Michael Corral: Vector Calculus

1.2 Vector Algebra 13
2
z
z
v=(a,b,c)
y
2
1
j
x
0 i 1 2
(a) 2
y
v=(a,b)
bj
x
0 ai
(b) v=ai+bj
2
x
1
k
i 0
1
j
(c) 3
1 2
y
ck
ai 0
x bj
(d) v=ai+bj+ck
y
Figure 1.2.7 Basis vectors in different dimensions
Whenavectorv=(a,b,c)iswrittenasv=ai+bj+ck,wesaythatvisincomponent
form, and that a, b, and c are the i, j, and k components, respectively, of v. We have:
v=v 1 i+v 2 j+v 3 k,k a scalar=⇒ kv=kv 1 i+kv 2 j+kv 3 k
v=v 1 i+v 2 j+v 3 k,w=w 1 i+w 2 j+w 3 k=⇒ v+w=(v 1 +w 1 )i+(v 2 +w 2 )j+(v 3 +w 3 )k
√
v=v 1 i+v 2 j+v 3 k=⇒‖v‖= v 2 1+v 2 2+v 2 3
Example 1.4. Let v=(2,1,−1) and w=(3,−4,2) in 3 .
(a) Find v−w.
Solution: v−w=(2−3,1−(−4)−1−2)=(−1,5,−3)
(b) Find 3v+2w.
Solution: 3v+2w=(6,3,−3)+(6,−8,4)=(12,−5,1)
(c) Write v and w in component form.
Solution: v=2i+j−k, w=3i−4j+2k
(d) Find the vector u such that u+v=w.
Solution: ByTheorem1.5,u=w−v=−(v−w)=−(−1,5,−3)=(1,−5,3),bypart(a).
(e) Find the vector u such that u+v+w=0.
Solution: By Theorem 1.5, u=−w−v=−(3,−4,2)−(2,1,−1)=(−5,3,−1).
(f) Find the vector u such that 2u+i−2j=k.
Solution: 2u=−i+2j+k=⇒ u=− 1 2 i+j+ 1 2 k
(g) Find the unit vector v
‖v‖ .
(
v
Solution:
‖v‖ = √
1
(2,1,−1)=
2 2 +1 2 +(−1) 2
)
√
2 1
, √6 , −1 √
6 6