A good answer might be:

The projection of w = (4, 2.5, 2.5)T onto v = (5, 0, 3.1)T is kv = (4, 0, 2.5)T


Collapsing one Dimension

In this case, projecting w onto v involved "collapsing" the y dimension of w. Lets see if the same result comes out mathematically. The same steps work for 3D as for 2D:

  1. Compute the lengths:
    | w |  =   (keep it symbolic)
    | v |2  =   ((5, 0, 3.1)T · (5, 0, 3.1)T)   =   34.61
  2. Compute the unit vectors:
    wu  =   (4, 2.5, 2.5)T / | w |
    vu  =   (5, 0, 3.1)T / | v |
  3. Compute the cosine of the angle between the vectors:
    wu · vu
     =   (4, 2.5, 2.5)T / | w | · (5, 0, 3.1)T / | v |
     =   27.75/( | w || v |)
  4. Assemble the projection:
    kv  =   | w | (wu · vu) vu
    kv  =   | w | (27.75 / (| w || v |))  (5, 0, 3.1)T / | v |
    kv  =   (27.75 / (| v |))  ( (5, 0, 3.1)T / | v | )
    kv  =   (27.75 / | v |2)   (5, 0, 3.1)T
    kv  =   (27.75 / 34.61)   (5, 0, 3.1)T
    kv  =   ((27.75*5)/34.61, 0, (27.75*3.1)/34.61T
    kv  =   ( 4.00, 0, 2.49)T

So the visual estimate kv = (4, 0, 2.5)T matches the mathematical result, ( 4.00, 0, 2.49)T. The orthogonal vector can easily be computed:

u  =   w - kv
u  =   (4, 2.5, 2.5)T - (4, 0, 2.5)T  =   (0, 2.5, 0)T

QUESTION 15:

Keep the same vector w as in the diagram. Visualize projecting w onto various vectors v'. Can you think of a vector v' that results in a projection kv' that collapses both the y and z dimensions of w ?