Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

126 0 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
27. By plotting the vertices, we can see that the parallelogram is determined by the
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vectors AB = (2, 3) and AD= (4, -2}. We know that the area of the parallelogram
determined by two vectors is equal to the length of the cross product of these vectors.
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In order to compute the cross product, we consider the vector AB as the three-
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dimensional vector (2, 3, 0) (and similarly for AD), and then the area of
parallelogram ABCD is
y
j
k
IA13 x ml= 2 3 o =l(o)i- (o) j+ (-4 - 12)kl=l-16kl = 16
4 - 2 0
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29. (a) Because the plane through P, Q, and R contains the vectors PQ and P R, a .vector orthogonal to both of these vectors
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(such as their cross product) is also orthogonal to the plane. Here PQ = ( - 3, 1, 2} and P R = (3, 2, 4}, so
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PQ x PR = ((1)(4)- (2)(2), (2)(3)- (-3)(4), (-3)(2)- (1)(3)) = (0, 18, - 9)
Therefore, (0, 18, - 9) (or any nonzero scalar multiple thereof, such as (0, 2, - 1)) is orthogonal to the plane through P , Q,
andR.
(b) Note that the area of the tr ~angle determined by P, Q, and R ls equal to half of the area of the
parallelogram determined by the three points. From part (a), the area of the parallelogram is
I.PQ x PR I = I (0, 18, -9) I = y'O + 324 + 81 = v'4o5 = 9v'5, so the area of the triangle is ~ · 9v'5 = ~ .J5.
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31 . (a) PQ = (4, 3, - 2) and P R = (5, 5, 1}, so a vector orthogonal to the plane through P, Q, and R is
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PQ x P R = ((3)(1) - ( -2)(5), ( -2)(5) - (4)(1), (4)(5) - (3) (5)) = (13, - 14, 5) [or any scalar mutiple thereof].
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(b) The area of the parallelogram determined by PQ and PR is
IPQ x P'RI = 1(13,-14, 5)1 = yl13 2 + (- 14) 2 +5 2 = v'395,sotheareaoftriangle PQR is ~v'39Q.
33. By Equation 14, the volume of the parallelepiped detennined by a , b , and cis the magnitude of their scalar triple product,
1 2 3 11 21 1- 1
which is a · (b x c) = -1 1 2 = 1 - 2
. 1 4 2
2 1 4
Thus the volume of the parallelepiped is 9 cubic units.
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35. a= PQ = (4, 2, 2}, b = PR = (3, 3, -1), and c = P S = (5,5, 1).
'
24 1 + 31-1 2 11
= 1(4-2)-2(- 4-4) + 3(-1-2) = 9.
so the volume of the parallelepiped is 16 cubic units.
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