PRACTICE QUIZ #1 for math127  Warren D. Smith    12 Feb 2003
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ANSWERS  [starting 9:52  done 10:03 (but answers unchecked)
  now doing checks or sanity checks... starting 10:04...
  am done 10:13 - 1 error found & corrected.]
-------------------------------------THE PRACTICE QUIZ----

Find  (1,2,5) X (8,-2,1).
  ANS    (12, 39, -18)  sanity check: dotprod with (1,2,5)=0? yes.
What is the area of the parallelogram these two vectors determine?
  ANS   squareroot(12*12+39*39+18*18) = squareroot(1989).
Find  (1,2,5) . (8,-2,1).
  ANS   8-4+5=9.
What is the angle between these two vectors (write as a simplified formula
involving arcsin or arccos, but do not need to then use calculator to deduce
approximate numerical value of arcsin(3/5) or whatever).
  ANS   arccos( 9 / squareroot( (1+4+25)*(64+4+1) ) )
       =arccos( 9 / squareroot( 30*69 ) )
       =arccos( 9 / squareroot( 2070 ) )
       =arccos( 3 / squareroot(230) )
sanity check: alternate answer is
        arcsin( squareroot(1989)/squareroot( 30*69 ) ) 
       =arcsin( squareroot(1989/2070) )
       =arcsin( squareroot(221/230) )
both answers yield numerical approx value 1.371669 radians.      

The plane  1x+2y+5z=1  and the line  (x,y,z) = (8t, -2t, t+7)
make what angle?  (write as a simplified formula
involving arcsin or arccos, but do not need to then use calculator to deduce
approximate numerical value of arcsin(3/5) or whatever).
  ANS normal to plane is (1,2,5), direction vector of line is (8,-2,1)
    so  angle = 90degrees - arccos( 3 / squareroot(230) )
              = arcsin( 3 / squareroot(230) )    [since sin(90deg-x)=cos(x)]
    using angle answer from previous problem.

If A,B,C are vectors, simplify the formula
   (A X B) . (5A + 7B + 2C) 
as much as possible.
   ANS  2 (A X B) . C.    since AXB is perp to A and to B.

Is there something wrong with this formula-fragment? What?  
5 (A.B) X C - D     (where A,B,C,D vectors and X means cross product,
and . means dot product)
  ANS   scalar X vector   is illegal! undefined!
         (A.B)     C

Consider the curve (x,y,z)=F(t) where F(t)=(cos(t), sin(t), 7t+2)
and t is time.
What is the velocity vector at time t?
   ANS     (-sin(t), cos(t), 7)
What is the arc length from t=0 to t=7?               2        2    2
   ANS  integral(from t=0 to 7)of   squareroot( sin(t) + cos(t)  + 7 ) dt
       =integral(from t=0 to 7)of   squareroot( 50 ) dt = 7*squareroot(50).

Consider the function
                     2           2
F(x,y,z) = sin(x) + y  + yz + 3 z  .

What is  grad F?
   ANS     (cos(x),  2y+z,  y+6z)
What is the unit-normalized version of the vector (2,1,1)?
   ANS      (2,1,1) / squareroot(4+1+1) = (2,1,1)/squareroot(6).
What is  F's directional derivative in that (unit) direction?
   ANS     [2cos(x) + 2y+z + y+6z]/squareroot(6).
Find conditions obeyed by a local minimum of F(x,y,z).
   ANS    grad F = 0  which is x,y,z must obey
     cos(x)=0,  2y+z=0,  y+6z=0  simultaneously.
Solve those conditions to actually find the location x,y,z of
some local minimum of F(x,y,z).
   ANS    x=3pi/2    y=0   z=0.
      sanity checks: x=3pi/2 makes sin(x) = -1 = min possible.
    (x=pi/2 makes sin(x) = 1 maximal, would be mistake to go for max
   not min.)  y=z=0 causes   2y+z=y+6z=0  and looks like a min of F...
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