Resolução do Livro Vetores e Geometria Analítica - Paulo Winterle (cápitulo 7, DISTÂNCIAS) PARTE I

Prévia do material em texto

Achar a distância de P1 à P2, nos casos: 
1º) P1 (-2, 0, 1) e P2 (1, -3, 2) 
𝑃1𝑃2ሬሬሬሬሬሬሬሬԦ = P2 – P1 → (1, -3, 2) – (-2, 0, 1) = (3, -3, 1) 
d (P1, P2) = ඥ(𝑥2 − 𝑥1)
2 + (𝑦2 − 𝑦1)
2 + (𝑧2 − 𝑧1)
2 
d (P1, P2) = ඥ32 + (−3)2 + 12 
d (P1, P2) = ξ9 + 9 + 1 
d (P1, P2) = ξ19 
2º) P1 (1, 0, 1) e P2 (2, -1, 0) 
𝑃1𝑃2ሬሬሬሬሬሬሬሬԦ = P2 – P1 → (2, -1, 0) – (1, 0, 1) = (1, -1, -1) 
d (P1, P2) = ඥ(𝑥2 − 𝑥1)
2 + (𝑦2 − 𝑦1)
2 + (𝑧2 − 𝑧1)
2 
d (P1, P2) = ඥ12 + (−1)2 + (−1)2 
d (P1, P2) = ξ1 + 1 + 1 
d (P1, P2) = ξ3 
Achar a distância do ponto P à reta r, nos casos: 
3º) P (2, 3, -1) r: x = 3 + t y = -2t z = 1 – 2t 
A (3, 0, 1) 
𝑣Ԧ = (1, -2, -2) 
RESOLUÇÃO DO LIVRO VETORES E 
GEOMETRIA ANALÍTICA 
𝐴𝑃ሬሬሬሬሬԦ = P – A → (2, 3, -1) – (3, 0, 1) = (-1, 3, -2) 
𝑣Ԧ x 𝐴𝑃ሬሬሬሬሬԦ = 
𝑖 𝑗Ԧ 𝑘ሬԦ 
1 −2 −2
−1 3 −2
 = 
−2 −2
3 −2
 𝑖Ԧ - 
1 −2
−1 −2
 𝑗Ԧ + 
1 −2
−1 3
 𝑘ሬԦ = 
(4 + 6) 𝑖Ԧ – (-2 – 2) 𝑗Ԧ + (3 – 2) 𝑘ሬԦ = (10, 4, 1) 
d (P, r) = 
|𝑣ሬԦ x 𝐴𝑃ሬሬሬሬሬԦ |
|𝑣ሬԦ |
 = 
|(10,4,1) |
|(1,−2,−2)|
 = 
ξ102+ 42+ 12
ඥ12+(−2)2+(−2)2
 = 
ξ117
ξ9
 = 
ξ117
3
 
4º) P (1, -1, 0) r: x = 2 – t y = 0 z = t 
A (2, 0, 0) 
𝑣Ԧ = (-1, 0, 1) 
: -x + z + d = 0 
Como P (1, -1, 0)  r: 
-x + z + d = 0 
-1 + 0 + d = 0 
d = 1 
Equação Geral: : -x + z + 1 = 0 → x – z – 1 = 0 
Para determinar I  r  , deve-se resolver a equação: 
x – z – 1 = 0 
2 – t – t – 1 = 0 
-2t + 1 = 0 
-2t = -1 
2t = 1 
t = ½ 
Substituindo t na equação r: 
I (2 – t, 0, t) → I (2 – ½, o, ½) → I (3/2, 0, ½) 
Por fim, temos que d (P, r) = d (P, I) = 
ට(
3
2
− 1)2 + (0 − 1)2 + (
1
2
− 0)2 = 
ට(
1
2
− 1)2 + (−1)2 + (
1
2
)2 = 
ට
1
4
+ 1 +
1
4
 = 
ට
1+4+1
4
 = 
ට
6
4
 = 
ξ6
2
 
5º) P (3, 2, 1) r: y = 2x z = x + 3 
A (0, 0, 3) 
𝑣Ԧ = (1, 2, 1) 
𝐴𝑃ሬሬሬሬሬԦ = P – A → (3, 2, 1) – (0, 0, 3) = (3, 2, -2) 
𝑣Ԧ x 𝐴𝑃ሬሬሬሬሬԦ = 
𝑖 𝑗Ԧ 𝑘ሬԦ 
1 2 1
3 2 −2
 = 
2 1
2 −2
 𝑖Ԧ - 
1 1
3 −2
 𝑗Ԧ + 
1 2
3 2
 𝑘ሬԦ = 
(-4 - 2) 𝑖Ԧ – (-2 – 3) 𝑗Ԧ + (2 – 6) 𝑘ሬԦ = (-6, 5, -4) 
d (P, r) = 
|𝑣ሬԦ x 𝐴𝑃ሬሬሬሬሬԦ |
|𝑣ሬԦ |
 = 
ඥ(−6)2+ 52+ (−4)2
ξ12+22+ 12
 = 
ξ77
ξ6
 
6º) P (0, 0, 0) r: 2x – y + z – 3 = 0 
 x + y – 2z + 1 = 0 
 
r: z = -2x + y + 3 
 y = -x + 2z – 1 
y = -x + 2 (-2x + y + 3) – 1 
y = -x – 4x + 2y + 6 – 1 
y = -5x + 2y + 5 
 
z = -2x – x + 2z – 1 + 3 
z = -3x + 2z + 2 
 
r: y = -5x + 5 (-1) → y = 5x - 5 
 z = -3x + 2 (-1) → z = 3x – 2 
 
Equação Paramétrica 
 x = t 
r: y = 5t – 5 
 z = 3t – 2 
𝑣Ԧ = (1, 5, 3) 
Q = (0, -5, -2) 
𝑃𝑄ሬሬሬሬሬԦ = Q – P → (0, -5, -2) – (0, 0, 0) = (0, -5, -2) 
𝑣Ԧ x 𝑃𝑄ሬሬሬሬሬԦ = 
𝑖 𝑗Ԧ 𝑘ሬԦ 
0 −5 −2
1 5 3
 = 
−5 −2
5 3
 𝑖Ԧ - 
0 −2
1 3
 𝑗Ԧ + 
0 −5
1 5
 𝑘ሬԦ = 
(-15 + 10) 𝑖Ԧ – (0 + 2) 𝑗Ԧ + (0 + 5) 𝑘ሬԦ = (-5, -2, 5) 
|𝑣Ԧ x 𝑃𝑄ሬሬሬሬሬԦ| = ඥ(−5)2 + (−2)2 + 52 = ξ54 
|𝑣Ԧ| = ξ12 + 52 + 32 = ξ35 
d (P, r) = 
|𝑣ሬԦ x 𝑃𝑄ሬሬሬሬሬԦ |
|𝑣ሬԦ |
 = 
ξ54
ξ35
 
 
7º) P (3, -1, 1) r: (x, y, z) = (2, 3, -1) + t (1, -4, 2) 
A (2, 3, -1) 
𝑣Ԧ = (1, -4, 2) 
𝐴𝑃ሬሬሬሬሬԦ = P – A → (3, -1, 1) – (2, 3, -1) = (1, -4, 2) 
𝑣Ԧ x 𝐴𝑃ሬሬሬሬሬԦ = 
𝑖 𝑗Ԧ 𝑘ሬԦ 
1 −4 2
1 −4 2
 = 
−4 2
−4 2
 𝑖Ԧ - 
1 2
1 2
 𝑗Ԧ + 
1 −4
1 −4
 𝑘ሬԦ = 
(-8 + 8) 𝑖Ԧ – (2 - 2) 𝑗Ԧ + (-4 + 4) 𝑘ሬԦ = (0, 0, 0) 
|𝑣Ԧ x 𝑃𝑄ሬሬሬሬሬԦ| = ඥ(−5)2 + (−2)2 + 52 = ξ54 
|𝑣Ԧ| = ξ12 + 52 + 32 = ξ35 
d (P, r) = 
|𝑣ሬԦ x 𝐴𝑃ሬሬሬሬሬԦ |
|𝑣ሬԦ |
 = 
|(0,0,0)|
|(1,−4,2)|
 = 
 ඥ(0)2+(0)2+ 02
 ඥ(−1)2+(−4)2+ 22
 = 
 ξ0
 ξ21
 = 0 
8º) P (1, 2, 3) r: eixo Ox 
𝑃ሬԦ x 𝑂𝑥 = 
𝑖 𝑗Ԧ 𝑘ሬԦ 
1 2 3
1 0 0
 = 
2 3
0 0
 𝑖Ԧ - 
1 3
1 0
 𝑗Ԧ + 
1 2
1 0
 𝑘ሬԦ = 
(0 - 0) 𝑖Ԧ – (0 - 3) 𝑗Ԧ + (0 - 2) 𝑘ሬԦ = (0, 3, -2) 
d (P, r) = 
|𝑃 x 𝑂𝑥 |
|𝑂𝑥|
 = 
|(0,3,−2)|
|(1,0,0)|
 = 
 ඥ(0)2+(3)2+ (−2)2
 ඥ(1)2+(0)2+ 02
 = 
 ξ13
 ξ1
 = ξ13