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

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126 0 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE<br />

27. By plotting the vertices, we can see that the parallelogram is determin<strong>ed</strong> by the<br />

---+ ---+<br />

vectors AB = (2, 3) and AD= (4, -2}. We know that the area of the parallelogram<br />

determin<strong>ed</strong> by two vectors is equal to the length of the cross product of these vectors.<br />

. ---+<br />

In order to compute the cross product, we consider the vector AB as the three-<br />

---+<br />

dimensional vector (2, 3, 0) (and similarly for AD), and then the area of<br />

parallelogram ABCD is<br />

y<br />

j<br />

k<br />

IA13 x ml= 2 3 o =l(o)i- (o) j+ (-4 - 12)kl=l-16kl = 16<br />

4 - 2 0<br />

---+ ---+<br />

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<br />

---+ ---+<br />

(such as their cross product) is also orthogonal to the plane. Here PQ = ( - 3, 1, 2} and P R = (3, 2, 4}, so<br />

---+ ---+<br />

PQ x PR = ((1)(4)- (2)(2), (2)(3)- (-3)(4), (-3)(2)- (1)(3)) = (0, 18, - 9)<br />

Therefore, (0, 18, - 9) (or any nonzero scalar multiple thereof, such as (0, 2, - 1)) is orthogonal to the plane through P , Q,<br />

andR.<br />

(b) Note that the area of the tr ~angle determin<strong>ed</strong> by P, Q, and R ls equal to half of the area of the<br />

parallelogram determin<strong>ed</strong> by the three points. From part (a), the area of the parallelogram is<br />

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.<br />

---t ---+<br />

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<br />

---+ ---+<br />

PQ x P R = ((3)(1) - ( -2)(5), ( -2)(5) - (4)(1), (4)(5) - (3) (5)) = (13, - 14, 5) [or any scalar mutiple thereof].<br />

---+ ---+<br />

(b) The area of the parallelogram determin<strong>ed</strong> by PQ and PR is<br />

IPQ x P'RI = 1(13,-14, 5)1 = yl13 2 + (- 14) 2 +5 2 = v'395,sotheareaoftriangle PQR is ~v'39Q.<br />

33. By Equation 14, the <strong>vol</strong>ume of the parallelepip<strong>ed</strong> detennin<strong>ed</strong> by a , b , and cis the magnitude of their scalar triple product,<br />

1 2 3 11 21 1- 1<br />

which is a · (b x c) = -1 1 2 = 1 - 2<br />

. 1 4 2<br />

2 1 4<br />

Thus the <strong>vol</strong>ume of the parallelepip<strong>ed</strong> is 9 cubic units.<br />

---t --+ --+<br />

35. a= PQ = (4, 2, 2}, b = PR = (3, 3, -1), and c = P S = (5,5, 1).<br />

'<br />

24 1 + 31-1 2 11<br />

= 1(4-2)-2(- 4-4) + 3(-1-2) = 9.<br />

so the <strong>vol</strong>ume of the parallelepip<strong>ed</strong> is 16 cubic units.<br />

® 2012 Cengage LC1tning. All Richts Rescrn-d. M.:ty not be scnnncd. copi<strong>ed</strong>, or duplicntcd, or posccd to u publicly accessible website, in whole or in p:lrt.

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