Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 13.3 THE DOT PRODUCT ET SECTION 12.3 ¤ 111
43. a +(b + c) =ha 1,a 2i +(hb 1,b 2i + hc 1,c 2i) =ha 1,a 2i + hb 1 + c 1,b 2 + c 2i
= ha 1 + b 1 + c 1 ,a 2 + b 2 + c 2 i = h(a 1 + b 1 )+c 1 , (a 2 + b 2 )+c 2 i
= ha 1 + b 1 ,a 2 + b 2 i + hc 1 ,c 2 i =(ha 1 ,a 2 i + hb 1 ,b 2 i)+hc 1 ,c 2 i
=(a + b)+c
45. Consider triangle ABC, whereD and E are the midpoints of AB and BC. We know that −→
AB + −→
BC =
−→
AC (1) and
−−→
DB + −→ −−→
−−→ −→
BE = DE (2). However, DB =
1
AB, and −→ −→
BE =
1
BC. Substituting these expressions for DB −−→
and −→
BE into
2
−→ −→
(2)gives 1 AB + 1 BC = −−→
DE. Comparing this with (1)gives −−→ −→
DE = 1 AC. Therefore −→
AC and −−→
DE are parallel and
2 2
2
−−→
DE = 1 2
−→
AC.
2
13.3 The Dot Product ET 12.3
1. (a) a · b is a scalar, and the dot product is defined only for vectors, so (a · b) · c has no meaning.
(b) (a · b) c is a scalar multiple of a vector, so it does have meaning.
(c) Both |a| and b · c are scalars, so |a| (b · c) is an ordinary product of real numbers, and has meaning.
(d) Both a and b + c are vectors, so the dot product a · (b + c) has meaning.
(e) a · b is a scalar, but c is a vector, and so the two quantities cannot be added and a · b + c has no meaning.
(f ) |a| is a scalar, and the dot product is defined only for vectors, so |a|·(b + c) has no meaning.
3. a · b = −2, 1 3
·h−5, 12i =(−2)(−5) +
1
3
(12) = 10 + 4 = 14
5. a · b = 4, 1, 1 4
·h6, −3, −8i = (4)(6) + (1)(−3) +
1
4
(−8) = 19
7. a · b =(i − 2 j +3k) · (5 i +9k) = (1)(5) + (−2)(0) + (3)(9) = 32
9. a · b = |a||b| cos θ =(6)(5)cos 2π 3 =30 − 1 2
= −15
11. u, v, and w are all unit vectors, so the triangle is an equilateral triangle. Thus the angle between u and v is 60 ◦ and
u · v = |u||v| cos 60 ◦ = (1)(1)
1
2 =
1
. If w is moved so it has the same initial point as u, we can see that the angle
2
between them is 120 ◦ and we have u · w = |u||w| cos 120 ◦ =(1)(1)
− 1 2 = −
1
. 2
13. (a) i · j = h1, 0, 0i·h0, 1, 0i = (1)(0) + (0)(1) + (0)(0) = 0. Similarly, j · k = (0)(0) + (1)(0) + (0)(1) = 0 and
k · i = (0)(1) + (0)(0) + (1)(0) = 0.
Another method: Because i, j,andk are mutually perpendicular, the cosine factor in each dot product (see Theorem 3)
is cos π 2 =0.
(b) By Property 1 of the dot product, i · i = |i| 2 =1 2 =1since i is a unit vector. Similarly, j · j = |j| 2 =1and
k · k = |k| 2 =1.
15. |a| = (−8) 2 +6 2 =10, |b| =
√7 2
+32 =4,anda · b =(−8) √ 7 + (6)(3) = 18 − 8 √ 7. From Corollary 6,
we have cos θ = a · b
|a||b| = 18 − 8 √ 7
10 · 4
= 9 − 4 √ √
7
9 − 4 7
. So the angle between a and b is θ =cos −1 ≈ 95 ◦ .
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