^2.
We need to find those values for t such that the x- y- and z-coordinates of the curve, viz. x(t), y(t) and z(t), satisfy the eqn of the plane, 8x + 15y + 3z = 20.
Substituting for x(t), y(t) and z(t) into the plane equation,
8t + 15t^2 + 9t^2 = 20
24t^2 + 8t - 20 = 0
6t^2 + 2t - 5 = 0
Using the quadratic formula,
t = [-2 ± √(4 + 4*6*5)]/(2*6) = [-2 ± √(4 + 120)]/(2*6) = [-2 ± √(124)]/(12) = [-1 ± √(31)]/(6)
t1 = [-1 + √(31)]/(6), t2 = [-1 - √(31)]/(6)
t1 = 0.7613, t2 = -1.0946
These values for t give the coords of intersection, using r(t) = <t, t^2, 3t^2>, as:
P1: (0.7613, 0.5796, 1.7387), P2: (-1.0946, 1.1982, 3.5946)