Thomas Calculus 13th [Solutions]

896 Chapter 12 Vectors and the Geometry of Space
(d) true, ( cu) v ( cu1 ) v1 ( cu2 ) v2 ( cu3) v3 u1 ( cv1 ) u2( cv2) u3( cv3 ) u ( cv)
c u1v 1 u2v2 u3v3
c( u v)
i j k i j k i j k
(e) true, c( u v) c u1 u2 u3 cu1 cu2 cu3 ( cu) v u1 u2 u3
u ( cv)
v1 v2 v3 v1 v2 v3 cv1 cv2 cv3
(f ) true, u u
(g) true, ( u u) u 0 u 0
2
2 2 2 2 2 2 2
u1 u2 u3 u1 u2 u3 | u|
(h) true, u v u and u v v ( u v) u v ( u v) 0
29. (a)
proj u v
v u v (b) ( u v ) (c) ( u v)
w (d) ( u v)
w
| v|| v|
(e) ( u v) ( u w ) (f )
| u| v
| v|
30. ( i j) j k j i ; i ( j j) i 0 0 . The cross product is not associative.
31. (a) yes, u v and w are both vectors (b) no, u is a vector but v w is a scalar
(c) yes, u and u w are both vectors (d) no, u is a vector but v w is a scalar
32. ( u v)
w is perpendicular to u v , and u v is perpendicular to both u and v ( u v)
w is parallel to
a vector in the plane of u and v which means it lies in the plane determined by u and v . The situation is
degenerate if u and v are parallel so u v 0 and the vectors do not determine a plane. Similar reasoning
shows that u ( v w ) lies in the plane of v and w provided v and w are nonparallel.
33. No, v need not equal w . For example, i j i j , but i ( i j)
i i i j 0 k k and
i ( i j) i ( i) i j 0 k k.
34. Yes. If u v u w and u v u w , then u ( v w)
0 and u ( v w ) 0. Suppose now that v w.
Then u ( v w)
0 implies that v w ku for some real number k 0. This in turn implies that
2
u ( v w) u ( ku) k | u | 0, which implies that u 0 . Since u 0 , it cannot be true that v w, so v w.
i j k
35. AB i j and AD i j AB AD 1 1 0 2k area AB AD 2
1 1 0
i j k
36. AB 7i 3j and AD 2i 5j AB AD 7 3 0 29k area AB AD 29
2 5 0
i j k
37. AB 3i 2j and AD 5i j AB AD 3 2 0 13k area AB AD 13
5 1 0
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