Solução_Calculo_Stewart_6e

F.
114 ¤ CHAPTER 13 VECTORSANDTHEGEOMETRYOFSPACE ET TX.10 CHAPTER 12
of −−→
P 1 P 2 onto n.
−−→
|n ·hx2 − x1,y2 − y1i| |ax2 − ax1 + by2 − by1| |ax1 + by1 + c|
comp n P 1 P 2 = = √ = √
|n|
a2 + b 2 a2 + b 2
since ax 2 + by 2 = −c. The required distance is
|3 · −2+−4 · 3+5|
√
32 +4 2 = 13 5 .
51. For convenience, consider the unit cube positioned so that its back left corner is at the origin, and its edges lie along the
coordinate axes. The diagonal of the cube that begins at the origin and ends at (1, 1, 1) has vector representation h1, 1, 1i.
The angle θ between this vector and the vector of the edge which also begins at the origin and runs along the x-axis [that is,
h1, 1, 1i·h1, 0, 0i
h1, 0, 0i]isgivenbycos θ =
|h1, 1, 1i| |h1, 0, 0i| = √ 1
⇒ θ =cos −1 √
1
3
3
≈ 55 ◦ .
53. Consider the H — C — H combination consisting of the sole carbon atom and the two hydrogen atoms that are at (1, 0, 0) and
(0, 1, 0) (or any H — C — H combination, for that matter). Vector representations of the line segments emanating from the
carbon atom and extending to these two hydrogen atoms are 1 − 1 2 , 0 − 1 2 , 0 − 1 2
=
1
2 , − 1 2 , − 1 2
and
0 −
1
, 1 − 1 , 0 − 1
2 2 2 = −
1
, 1 , − 1
2 2 2 . The bond angle, θ, is therefore given by
1 , − 1 , −
1
2 2 2 · −
1
2
cos θ =
, 1 , − 1 2 2
1
, − 1 , − 1 2 2 2 −
1
, 1 , − 1
= − 1 − 1 + 1 4 4 4
= − 1 3
2 2 2
55. Let a = ha 1,a 2,a 3i and = hb 1,b 2,b 3i.
Property 2: a · b = ha 1,a 2,a 3i·hb 1,b 2,b 3i = a 1b 1 + a 2b 2 + a 3b 3
3
4
= b 1a 1 + b 2a 2 + b 3a 3 = hb 1,b 2,b 3i·ha 1,a 2,a 3i = b · a
3
4
⇒ θ =cos −1 − 1 3
≈ 109.5 ◦ .
Property 4: (c a) · b = hca 1,ca 2,ca 3i·hb 1,b 2,b 3i =(ca 1)b 1 +(ca 2)b 2 +(ca 3)b 3
= c (a 1 b 1 + a 2 b 2 + a 3 b 3 )=c (a · b) =a 1 (cb 1 )+a 2 (cb 2 )+a 3 (cb 3 )
= ha 1 ,a 2 ,a 3 i·hcb 1 ,cb 2 ,cb 3 i = a · (c b)
Property 5: 0 · a = h0, 0, 0i·ha 1,a 2,a 3i =(0)(a 1)+(0)(a 2)+(0)(a 3)=0
57. |a · b| = |a||b| cos θ = |a||b||cos θ|. Since|cos θ| ≤ 1, |a · b| = |a||b||cos θ| ≤ |a||b|.
Note: We have equality in the case of cos θ = ±1,soθ =0or θ = π, thus equality when a and b are parallel.
59. (a) The Parallelogram Law states that the sum of the squares of the
lengths of the diagonals of a parallelogram equals the sum of the
squares of its (four) sides.
(b) |a + b| 2 =(a + b) · (a + b) =|a| 2 +2(a · b)+|b| 2 and |a − b| 2 =(a − b) · (a − b) =|a| 2 − 2(a · b)+|b| 2 .
Adding these two equations gives |a + b| 2 + |a − b| 2 =2|a| 2 +2|b| 2 .