Michael Corral: Vector Calculus

1.2 Vector Algebra 11
Theorem 1.3. Let v=(v 1 ,v 2 ), w=(w 1 ,w 2 ) be vectors in 2 , and let k be a scalar. Then
(a) kv=(kv 1 ,kv 2 )
(b) v + w=(v 1 +w 1 ,v 2 +w 2 )
Proof: (a) Without loss of generality, we assume that v 1 ,v 2 > 0 (the other possibilities
are handled in a similar manner). If k=0 then kv=0v=0=(0,0)=(0v 1 ,0v 2 )=
(kv 1 ,kv 2 ), which is what we needed to show. If k0, then (kv 1 ,kv 2 ) lies on a line with
slope kv 2
kv 1
= v 2
v 1
,whichisthesameastheslopeofthelineonwhichv(andhencekv)lies,
and(kv 1 ,kv 2 )pointsinthesamedirectiononthatlineaskv. Also,byformula(1.3)the
magnitude of (kv 1 ,kv 2 ) is √ √ √ √
(kv 1 ) 2 +(kv 2 ) 2 = k 2 v 2 1+k 2 v 2 2= k 2 (v 2 1+v 2 2)=|k| v 2 1+v 2 2=
|k|‖v‖. So kv and (kv 1 ,kv 2 ) have the same magnitude and direction. This proves (a).
(b) Without loss of generality, we assume that
v 1 ,v 2 ,w 1 ,w 2 > 0(theotherpossibilitiesarehandled
in a similar manner). From Figure 1.2.5,
we see that when translating w to start at
the end of v, the new terminal point of w is
(v 1 + w 1 ,v 2 + w 2 ), so by the definition of v+w
this must be the terminal point of v+w. This
proves (b).
QED
y
v 2 +w 2
v
w 2
w w 2
w v+w
v 2
w v 1
0 w 1 v 1 v 1 +w 1
Figure 1.2.5
Theorem1.4. Letv=(v 1 ,v 2 ,v 3 ),w=(w 1 ,w 2 ,w 3 )bevectorsin 3 ,letkbeascalar. Then
(a) kv=(kv 1 ,kv 2 ,kv 3 )
(b) v + w=(v 1 +w 1 ,v 2 +w 2 ,v 3 +w 3 )
The following theorem summarizes the basic laws of vector algebra.
Theorem 1.5. For any vectors u, v, w, and scalars k,l, we have
(a) v+w=w+v Commutative Law
(b) u+(v+w)=(u+v)+w
(c) v+0=v=0+v
(d) v+(−v)=0
(e) k(lv)=(kl)v
(f) k(v+w)=kv+kw
(g) (k+l)v=kv+lv
Associative Law
Additive Identity
Additive Inverse
Associative Law
Distributive Law
Distributive Law
Proof: (a) We already presented a geometric proof of this in Figure 1.2.4(a).
(b)Toillustratethedifferencebetweenanalyticproofsandgeometricproofsinvector
algebra, we will present both types here. For the analytic proof, we will use vectors
in 3 (the proof for 2 is similar).
x