I'm studying for a final, and on my last midterm there was this question:
At what point on the following curve C is the normal plane parallel to 6x + 6y -8z =1, for C: <t^3, 3t, t^4>?
What do I do for this?
The normal plane is the plane at a point P on the curve C that consists of all lines that are orthogonal to the tangent vector. Because of that property, the tangent vector is a perfect vector to chose as the "normal vector" of our normal plane. . . Hopefully you got all that.
Therefore, to find the tangent vector you simply take the derivative of your curve C, and then plug in your point P:
We know the normal vector to our normal plane must be parallel to the plane 6x+6y-8z=1, which means our normal vector should be <6, 6, -8>.
See if you can take where it from here.
r'(t) = 3(t^2)i + (3)j + 4(t^3)k
[r'(t)] = sqrt[9+9+16] = sqrt
so I end up with
<6> = (3)j
So I still don't get what I am doing wrong
Err. My hind was that the vector <6,6,-8> is just a scaled version of <3,3,-4>, and we can use <3,3,-4> instead of <6,6,-8> (or you can equivalently just multiply T(t) by 2 to get . It doesn't matter which way you do it.
Or course none of this matters iy you do not know WHY you are doing these things. You know that the vector function T(t) at the point P will be orthgonal to your normal plane, as would ANY vector that is parallel to the vector T(t) at point P. <3,3,-4> is parallel to <6,6,-8>, so you can use <3,3,-4> instead since that 3 matches up with the 3 in T(t).
But now what do you do? It is asking for a POINT, not a vector. We'll we know the T(t) is parallel to <3,3,-4> or . See if you can take it from there.