The question is:
Find the Cartesian form of the equation for the plane that is perpendicular to the space curve
at the point P(32, -32, 16).
Now the simple part of course, once I have the normal (n) to the curve at P, I can construct a plane equation via where (the vector from the plane's origin (P) to any point Q).
Of course, I can find the derivative of r(t), which is:
That will give me the tangent vector at any time t.
I know that t = 16, from the 'k' component, so I can evaluate
But I need the normal vector (perpendicular to the curve).
So if I understand correctly, u(t) = r'(t)/|r'(t)| ? (Which is the unit tangent vector at 't').
Then u'(t) = r''(t)? (The derivative of r'(t)).
So you want me to find, ?
Yes, I know how to find a unit vector from a vector, but I'm interested in the perpendicular vector to the tangent, not the tangent.
You do not need a vector that is normal to the curve (such a vector would be of no use in this problem). You only need one that is normal to the desired plane.
The point corresponds to and
This vector is tangent to the curve, so it must be normal to the plane perpendicular to the curve. Using the point and normal vector the equation of the plane is
Oh my! Thank you so much! It was way easier than I thought. I misunderstood the question initially, as you can see.
I didn't realise that the derivative of r(t) was in fact going to give me normal vectors for planes at each 't' that are perpendicular to r(t). I see that now, thanks to you.
Once I understood that, the rest is easy!