This preview shows pages 1–2. Sign up to view the full content.
67. The three vectors given are
ˆˆ
ˆ
5.0 i 4.0 j 6.0 k
ˆ
2.0 i 2.0 j 3.0 k
ˆ
4.0 i 3.0 j 2.0 k
a
b
c
=+
−
=−
+
+
+
G
G
G
(a) The vector equation
rabc
=−+
G
GG
G
is
ˆ
[5.0 ( 2.0) 4.0]i (4.0 2.0 3.0)j ( 6.0 3.0 2.0)k
ˆ
=11i+5.0j 7.0k.
r
−+
+−
+
+
−
−
+
−
G
(b) We find the angle from +
z
by “dotting” (taking the scalar product)
G
r
with
#
k. Noting
that
22
2
=   =
(11.0) + (5.0) + ( 7.0) = 14,
rr
−
G
Eq. 320 with Eq. 323 leads to
()
(
)
k
7.0
14 1 cos
120 .
r
φφ
⋅=
− =
¡
=°
G
G
(c) To find the component of a vector in a certain direction, it is efficient to “dot” it (take
the scalar product of it) with a unitvector in that direction. In this case, we make the
desired unitvector by
2
ˆ
2.0i+2.0j+3.0k
ˆ
.

2.0
(2.0)
(3.0)
b
b
b
−
==
+
G
G
We therefore obtain
(
)
(
)
2
5.0
2.0
4.0 2.0
6.0 3.0
ˆ
4.9 .
2.0
(2.0)
(3.0)
b
aa
b
+
−
=⋅=
=
−
+
G
(d) One approach (if all we require is the magnitude) is to use the vector cross product, as
the problem suggests; another (which supplies more information) is to subtract the result
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Any
 Physics

Click to edit the document details