PROBLEM 1

(a)  We have ||v1|| = sqrt(1*1 + 3*3 + 3*3 + 9*9) = sqrt(100) = 10,
so  w1 = v1 / 10 = [1/10, 3/10, 3/10, 9/10].  Then
w1.v2 = (1*2 + 3*1 + 3*6 + 9*3) / 10 = 5, so the projection
of v2 to the orthogonal complement of the span of w1 is
v2 - (w1.v2) w1 = [3/2, -1/2, -9/2, -3/2]  (as a column vector).
We calculate that this has norm sqrt(25)=5, so w2 is obtained by
dividing it by 5 to obtain [3/10, -1/10, 9/10, -3/10].
Then w1.w2 = 3/100 - 3/100 + 27/100 - 27/100 = 0 as it should be.

(b) Either by doing the computation in (a) in reverse, or by
forming the dot products with w1 and w2 to obtain the coefficients
of v1 and v2, we find that  v1 = 10 w1  and  v2 = 5 w1 + 5 w2.
That is, the coordinates are

       [10]          [5]
  v1 = [  ]     v2 = [ ]
       [ 0]          [5].

(c) The dot product of e1 with any vector w is the first component of w.
Thus  e1.w1 = 1/10  and  e1.w2 = 3/10.  Therefore the projection of e1
onto V is  e1|| = (1/10)w1 + (3/10)w2 = [1/10, 0, 3/10, 0].
[A popular but harder route was the formula  Q * Q-transpose
(5.3.10 on p.216) for the projection to a subspace V with a known
orthonormal basis.]  The part of e1 perpendicular to V is then
e1 - e1|| = [9/10, 0, -3/10, 0], which indeed is orthogonal
to w1 (because (9/10)(1/10) - (3/10)(3/10) = 0)
and to w2 (because (9/10)(3/10) - (3/10)(9/10) = 0).