A good answer might be:

My visual estimate is that if you "drop" the arrow head of w along a perpendicular to v, you hit v at about 4/10 the length of v. So k is about 0.4 and kv   =   0.4 (6, 4, 2)T   =   (2.4, 1.6, 0.8)T

It turns out that I was grossly wrong. Hope you did better.


Checking the Answer

The visual approach to 3D problems is hard, unless you actually have a 3D image in front of you. The correct answer calls for some figuring:

  1. Compute the lengths:
    |w|  =   (keep it symbolic)
    |v|2  =   ((6, 4, 2)T · (6, 4, 2)T)   =   56
  2. Compute the unit vectors:
    wu  =   (4, 2, 3)T / |w|
    vu  =   (6, 4, 2)T / |v|
  3. Compute the cosine of the angle between the vectors:
    wu · vu  =   (4, 2, 3)T · (6, 4, 2)T / |w||v|
     =   38/( |w| |v|)
  4. Assemble the projection:
    kv  =   |w| (wu · vu) vu
    kv  =   |w| {38 / (|w||v|)}   {(6, 4, 2)T / |v|}
    kv  =   {38 /|v|}   {(6, 4, 2)T / |v|}
    kv  =   {38 /|v|2}   (6, 4, 2)T
    kv  =   {38/56}   (6, 4, 2)T
    kv  =   0.679   (6, 4, 2)T
    kv  =   ( 4.07, 2.7, 1.36)T

QUESTION 17:

What is u, where u + kv  =   w? (In the diagram u is the dotted black line).