#|
					Hw3
|#

					; section 12.2

;; (pick 3 (combine (iseq 1 7) 9 12 14 16 20 21))
(message "
Problems 6, 9, 20, section 12.2, page 84
")
(message "
	Problem 6: We did this one in class:

	The longest acceleration vector occurs at the point where the curve is
the tightest (has greatest curvature) (i.e., at P).
	
	It is pointing roughly in the direction of R, from P.
")

(message "
	Problem 9: the faster car will have the greater magnitude. The car with
velocity vector (21,35) is moving slightly faster:
")
(def c1 (vector 21 35))
(def c2 (vector 40 0))
(norm c1)
(norm c2)

(message "
	Problem 20: We're to demonstrate geometrically the property that

				(a + b) v = av + bv.

Draw a nice diagram: that's what I'm looking for.
")

					; section 12-3

;; (pick 3 (list 2 4 5 6 10 13 15 16 19 27))
(message "
Problems 2, 6, 13, page 92
")
(message "
	Problem 2: Answer: -7
")
(def a (vector 0 2 1))
(def z (vector 1 -3 -1))
(dot a z)

(message "
	Problem 6: The dot product of the two vectors gives t^2 - t - 2, which
can be factored as

				(t + 1)*(t - 2)

Thus either of t=-1, or t=2 would work.
")
(def a (vector 'x -1 1))
(def z (vector 'x 'x -2))
(func dot-az (x) (dot (vector x -1 1) (vector x x -2)))
(domain -2 2)
(plot dot-az)

(message "
	Problem 13: Find the equation of the plane perpendicular to -i+2j+k,
passing through (1,0,2).

	The dot product of the vector -i+2j+k and another vector, formed by a
point in the plane (x,y,z) and (1,0,2), is zero:

			-1(x-1) + 2(y-0) + 1(z-2) = 0

")

					; section 12-4
;; (pick 3 (list 1 2 3 5 7 10))
;; 1 3 10 
(message "
Problems 1, 3, 10, page 102
")
(message "
	Problem 1: Boy did you guys get lucky!
			kxj=-i.
")
(def u (vector 0 0 1))
(def v (vector 0 1 0))
(cross u v)

(message "
	Problem 3: not much harder: axb = #(-1 1 1)
")
(def a (vector 1 0 1))
(def b (vector 1 1 0))
(cross a b)

(message "
	Problem 10: axb = #(-2 -7 -13)
")
(def a (vector 3 1 -1))
(def b (vector 1 -4 2))
(def axb (cross a b))
(dot a axb)
(dot b axb)