082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

248 Chapter 7 Vectors
Example7.20 Simplify (a ×b) − (b ×a).
Solution Usethe resulta ×b = −(b ×a) toobtain
(a×b)−(b×a)=(a×b)−(−(a×b))
=(a×b)+(a×b)
=2(a×b)
Example7.21 (a) Ifa =a 1
i +a 2
j +a 3
kandb =b 1
i +b 2
j +b 3
k, show that
a×b = (a 2
b 3
−a 3
b 2
)i− (a 1
b 3
−a 3
b 1
)j+ (a 1
b 2
−a 2
b 1
)k
(b) Ifa=2i+j+3kandb=3i+2j+kfinda×b.
Solution (a) a×b=(a 1
i+a 2
j+a 3
k)× (b 1
i+b 2
j+b 3
k)
=a 1
i× (b 1
i+b 2
j+b 3
k)+a 2
j× (b 1
i+b 2
j+b 3
k)
+a 3
k× (b 1
i+b 2
j+b 3
k)
=a 1
b 1
(i×i)+a 1
b 2
(i×j)+a 1
b 3
(i×k)+a 2
b 1
(j×i)+a 2
b 2
(j×j)
+a 2
b 3
(j×k)+a 3
b 1
(k×i)+a 3
b 2
(k×j)+a 3
b 3
(k×k)
Using the results of Examples 7.18 and 7.19, we find that the expression for a ×b
simplifies to
a×b = (a 2
b 3
−a 3
b 2
)i− (a 1
b 3
−a 3
b 1
)j+ (a 1
b 2
−a 2
b 1
)k
(b) Usingthe resultof part(a) we have
a ×b = ((1)(1) − (3)(2))i − ((2)(1) − (3)(3))j + ((2)(2) − (1)(3))k
=−5i+7j+k
Verifyforyourself thatb ×a = 5i −7j −k.
7.6.1 Usingdeterminantstoevaluatevectorproducts
Evaluationofavectorproductusingthepreviousformulaisverycumbersome.Amore
convenient and easily remembered method is now described. The vectors a and b are
written inthe following pattern:
∣ i j k ∣∣∣∣∣
a 1
a 2
a 3
∣b 1
b 2
b 3
This quantity is called a determinant. A more thorough treatment of determinants is
given in Section 8.7. To find the i component of the vector product, imagine crossing
out the row and column containing i and performing the following calculation on what
isleft,thatis
a 2
b 3
−a 3
b 2
Theresultingnumberistheicomponentofthevectorproduct.Thejcomponentisfound
by crossing out the row and column containing j, performing a similar calculation, but
now changing the sign of the result.Thus thejcomponent equals
−(a 1
b 3
−a 3
b 1
)