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Properties of the Dot Product<br />

Commutative Property:<br />

p ! # q<br />

! q! # p<br />

! ,<br />

Distributive Property:<br />

Magnitudes Property: p ! #<br />

! ! p 0 p 0 2 ,<br />

p ! # 1q<br />

! r<br />

! 2 p! # q<br />

! p! # r<br />

! ,<br />

Associative Property with a scalar K: 1kp ! 2 # q<br />

! p! # 1kq<br />

! 2 k1 p! # q<br />

! 2<br />

EXAMPLE 3<br />

Selecting a strategy to determine the angle between two<br />

geometric vectors<br />

If the vectors and 4a ! b !<br />

are perpendicular, and 0 a ! 0 2@ b ! @ , determine<br />

the angle (to the nearest degree) between the nonzero vectors a ! and b ! .<br />

a ! 3b ! CHAPTER 7<br />

Solution<br />

Since the two given vectors are perpendicular, .<br />

Multiplying,<br />

(Distributive property)<br />

Simplifying,<br />

(Commutative property)<br />

Since 0 a ! 4 0 a ! 0 2 <br />

(Squaring both sides)<br />

Substituting, 414 @b ! 0 2@ b ! 11a ! ! !<br />

# b 3@ b @ 2<br />

@ , 0 a ! 0<br />

2<br />

@ 2 2 11112 @b ! 12@<br />

@ 21@b ! b ! 0<br />

@ 2 2 4@ b ! @<br />

2<br />

@ 2cos u2 3 @b ! @ 2 0<br />

Solving for cos u,<br />

cos u 13 @b! @<br />

2<br />

4a ! # a<br />

! a! # b<br />

!<br />

12b ! # a<br />

! 3b ! # b<br />

!<br />

0<br />

22 @b ! @<br />

2<br />

cos u 13<br />

22 , @ b! @ 2 0<br />

a ! # 14a<br />

! b<br />

!<br />

2 3b ! # 14a<br />

! b<br />

!<br />

2 0<br />

1a ! 3b ! 2 # 14a<br />

! b<br />

!<br />

2 0<br />

Thus, cos 1 a 13 b u, u 126.2°<br />

22<br />

Therefore, the angle between the two vectors is approximately 126.2°.<br />

It is often necessary to square the magnitude of a vector expression. This is<br />

illustrated in the following example.<br />

EXAMPLE 4<br />

Proving that two vectors are perpendicular using the dot product<br />

If 0 x ! y ! 0 0 x ! y ! 0 , prove that the nonzero vectors, x ! and y ! , are perpendicular.<br />

NEL<br />

375

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