Math 127(3) - Warren D. Smith - Homework #2 - due Tues 4 Feb 2003
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ANSWERS

1. Do exercises 20, 32, 38, and 48 
of section 12.3, pages 803-804.

20 ANS: arccos(12/sqrt(13*52)) = 1.09 radians = 62.5 degrees.
This is the angle between ( (-2,-3,0) and (-6,0,4).
Here 12 = 2*6 - 3*0 + 0*4, 13= 2*2+3*3, 52 = 6*6+4*4.
[Also: 12/sqrt(13*52) = 12/26 = 6/13 simplifies.]

32 ANS: |a.b| = |a| |b|  if a and b are either parallel or antiparallel.
[Since  |cos(angle)| = LeftHandSide/RightHandSide = 1
so angle must be 0 or 180 so that the cosine is either +1 or -1.]

38 ANS: a.b=0 and a.c=0 means that a.(pb+qc)=p0+q0=0 for any scalars p
and q.  I.e. all vectors of the form pb+qc, (p,q, scalar) are 
perpendicular to a if b and c are.

whoops actually that was 37 ans, 38 is almost same:
38 ANS: aXb=0 and aXc=0 means that aX(pb+qc)=p0+q0=0 for any scalars p
and q.  I.e. all vectors of the form pb+qc, (p,q, scalar) are 
parallel to a if b and c are.

48 ANS: (2,3,0) is perpendicular to the unit vector (0,0,1)
obviously (since dot product is 0). A second vector perpendicular to
both of them is their cross product, namely
(3,-2,0) and the unit-normalized version of this is 
(3,-2,0)/sqrt(13).

WDS2. There is a plane pictured on the top left of page 834.
It goes thru the 3 points (006), (400) and (030).
Find the angle between this plane and the XY plane
(in degrees, do it accurate to 0.1 degree or better).

ANS
The equation of this plane is  (x,y,z) . (1/4, 1/3, 1/6)=1
i.e.   x/4 + y/3 + z/6 = 1.
Its normal vector is  (1/4, 1/3, 1/6), or (rescaling by 12)
a simpler vector normal (i.e. perpendicular) to the plane is  (3,4,2).
The ANGLE to the xy plane (whose normal vector is (0,0,1))
is the same as the angle between the normal vectors (that is the
way to find angles beteern planes) namely
    arccos(  (3,4,2).(0,0,1) / [sqrt(3*3+4*4+2*2)*sqrt(1*1+0+0)] ) =
    arccos( 2 / sqrt(1*29) ) = 68.2 degrees.

3. Compute   (1,5,7) X (6,2,3).
ANS
(1, 39, -28).
[sanity checks: dot prods with them both are both 0.]

4. In section 12.7 pages 835-836 of the book,  they list 7 propositions
in geometry which they'd like you to prove by vector-algebra methods.
Choose one of them (but not prop #1 - that is too easy), and prove it.

ANS prop2. Angle inscribed in a semicircle is right angle.

Well, we have in their figure that
  |c-a| = radius = |b| = |d-a|.
and
  c = a+b  and d = a-b.
(All letters are vectors.)         2     2
c.d = (a+b).(a-b) = a.a - b.b = |a| - |b|

          2        2
  = radius - radius = 0

so c,d perpendicular, QED.


ANS prop3. In a parallelogram the sum of squares of diagonals
equals sum of squares of the 4 side lengths:
This is just to prove

     2        2       2      2
|a+b|  + |a-b|  = 2|a| + 2|b|.

expand out 

(a+b).(a+b) + (a-b).(a-b) = 
a.a + b.b + 2a.b + a.a + b.b - 2a.b =
2a.a + 2b.b.
QED.

ANS law of sines for a triangle [solution by Sung-il Hong]
if a,b,c are the side vectors round the triangle 
(all anticlockwise-pointing)  then  a+b+c=0  as a vector equation.
So  a X (a+b+c) = 0   and  b X (a+b+c) = 0.
Since aXa=0 and bXb=0 by anticommutativity (or by: degenerate flat
parallelogram, or by: algebraic formula cancellation)
we have
   a X b = -a X c        a X b = -b X a = b X c
now use (since length of cross product is area of parallelogram)
   |aXb| = |a|  |b|  sin C
   |aXc| = |a|  |c|  sin B
   |bXc| = |b|  |c|  sin A
so    sinC / |c| = sinB / |b|   and so on, proving law of sines, Q.E.D.