Vectors Help Needed.?

The intersection point of has a single value for each coordinate

k: t = 1

j: s = t = 1

i: 2t = 2 = s^2 + c, so c = 1

The intersection point has position vector (2,1,1).

The parameters of the first are

x = s^2 + 1, y = s, z = 1

x = y^2 + 1, y = y, and, z = 1

Parabola is x = y^2 + 1 on the plane z = 1

The parameters of the second are

x = 2t, y = t, z = t

2z = x

z = y

z = x – y, a straight line in 3D

The line intersects the plane z = 1

at the point (2, 1, 1), and we find that the point (2, 1)

is on the curve x = y^2 +1,

When you have 2 position vectors, and want to find where their paths intersect, you can set each component in the first vector equal to each component in the 2nd vector, and solve for "t". This will give you the x, y, and z components where the vectors intersect, which is the vector you are looking for

r1(t) = <R1x(t), R1y(t), R1z(t)>

r2(t) = <R2x(t), R2y(t), R2z(t)>

Solve R1z(t) = R2z(t) for t. This should give you the time when the 2 paths' z coordinate is the same

plug this t into R1z(t) or R2z(t) to find this z coordinate.

Solve R1y(t) = R2y(t) for t. This should give you the time when the 2 paths' y coordinate is the same

plug this t into R1y(t) or R2y(t) to find this y coordinate.

Solve R1x(t) = R2x(t) for t. This should give you the time when the 2 paths' x coordinate is the same

plug this t into R1x(t) or R2x(t) to find this x coordinate. (Hint: You don't know what "c" will be until after solving for "t" where the Y and Z components intersect. Sol solve for the t where the Y and Z components intersect, and then choose "c" so that the X component intersects there)