Resolução do Livro Vetores e Geometria Analítica - Paulo Winterle (cápitulo 6, O PLANO) PARTE IV

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34º) Determinar m de modo que os planos 1 e 2 sejam perpendiculares: 
a) 1: mx + y – 3z – 1 = 0 e 2: 2x – 3my + 4z + 1 = 0 
�⃗� 1 = (m, 1, -3) e �⃗� 2 = (2, -3m, 4) 
1 ⊥ 2 → �⃗� 1 . �⃗� 2 = 0 → (m, 1, -3) . (2, -3m, 4) → 2m – 3m - 12 = 0 → -m = 12 → m = -12 
 
b) x = 2 – h + 2t 
 1: y = 2h + 3 e 2: 2mx + 4y – z + 1 = 0 
 z = t – 2h + 1 
�⃗� 1 = ? e �⃗� 2 = (2m, 4, -1) 
�⃗� = h = (-1, 2, -2) e 𝑣 = t = (2, 0, 1) 
�⃗� 1 = �⃗� x 𝑣 = 
𝑖 𝑗 �⃗� 
−1 2 −2
2 0 1
 = 
2 −2
0 1
 𝑖 - 
−1 −2
2 1
 𝑗 + 
−1 2
2 0
 �⃗� = 
= (2 + 0) 𝑖 – (-1 + 4) 𝑗 + (-0 – 4) �⃗� = (2, -3, -4) 
1 ⊥ 2 → �⃗� 1 . �⃗� 2 = 0 
(2, -3, -4) . (2m, 4, -1) = 0 
4m – 12 + 4 = 0 
RESOLUÇÃO DO LIVRO VETORES E 
GEOMETRIA ANALÍTICA 
4m – 8 = 0 
4m = 8 
m = 2 
 35º) Dados a reta r e o plano , determinar o valor de m para que se tenha I) r// e II) r ⊥ , nos casos: 
a) r: x = -3 + t, y = -1 + 2t, z = 4t e : mx – y – 2z – 3 = 0 
I → r //  → 𝑣 . �⃗� = 0 
(1, 2, 4) . (m, -1, -2) = 0 
m – 2 – 8 = 0 
m = 10 
II → r ⊥  → 𝑣 = 𝛂 �⃗� 
(1, 2, 4) = α . (m, -1, -2) 
(1, 2, 4) = (αm, - α, -2α) 
1 = αm → 1 = -2m → m = -1/2 
2 = -α → α = -2 
4 = -2 α → α -2 
Logo temos que m = 10 e m = -1/2 
b) r: (x, y, z) = (1, 2, 0) + t(2, m, -1) e : 3x + 2y + mz = 0 
I → r //  → 𝑣 . �⃗� = 0 
(2, m, -1) . (3, 2, m) = 0 
6 + 2m – m = 0 
m = -6 
II → r ⊥  → 𝑣 = α �⃗� 
(2, m, -1) = α (3, 2, m) 
(2, m, -1) = (3α, 2α, mα) 
2 = 3α → α = 2/3 
m = 2𝛼 → m = 2 . (2/3) → m = 4/3 
-1 = m α → -1 = m . (2/3) → m = -3/2 
Logo m = -6 e não existe valor para m. 
36º) Verificar se a reta r está contida no plano do : 
a) r: y = 4x + 1 e : 2x + y – 3z – 4 = 0 
 z = 2x – 1 
𝑣 = (1, 4, 2) e �⃗� = (2, 1, -3) 
𝑣 . �⃗� = 0 
(1, 4, 2) . (2, 1, -3) = 0 
2 + 4 – 6 = 0 
0 = 0 
Logo temos que a reta está contida no plano. 
 x = h + t 
b) r: x – 2 = 
𝒚+𝟐
𝟐
 = z + 3 e 1: y = -1 + 2h – 3t 
 z = -3 + h – t 
�⃗� = (1, 2, 1) e 𝑣 = (1, -3, -1) 
�⃗� = �⃗� x 𝑣 = 
𝑖 𝑗 �⃗� 
1 2 1
1 −3 −1
 = 
2 1
−3 −1
 𝑖 - 
1 1
1 −1
 𝑗 + 
1 2
1 −3
 �⃗� = 
= (-2 + 3) 𝑖 – (-1 – 1) 𝑗 + (-3 – 2) �⃗� = (1, 2, -5) 
�⃗� = (1, 2, -5) 
𝑣 . �⃗� = 0 
(1, -3, -1) . (1, 2, -5) = 0 
1 – 6 + 5 = 0 
0 = 0 
Logo a reta está contida no plano. 
Nos problemas de 37 a 39, calcular os valores de m e n para que a reta r esteja contida no plano : 
37º) x = -2 + t 
 r: y = 3 – 2t e : mx + 2y – 3z + n = 0 
 z = 2t 
𝑣 = (1, -2, 2) e �⃗� = (m, 2, -3) 
𝑣 . �⃗� = 0 
(1, -2, 2) . (m, 2, -3) = 0 
m – 4 – 6 = 0 
m = 10 
A (-2, 3, 0) e m = 10 
: mx + 2y – 3z + n = 0 
10 (-2) + 2 . 3 – 3 . 0 + n = 0 
-20 + 6 – 0 + n = 0 
n = 14 
38º) r: y = 2x – 1 e : 5x – ny + z + 2 = 0 
 z = -x + m 
𝑣 = (1, 2, -1) e �⃗� = (5, -n, 1) 
𝑣 . �⃗� = 0 
(1, 2, -1) . (5, -n, 1) = 0 
5 – 2n – 1 = 0 
-2n = -4 
n = 2 
A (0, -1, m) e n = 2 
: 5x – ny + z + 2 = 0 
5 . 0 – 2 (-1) + m + 2 = 0 
0 + 2 + m + 2 = 0 
m = -4 
 x = 1 + 3t 
39º) y = -2 + mt e : 3x – 3y + z – 7 = 0 
 z = n – 4t 
𝑣 = (3, m -4) e �⃗� = (3, -3, 1) 
𝑣 . �⃗� = 0 
(3, m, -4) . (3, -3, 1) = 0 
9 – 3m – 4 = 0 
-3m = -5 
m = 5/3 
A (1, -2, n) 
: 3x – 3y + z – 7 = 0 
3 . 1 – 3 (-2) + n – 7 = 0 
3 + 6 + n – 7 = 0 
n = -2 
Nos problemas de 40 a 42, estabelecer equações reduzidas na variável x da reta de interseção dos 
planos: 
40º) 1: 3x – y + 2z – 1 = 0 e 2: x + 2y – 3z – 4 
Vamos usar o x = 0 
3 (0) – y + 2z – 1 = 0 → -y + 2z – 1 = 0 x (2) 
0 + 2y – 3z – 4 = 0 → 2y – 3z – 4 = 0 
-2y + 4z – 2 = 0 
2y – 3z – 4 = 0 
z – 6 = 0 
z = 6 
 
- y + 2z – 1 = 0 
-y + 2 . 6 – 1 = 0 
-y + 12 – 1 = 0 
-y = -11 
y = 11 
A (x, y, z) → (0, 11, 6) 
�⃗� 1 = (3, -1, 2) e �⃗� 2 = (1, 2, -3) 
𝑣 = �⃗� 1 x �⃗� 2 = 
𝑖 𝑗 �⃗� 
3 −1 2
1 2 −3
 = 
−1 2
2 −3
 𝑖 - 
3 2
1 −3
 𝑗 + 
3 −1
1 2
 �⃗� 
(3 – 4) 𝑖 – (-9 – 2) 𝑗 + (6 + 1) �⃗� = (-1, 11, 7) 
 x = -t 
r: y = 11 – 11t → r: y = -11x + 11 
 z = 6 + 7t z = -7x + 6 
41º) 1: 3x – 2y – z – 1 = 0 e 2: x + 2y – z – 7 = 0 
Vamos usar o x = 0 
3 . 0 – 2y – z – 1 = 0 → -2y – z – 1 = 0 
0 + 2y – z – 7 = 0 2y – z – 7 = 0 
-2y – (-4) – 1 = 0 -2z – 8 = 0 
-2y = -3 2z = -8 
y = 3/2 z = -4 
A (x, y, z) → (0, 3/2, -4) 
�⃗� 1 = (3, -2, -1) e �⃗� 2 = (1, 2, -1) 
𝑣 = �⃗� 1 x �⃗� 2 = 
𝑖 𝑗 �⃗� 
3 −2 −1
1 2 −1
 = 
−2 −1
2 −1
 𝑖 - 
3 1
1 −1
 𝑗 + 
3 −2
1 2
 �⃗� = 
= (2 + 2) 𝑖 – (-3 + 1) 𝑗 + (6 + 2) �⃗� = (4, 2, 8) 
 x = 4 
r: y = 3/2 + 2t 
 z = -4 + 8t 
𝑥
4
 = 
𝑦−3/2
2
 = 
𝑧+4
8
 
𝑥
4
 = 
𝑦−3/2
2
 
4y – 6 = 2x 
4y = 2x – 6 
y = 
2𝑥−6
4
 
y = 
1
2
x - 
3
2
 
 
𝑥
4
 = 
𝑧+4
8
 
4z + 16 = 8x 
4z = 8x – 16 
z = 2x – 4 
r: y = ½x + 3/2 
 z = 2x - 4 
42º) 1: x + y – z + 2 = 0 e 2: x + y + 2z – 1 = 0 
Usando x = 0 
0 + y – z + 2 = 0 x (-1) → -y + z – 2 = 0 
0 + y + 2z – 1 = 0 y + 2z – 1 = 0 
y – 1 + 2 = 0 3z – 3 = 0 
y = -1 z = 1 
A (x, y, z) → (0, -1, 1) 
�⃗� 1 = (1, 1, -1) e �⃗� 2 = (1, 1, 2) 
𝑣 = �⃗� 1 x �⃗� 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) �⃗� = (3, -3, 0) 
 x = 3t 
r: y = -1 – 3t 
 z = 1 
𝑥
3
 = 
𝑦+1
−3
 ; z = 1 
𝑥
3
 = 
𝑦+1
−3
 
3y + 3 = -3x 
3y = -3x – 3 
y = 
−3𝑥−3
3
 
y = -x – 1 
r: y = -x – 1 
 z = 1 
Nos problemas 43 e 44, encontrar equações paramétricas da reta interseção dos planos: 
43º) 1: 3x + y – 3z – 5 = 0 e 2: x – y – z – 3 = 0 
3 . 0 + y – 3z – 5 = 0 → y – 3z – 5 = 0 
0 – y – z – 3 = 0 -y – z – 3 = 0 
y – 3 (-2) – 5 = 0 -4z – 8 = 0 
y + 6 – 5 = 0 z = -2 
y = -1 
A (x, y, z) → (0, -1, -2) 
�⃗� 1 = (3, 1, -3) e �⃗� 2 = (1, -1, -1) 
𝑣 = 𝑛1⃗⃗ ⃗⃗ x �⃗� 2 = 
𝑖 𝑗 �⃗� 
3 1 −3
1 −1 −1
 = 
1 −3
−1 −1
 𝑖 - 
3 −3
1 −1
 𝑗 + 
3 1
1 −1
 �⃗� = 
= (-1 – 3) 𝑖 – (-3 + 3) 𝑗 + (-3 – 1) �⃗� = (-4, 0, -4) 
-1/4 . (-4, 0, -4) = (1, 0, 1) 
 
 x = t 
r: y = -1 
 z = t - 2 
44º) 1: 2x + y – 4 = 0 e 2: z = 5 
Usando o x = 0 
2 . 0 + y – 4 = 0 → y = 4 
z = 5 z = 5 
A (x, y, z) → (0, 4, 5) 
�⃗� 1 = (2, 1, 0) e �⃗� 2 = (1, 1, 5) 
𝑣 = 𝑛1⃗⃗ ⃗⃗ x �⃗� 2 = 
𝑖 𝑗 �⃗� 
2 1 0
1 1 5
 = 
1 0
1 5
 𝑖 - 
2 0
1 5
 𝑗 + 
2 1
1 1
 �⃗� = 
= (5 – 0) 𝑖 – (10 – 0) 𝑗 + (2 – 1) �⃗� = (5, -10, 1) 
1/5 . (5, -10, 1) = (1, -2, 1/5) 
 x = t 
r: y = 4 – 2t 
 z = 5