1 
  T = v(t)/(|v(t)|)
  N = T'(t)/(|T'(t)|)
  B = T x N
  a_T = a . T
  a_N = |a x T|
  a = a_T T + a_N N
  k = |v x a| / |v|^3

2
a) N = T'(t)/(|T'(t)|), N = (a - a_T T) / a_N,  N = (T x a)/a_N  x T
b) a_T = a . T, a_T = d/dt |v(t)|
c) a_N = |a x T|, a_N = |v(t)|^2 k(t)

3 (+ 4...my mistake)
(1/135) (91^(3/2) - 1)

5
T = <0, 0, 1>
N = <1, 0, 0>
B = <0, 1, 0>
k = infinity

6
a) u is perpendicular to v
b) u = 0
c) u is parallel to v
d) 0

7
I got 37*sqrt(14) / 28

8
|u + v|^2 = (u1 + v1)^2 + ... + (un + vn)^2 = u1^2 + 2u1v1 + v1^2 +
... + un^2 + 2unvn + vn^2 
|u - v|^3 = (u1 - v1)^2 + ... + (un - vn)^2 = u1^2 - 2u1v1 + v1^2 +
... + un^2 - 2unvn + vn^2

thus |u + v|^2 - |u - v|^2 = 4u1v1 + 4u2v2 + .... = 4(u . v)

thus u . v = (1/4)|u + v|^2 - (1/4)|u - v|^2 

9
a) 2x +4y - z = 3
b) -4x - y - z = -1

10
a x b and c x d are both perpendicular to the plane. Thus, since we
are in 3d, these two vectors must be parallel. Thus their cross
product is 0

11
(x - 6) - 10(y - 2) - 23(z + 1) = 0

12
We get collisions if
k = 2n*pi, where n is any positive integer

13
Check book for these pictures

14
z^2 - 2z + r^2 = 0

15
Cylindrical: r^2 - z = 0 

Spherical: rho^2 - rho^2 cos^2(phi) - rho cos(phi) = 0

16
r(t) = <t, 1, 3t - 9/2>

17 (old #18)
a) 1
b) infinity

18 (old #19)
a) 6x + 3y - 2z = 12
b) 38x + 6y - z = -37

19 (old #20)
a) grad f = [zln(x + y + z) + xz/(x + y + z), xz/(x+y+z),xln(x + y +
z) + xz/(x + y + z)]
b) grad f = [6xy sin^2(x^2 y) cos(x^2 y), 3x^2 sin^2(x^2 y) cos(x^2
y), 0]