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

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Achar a distância da reta r ao plano , nos casos: 
17º) r: x = 4 + 3t y = -1 + t z = t e : x – y – 2z + 4 = 0 
�⃗⃗� = (3, 1, 1) 
�⃗� = (1, -1, -2) 
P (4, -1, 0) 
d (r, ) = d (P, ) = 
|𝑎𝑥0+𝑏𝑦0+𝑐𝑧0+𝑑|
√𝑎2+ 𝑏2+ 𝑐2
 
d (P, ) = 
|1 . 4−1 .(−1)−2 . 0+4|
√12+ (−1)2+(−2)2
 = 
|4+1+ 4|
√6
 = 
9
√6
 
18º) r: x = 3 e : x + y – 12 = 0 
 y = 4 
P (3, 4, 0) 
�⃗� = (1, 1, 0) 
d (r, ) = 
|𝑎𝑥0+𝑏𝑦0+𝑐𝑧0+𝑑|
√𝑎2+ 𝑏2+ 𝑐2
 
d (r, ) = 
|1 . 3+1 . 4+0 . 0−12|
√12+ (1)2+(0)2
 = 
|3+4−12|
√2
 = 
5
√2
 
 
 
RESOLUÇÃO DO LIVRO VETORES E 
GEOMETRIA ANALÍTICA 
19º) r: x = 3 e : y = 0b 
 y = 4 
P (3, 4, 0) 
�⃗� = (0, 1, 0) 
d (r, ) = 
|𝑎𝑥0+𝑏𝑦0+𝑐𝑧0+𝑑|
√𝑎2+ 𝑏2+ 𝑐2
 
d (r, ) = 
|0 . 3+1 . 4+0 . 0|
√02+ (1)2+(0)2
 = 
|4|
√1
 = 4 
Achar a distância entre r1 e r2, nos casos: 
20º) r1: x = 2 – t y = 3 + t z = 1 – 2t 
 r2: x = t y = -1 – 3t z = 2t 
A1 (2, 3, 1) e 𝑣 1 = (-1, 1, -2) 
A2 (0, -1, 0) e 𝑣 2 = (1, -3, 2) 
𝐴1𝐴2⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ = (A2 – A1) → (0, -1, 0) – (2, 3, 1) = (-2, -4, -1) 
(𝑣 1, 𝑣 2, 𝐴1𝐴2⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ) = 
−1 1 −2
1 −3 2
−2 −4 −1
 = 
−3 2
−4 −1
 . (−1) - 
1 2
−2 −1
 . 1 + 
1 −3
−2 −4
 . (-2)= 
= (3 + 8) (-1) – (-1 + 4). 1 + (-4 – 6) . (-2) = -11 – 3 + 20 = 6 
𝑣 1 𝑥 𝑣 2 = 
𝑖 𝑗 �⃗� 
−1 1 −2
1 −3 2
 = 
1 −2
−3 2
 𝑖 - 
−1 −2
1 2
 𝑗 + 
−1 1
1 −3
 �⃗� = 
= (2 – 6) 𝑖 – (-2 + 2) 𝑗 + (3 – 1) �⃗� = (-4, 0, 2) 
d (r1, r2) = 
|(𝑣1⃗⃗ ⃗,𝑣2⃗⃗ ⃗,𝐴1𝐴2⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ )|
|𝑣1⃗⃗ x 𝑣2⃗⃗ |
 = 6
√(−4)2+ (0)2+(2)2
 = 
6
√20
 = 
3
√5
 
21º) r1: x = y = z r2 = x + 1 z = 2x – 1 
A (0, 0, 0) 
A2 (0, 1, -1) 
𝑣 1 = (1, 1, 1) 
𝑣 2 = (1, 1, 2) 
𝐴1𝐴2⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ = (A2 – A1) → (0, 1, -1) – (0, 0, 0) = (0, 1, -1) 
(𝑣 1, 𝑣 2, 𝐴1𝐴2⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ) = 
1 1 1
1 1 2
0 1 −1
 = 
1 2
1 −1
 . (1) - 
1 2
0 −1
 . 1 + 
1 1
0 1
 . (1)= 
= (-1 – 2) (1) – (-1 – 0) (1) + (1 – 0) (1) = -3 + 1 + 1 = -1 
𝑣 1 𝑥 𝑣 2 = 
𝑖 𝑗 �⃗� 
1 1 1
1 1 2
 = 
1 1
1 2
 𝑖 - 
1 1
1 2
 𝑗 + 
1 1
1 1
 �⃗� = 
= (2 – 1) 𝑖 – (2 – 1) 𝑗 + (1 – 1) �⃗� = (1, -1, 0) 
d (r1, r2) = 
|(𝑣1⃗⃗ ⃗,𝑣2⃗⃗ ⃗,𝐴1𝐴2⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ )|
|𝑣1⃗⃗ x 𝑣2⃗⃗ |
 = |−1|
√(1)2+ (−1)2+(0)2
 = 
1
√2
 
22º) r1: y = 2x z = 3 r2: (x, y,z) = (2, -1, 2) + t (1, -1, 3) 
A1 (0, 0, 3) 
A2 (2, -1, 2) 
𝑣1⃗⃗⃗⃗ = (1, 2, 0) 
𝑣2⃗⃗⃗⃗ = (1, -1, 3) 
𝐴1𝐴2⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ = (A2 – A1) → (2, -1, 2) – (0, 0, 3) = (2, -1, -1) 
(𝑣 1, 𝑣 2, 𝐴1𝐴2⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ) = 
1 2 0
1 −1 3
2 −1 −1
 = 
−1 3
−1 −1
 . (1) - 
1 3
2 −1
 . 2 + 
1 −1
2 −1
 . (0) = 
= (1 + 3) (1) – (-1 – 6) (2) + (-1 + 2) (0) = 4 + 14 + 0 = 18 
𝑣 1 𝑥 𝑣 2 = 
𝑖 𝑗 �⃗� 
1 2 0
1 −1 3
 = 
2 0
−1 3
 𝑖 - 
1 0
1 3
 𝑗 + 
1 2
1 −1
 �⃗� = 
= (6 + 0) 𝑖 – (3 – 0) 𝑗 + (-1 – 2) �⃗� = (6, -3, -3) 
d (r1, r2) = 
|(𝑣1⃗⃗ ⃗,𝑣2⃗⃗ ⃗,𝐴1𝐴2⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ )|
|𝑣1⃗⃗ x 𝑣2⃗⃗ |
 = 18
√(6)2+ (−3)2+(−3)2
 = 
18
√54
 = 
18
3√6
 = 
6
√6
 = 
6
√6
 . 
√6
√6
 = √6 
23º) r1: x = t + 1 y = t + 2 z = -2t – 2 
 r2: y = 3x – 1 z = -4x 
A1 (1, 2, -2) 
A2 (0, 1, 0) 
𝑣1⃗⃗⃗⃗ = (1, 1, -2) 
𝑣2⃗⃗⃗⃗ = (1, 3, -4) 
𝐴1𝐴2⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ = (A2 – A1) → (0, 1, 0) – (1, 2, -2) = (-1, -1, 2) 
(𝑣 1, 𝑣 2, 𝐴1𝐴2⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ) = 
1 1 −2
1 3 −4
−1 −1 2
 = 
3 −4
−1 2
 . (1) - 
1 −4
−1 2
 . 1 + 
1 3
−1 −1
 . (-2) = 
= (6 - 4) (1) – (2 – 4) (1) + (-1 + 3) (-2) = 2 + 2 - 4 = 0 
𝑣 1 𝑥 𝑣 2 = 
𝑖 𝑗 �⃗� 
1 1 −2
1 3 −4
 = 
1 −2
3 −4
 𝑖 - 
1 −2
1 −4
 𝑗 + 
1 1
1 3
 �⃗� = 
= (-4 + 6) 𝑖 – (-4 + 2) 𝑗 + (3 – 1) �⃗� = (2, 2, 2) 
d (r1, r2) = 
|(𝑣1⃗⃗ ⃗,𝑣2⃗⃗ ⃗,𝐴1𝐴2⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ )|
|𝑣1⃗⃗ x 𝑣2⃗⃗ |
 = 0
√(2)2+ (2)2+(2)2
 = 
0
√12
 = 0 
24º) r1: x = 3 y = 2 r2: x = 1 y = 4 
A (3, 2, 0) 
B (1, 4, 0) 
√d (r1, r2) = d (A, B) = (1 − 3)2 + (4 − 2)2 + (0 − 0)2 = √(2)2 + (2)2 = √8 = 2√2 u.c.