Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 13.4 THE CROSS PRODUCT ET SECTION 12.4 ¤ 117
25. a × (b + c) =a ×hb 1 + c 1,b 2 + c 2,b 3 + c 3i
= ha 2 (b 3 + c 3 ) − a 3 (b 2 + c 2 ) , a 3 (b 1 + c 1 ) − a 1 (b 3 + c 3 ) , a 1 (b 2 + c 2 ) − a 2 (b 1 + c 1 )i
= ha 2 b 3 + a 2 c 3 − a 3 b 2 − a 3 c 2 , a 3 b 1 + a 3 c 1 − a 1 b 3 − a 1 c 3 , a 1 b 2 + a 1 c 2 − a 2 b 1 − a 2 c 1 i
= h(a 2 b 3 − a 3 b 2 )+(a 2 c 3 − a 3 c 2 ) , (a 3 b 1 − a 1 b 3 )+(a 3 c 1 − a 1 c 3 ) , (a 1 b 2 − a 2 b 1 )+(a 1 c 2 − a 2 c 1 )i
= ha 2b 3 − a 3b 2,a 3b 1 − a 1b 3,a 1b 2 − a 2b 1i + ha 2c 3 − a 3c 2,a 3c 1 − a 1c 3,a 1c 2 − a 2c 1i
=(a × b)+(a × c)
27. By plotting the vertices, we can see that the parallelogram is determined by the
vectors −→
AB = h2, 3i and −→
AD = h4, −2i. We know that the area of the parallelogram
determined by two vectors is equal to the length of the cross product of these vectors.
In order to compute the cross product, we consider the vector −→
AB as the threedimensional
vector h2, 3, 0i (and similarly for −→
AD), and then the area of
parallelogram ABCD is
−→
AB × −→
i j k
AD = 2 3 0
= |(0) i − (0) j +(−4 − 12) k| = |−16 k| =16
4 −2 0
29. (a) Because the plane through P , Q, andR contains the vectors −→ −→
PQand PR, a vector orthogonal to both of these vectors
(such as their cross product) is also orthogonal to the plane. Here −→
−→
PQ = h−1, 2, 0i and PR = h−1, 0, 3i,so
−→
PQ× −→
PR = h(2)(3) − (0)(0), (0)(−1) − (−1)(3), (−1)(0) − (2)(−1)i = h6, 3, 2i
Therefore, h6, 3, 2i (or any scalar multiple thereof) is orthogonal to the plane through P , Q,andR.
(b) Note that the area of the triangle determined by P , Q,andR is equal to half of the area of the parallelogram determined by
the three points. From part (a), the area of the parallelogram is −→ −→ √ PQ× PR = |h6, 3, 2i| = 36 + 9 + 4 = 7, so the area
of the triangle is 1 2 (7) = 7 2 .
31. (a) −→
PQ = h4, 3, −2i and
−→
PR = h5, 5, 1i, so a vector orthogonal to the plane through P , Q, andR is
−→
PQ× −→
PR = h(3)(1) − (−2)(5), (−2)(5) − (4)(1), (4)(5) − (3)(5)i = h13, −14, 5i [or any scalar mutiple thereof].
(b) The area of the parallelogram determined by −→ −→
PQ and PR is
−→ −→
PQ× PR = |h13, −14, 5i| = 132 +(−14) 2 +5 2 = √ √
390, so the area of triangle PQRis 1 2 390.
33. We know that the volume of the parallelepiped determined by a, b,andc is the magnitude of their scalar triple product, which
6 3 −1
1 2
is a · (b × c) =
0 1 2
=6
4 −2 5
−2 5 − 3 0 2
4 5 + (−1) 0 1
=6(5+4)− 3(0 − 8) − (0 − 4) = 82.
4 −2
Thus the volume of the parallelepiped is 82 cubic units.