Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 13.4 THE CROSS PRODUCT ET SECTION 12.4 ¤ 115
13.4 The Cross Product ET 12.4
i j k
0 −2
1. a × b =
6 0 −2
=
0 8 0
8 0 i − 6 −2
0 0 j + 6 0
0 8 k
=[0− (−16)] i − (0 − 0) j +(48− 0) k =16i +48k
Now (a × b) · a = h16, 0, 48i·h6, 0, −2i =96+0− 96 = 0 and (a × b) · b = h16, 0, 48i·h0, 8, 0i =0+0+0=0,so
a × b is orthogonal to both a and b.
i j k
3 −2
3. a × b =
1 3 −2
=
−1 0 5
0 5 i − 1 −2
−1 5 j + 1 3
−1 0 k
=(15− 0) i − (5 − 2) j +[0− (−3)] k =15i − 3 j +3k
Since (a × b) · a =(15i − 3 j +3k) · (i +3j − 2 k) =15− 9 − 6=0, a × b is orthogonal to a.
Since (a × b) · b =(15i − 3 j +3k) · (−i +5k) =−15 + 0 + 15 = 0, a × b is orthogonal to b.
i j k
−1
5. a × b =
1 −1 −1
=
1
1
1
2
1
2
−1
i − 1
1
1
2
2
−1
j + 1
1
1
2
−1
k
2
1
= − 1 2 − (−1) i − 1
2 − (− 1 2 ) j + 1 − (− 1 2 ) k = 1 2 i − j + 3 2 k
Now (a × b) · a = 1
2 i − j + 3 2 k · (i − j − k) = 1 2 +1− 3 2 =0and
(a × b) · b = 1
i − j + 3 k 2 2 · 1
i + j + 1 k 2 2
= 1 − 1+ 3 4 4
=0,soa × b is orthogonal to both a and b.
i j k 7. a × b =
t t 2 t 3
t 2 t 3 t t 3 t t 2
=
i −
j +
2t 3t 2 1 3t 2 1 2t k
1 2t 3t 2
=(3t 4 − 2t 4 ) i − (3t 3 − t 3 ) j +(2t 2 − t 2 ) k = t 4 i − 2t 3 j + t 2 k
Since (a × b) · a = t 4 , −2t 3 ,t 2 · t,
t 2 ,t 3 = t 5 − 2t 5 + t 5 =0, a × b is orthogonal to a.
Since (a × b) · b = t 4 , −2t 3 ,t 2 · 1, 2t, 3t 2 = t 4 − 4t 4 +3t 4 =0, a × b is orthogonal to b.
9. According to the discussion preceding Theorem 8, i × j = k,so(i × j) × k = k × k = 0 [by Example 2].
11. (j − k) × (k − i) =(j − k) × k +(j − k) × (−i) by Property 3 of Theorem 8
= j × k +(−k) × k + j × (−i)+(−k) × (−i) by Property 4 of Theorem 8
=(j × k)+(−1)(k × k)+(−1)(j × i)+(−1) 2 (k × i) by Property 2 of Theorem 8
= i +(−1) 0 +(−1)(−k)+j = i + j + k by Example 2 and the
discussion preceding Theorem 8