Find the points of intersection of the curve r(t) = ti+t2j-3tkand the plane 2x - y + z = -2.
To determine the points of intersection of the curve with the plane, we simply recognize that the points of intersection are where the x, y, and z components of the curve r(t) are equivalent to a point (x,y,z) on the plane.
We're going to start by determining what the x, y, and z component of each point on the curve is, which is equivalent to simply separating out the vector components i, j, and k:
`vec(r)(t) = <x, y, z> = <t, 2t, -3t>`
Now, recognizing that the x, y, and z must be equivalent, we can simply substitute the values above with the x, y, and z in the equation for the plane:
`2x - y + z = -2`
`2(t) - 2t + (-3t) = -2`
`-3t = -2`
Now, divide both sides by -3:
`t = 2/3`
Now, we have the parameter that we can use to find the point at which the curve and plane intersect. To find the point, itself, we just plug the value into the fuction `vec(r)(t)`:
`vec(r)(t) = <t, 2t, -3t> = <2/3, 4/3, -2>`
Now, let's just put this value into the equation for the plane to confirm that it is on the plane:
`2x-y+z = -2`
`2(2/3) - 4/3 - 2 = -2`
`-2 = -2`
Looks like the point is on the plane, as well as on the curve! This means it is the point at which the two intersect, and it means that we've found our solution!
`(x, y, z) = (2/3, 4/3, -2)`