Answer:
My eyeball-estimate of 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:
- Compute the lengths:
- | w | = (keep it symbolic)
- | v |2 =
((5, 0, 3.1)T · (5, 0, 3.1)T)
= 34.61
- Compute the unit vectors:
- wu = (4, 2.5, 2.5)T / | w |
- vu = (5, 0, 3.1)T / | v |
- 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 |)
- 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