A First Course in Linear Algebra, 2017a

553
4.7.16 ( ) ( )
⃗u •⃗v ⃗u •⃗v
⃗u −
‖⃗v‖ 2⃗v • ⃗u −
‖⃗v‖ 2⃗v = ‖⃗u‖ 2 − 2(⃗u •⃗v) 2 1
‖⃗v‖ 2 +(⃗u 1
•⃗v)2 ‖⃗v‖ 2 ≥ 0
And so
‖⃗u‖ 2 ‖⃗v‖ 2 ≥ (⃗u •⃗v) 2
You get equality exactly when ⃗u = proj ⃗v ⃗u = ⃗u•⃗v
‖⃗v‖ 2 ⃗v in other words, when ⃗u is a multiple of⃗v.
4.7.17
⃗w − proj ⃗v (⃗w)+⃗u − proj ⃗v (⃗u)
= ⃗w +⃗u − (proj ⃗v (⃗w)+proj ⃗v (⃗u))
= ⃗w +⃗u − proj ⃗v (⃗w +⃗u)
This follows because
proj ⃗v (⃗w)+proj ⃗v (⃗u) =
⃗u •⃗v ⃗w •⃗v
‖⃗v‖ 2⃗v +
‖⃗v‖ 2⃗v
=
(⃗u + ⃗w) •⃗v
‖⃗v‖ 2 ⃗v
= proj ⃗v (⃗w +⃗u)
( )
( ⃗v·u)
4.7.18 (⃗v − proj ⃗u (⃗v)) •⃗u =⃗v •⃗u − ⃗u •⃗u =⃗v •⃗u −⃗v •⃗u = 0. Therefore, ⃗v =⃗v − proj
‖⃗u‖ 2 ⃗u (⃗v)+proj ⃗u (⃗v).
The first is perpendicular to ⃗u and the second is a multiple of ⃗u so it is parallel to ⃗u.
4.9.1 If ⃗a ≠⃗0, then the condition says that ‖⃗a ×⃗u‖ = ‖⃗a‖sinθ = 0 for all angles θ. Hence ⃗a =⃗0 after all.
4.9.2
4.9.3
⎡
⎣
⎡
⎣
3
0
−3
3
1
−3
⎤
⎡
⎦ × ⎣
⎤
⎡
⎦ × ⎣
−4
0
−2
−4
1
−2
⎤
⎡
⎦ = ⎣
⎤
⎡
⎦ = ⎣
0
18
0
1
18
7
⎤
⎦. So the area is 9.
⎤
⎦. The area is given by
√
1
1 +(18) 2 + 49 = 1 374
2
2√
4.9.4 [ 1 1 1 ] × [ 2 2 2 ] = [ 0 0 0 ] . The area is 0. It means the three points are on the same
line.
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
1 3 8
4.9.5 ⎣ 2 ⎦ × ⎣ −2 ⎦ = ⎣ 8 ⎦. The area is 8 √ 3
3 1 −8
4.9.6
⎡
⎣
1
0
3
⎤
⎡
⎦ × ⎣
4
−2
1
⎤
⎡
⎦ = ⎣
6
11
−2
⎤
⎦. The area is √ 36 + 121 + 4 = √ 161