Resolução do Livro Vetores e Geometria Analítica - Paulo Winterle (cápitulo 3, PRODUTO VETORIAL) PARTE I

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1º) Se �⃗⃗� = 3 𝒊 – 𝒋 - 2 �⃗⃗� , �⃗⃗� = 2 𝒊 + 4 𝒋 - �⃗⃗� e �⃗⃗⃗� = - 𝒊 + �⃗⃗� , determinar: 
Informações importantes: �⃗� = (3, -1, -2), 𝑣 = (2, 4, -1) e �⃗⃗� = (-1, 0, 1) 
a) |�⃗⃗� x �⃗⃗� | 
|�⃗� x �⃗� | = 0, mas temos que �⃗� x �⃗� é: 
 𝑖 𝑗 �⃗� 
3 −1 −2
3 −1 −2
 
 𝑖 𝑗 
3 −1
3 −1
 = 2 𝑖 – 6 𝑗 – 3 �⃗� + 6 𝑗 – 2 𝑖 + 3 �⃗� = 0 
b) (2�⃗⃗� ) x (3�⃗⃗� ) 
 𝑖 𝑗 �⃗� 
4 8 −2
6 12 −3
 
 𝑖 𝑗 
4 8
6 12
 = -24 𝑖 – 10 𝑗 + 48 �⃗� + 12 𝑗 + 24 𝑖 – 48 �⃗� = 0 
c) (�⃗⃗� x �⃗⃗⃗� ) + (�⃗⃗⃗� x �⃗⃗� ) 
 𝑖 𝑗 �⃗� 
3 −1 −2
−1 0 1
 
 𝑖 𝑗 
3 −1
−1 0
 = - 𝑖 + 2 𝑗 – 3 𝑗 - �⃗� = (-1, -1, -1) 
�⃗⃗� x �⃗� = - �⃗� x �⃗⃗� = (1, 1, 1) 
(�⃗� x �⃗⃗� ) + (�⃗⃗� x �⃗� ) = 0 
d) (�⃗⃗� x �⃗⃗� ) x (�⃗⃗� x �⃗⃗� ) 
 
RESOLUÇÃO DO LIVRO VETORES E 
GEOMETRIA ANALÍTICA 
 𝑖 𝑗 �⃗� 
3 −1 −2
2 4 −1
 
 𝑖 𝑗 
3 −1
2 4
 = 𝑖 - 4 𝑗 + 12 �⃗� + 3 𝑗 + 8 𝑖 + 2 �⃗� = (9, -1, 14) 
𝑣 x �⃗� = - �⃗� x 𝑣 = (-9, 3, -14) 
(�⃗� x 𝑣 ) + (𝑣 x �⃗� ) = 0 
e) (�⃗⃗� - �⃗⃗� ) x �⃗⃗⃗� 
(�⃗� - 𝑣 ) = (1, -5, -1) 
 𝑖 𝑗 �⃗� 
1 −5 −1
−1 0 1
 
 𝑖 𝑗 
1 −5
−1 0
 = -5 𝑖 + 𝑗 - 𝑗 - 5 �⃗� = (-5, 0, -5) 
f) (�⃗⃗� x �⃗⃗� ) x �⃗⃗⃗� 
(�⃗� x 𝑣 ) = (9, -1, 14) 
 𝑖 𝑗 �⃗� 
9 −1 14
−1 0 1
 
 𝑖 𝑗 
9 −1
−1 0
 = - 𝑖 - 14 𝑗 - 3 𝑗 - �⃗� = (-1, -23, -1) 
g) �⃗⃗� x (�⃗⃗� x �⃗⃗⃗� ) 
(𝑣 x �⃗⃗� ) = 
 𝑖 𝑗 �⃗� 
2 4 −1
−1 0 1
 
 𝑖 𝑗 
2 4
−1 0
 = 4 𝑖 + 𝑗 - 2 𝑗 + 4 �⃗� = (4, -1, 4) 
 �⃗� x (𝑣 x �⃗⃗� ) = 
 
 𝑖 𝑗 �⃗� 
3 −1 −2
4 −1 4
 
 𝑖 𝑗 
3 −1
4 −1
 = (6, -20, 1) 
h) �⃗⃗� x (�⃗⃗� + �⃗⃗⃗� ) 
 
 𝑖 𝑗 �⃗� 
3 −1 −2
1 4 0
 
 𝑖 𝑗 
3 −1
1 4
 = (8, -2, 13) 
i) �⃗⃗� x �⃗⃗� + �⃗⃗� x �⃗⃗⃗� 
�⃗� x (𝑣 + �⃗⃗� ) = (-8, -2, 13) 
j) (�⃗⃗� x �⃗⃗� ) . �⃗⃗� 
(�⃗� x 𝑣 ) = (9, -1, 14) 
(�⃗� x 𝑣 ) . 𝑣 = (9, -1, 14) . (2, 4, -1) = 0 
k) (�⃗⃗� x �⃗⃗� ) . �⃗⃗⃗� 
(9, -1, 14) . (-1, 0, 1) = -9 + 14 = 5 
l) �⃗⃗� . (�⃗⃗� x �⃗⃗⃗� ) 
(3, -1, -2) . (4, -1, 4) = 12 + 1 – 8 = 5 
2º) Efetuar: 
a) 𝒊 𝒙 �⃗⃗� 
 
 𝑖 𝑗 �⃗� 
1 0 0
0 0 1
 
 𝑖 𝑗 
1 0
0 0
 = - 𝑗 
b) 𝒋 x (2 𝒊 ) 
2 . (𝑗 x 𝑖 ) = - �⃗� 
c) (3 𝒊 ) x (2 �⃗⃗� ) 
6 . (𝑖 x �⃗� ) = - 6 𝑗 
d) 𝒊 . (𝒋 x �⃗⃗� ) 
= 𝑖 . 𝑖 = 1 
e) (3 𝒊 ) . (2 𝒋 ) 
-6 (𝑖 . 𝑗 ) = 0 
f) (3 𝒊 ) x (2 𝒋 ) 
= 6 �⃗� 
g) 𝒊 . (𝒋 x 𝒊 ) 
= 0 
h) 𝒋 . (𝒋 . �⃗⃗� ) 
= 0 
i) (𝒊 x 𝒋 ) x �⃗⃗� 
= 0 
j) (𝒊 x 𝒋 ) x 𝒋 
= - 𝑖 
k) 𝒊 x (𝒋 x 𝒋 ) 
= 0 
l) (𝒋 x �⃗⃗� ) . 𝒊 
= 1 
3º) Dados os pontos A (2, 1, -1), B (3, 0, 1) e C (2, -1, -3), determinar o ponto D tal que 𝑨𝑫⃗⃗⃗⃗⃗⃗ = 𝑩𝑪⃗⃗⃗⃗⃗⃗ x 𝑨𝑪⃗⃗⃗⃗ ⃗. 
𝐴𝐷⃗⃗ ⃗⃗ ⃗ = 𝐵𝐶⃗⃗⃗⃗ ⃗ x 𝐴𝐶⃗⃗⃗⃗ ⃗ → D = (𝐵𝐶⃗⃗⃗⃗ ⃗ x 𝐴𝐶⃗⃗⃗⃗ ⃗) + A → D = (-6, -2, 2) + (2, 1, -1) → D = (-4, -1, 1), mas, 
𝐵𝐶⃗⃗⃗⃗ ⃗ = C – B → (-1, -1, -4) e 𝐴𝐶⃗⃗⃗⃗ ⃗ = C – A → (0, -2, -2) 
 
 𝑖 𝑗 �⃗� 
−1 −1 −4
0 −2 −2
 
 𝑖 𝑗 
−1 −1
0 0
 = (-6, -2, 2) 
4º) Determinar o vetor �⃗⃗� tal que �⃗⃗� . (1, 4, -3) = -7 e �⃗⃗� x (4, -2, 1) = (3, 5, -2). 
Tomando �⃗⃗� = (x, y, z) temos, 
�⃗⃗� . (1, 4, -3) = -7 
(x, y, z) . (1, 4, -3) = -7 
x + 4y – 3z = -7 
�⃗⃗� x (4, -2, 1) = (3, 5, -2), mas 
�⃗⃗� x (4, -2, 1) = 
 𝑖 𝑗 �⃗� 
𝑥 𝑦 𝑧
4 −2 1
 
 𝑖 𝑗 
𝑥 𝑦
4 −2
 = y 𝑖 + 4 z𝑗 – 2x �⃗� - x𝑗 + 2 z𝑖 – 4 y�⃗� = (y + 2z, 4z – x, -2x – 4y) 
(y + 2z, 4z – x, -2x – 4y) = (3, 5, -2) 
logo, 
y + 2z = 3 → y = 3 – 2z 
4z – x = 5 → -x = 5 – 4z → x = 4z – 5 
-2x – 4y = -2 → x + 2y = 1 
 
x + 4y – 3z = -7 
4z – 5 + 4 (3 – 2z) – 3z = -7 
4z – 5 + 12 – 8z – 3z = -7 
-7z + 7 = -7 
-7z = -14 
z = 2 
 
y = 3 – 2 (2) 
y = -1 
 
x + 2y = 1 
x + 2(-1) = 1 
x – 2 = 1 
x = 3 
Logo temos que 𝑣 = (3, -1, 2) 
5º) Resolver os sistemas: 
a) �⃗⃗� x 𝒋 = �⃗⃗� 
 �⃗⃗� . (4 𝒊 – 2 𝒋 + �⃗⃗� ) = 10 
 
 
𝑥 x 𝑗 = 
 𝑖 𝑗 �⃗� 
𝑥 𝑦 𝑧
0 1 0
 
 𝑖 𝑗 
𝑥 𝑦
0 1
 = x �⃗� + z 𝑖 = (z, 0, x) 
𝑥 x 𝑗 = �⃗� → (z, 0, x) = (0, 0, 1) = z = 0 e x = 1 
𝑥 . (4 𝑖 – 2 𝑗 + �⃗� ) = 10 → 4x – 2y + z = 10 
4x – 2y + 0 = 10 
4 – 2y + 0 = 10 
-2y = 10 – 4 
y = -3 
Logo 𝑥 = (1, -3, 0) 
 
b) �⃗⃗� x (2 𝒊 - 𝒋 + 3 �⃗⃗� ) = �⃗⃗� 
�⃗⃗� . (𝒊 + 2 𝒋 – 2 �⃗⃗� ) = 12 
𝑥 x (2, -1, 3) = 
 𝑖 𝑗 �⃗� 
𝑥 𝑦 𝑧
2 −1 3
 
 𝑖 𝑗 
𝑥 𝑦
2 −1
 = 3 y𝑖 + 2z 𝑗 - x�⃗� – 3x 𝑗 + z𝑖 – 2y �⃗� = (3y + z, 2z – 3x, -2x – 2y) 
𝑥 x (2, -1, 3) = 0⃗ 
3y = z = 0 → 3y = -z → y = -z/3 
2z – 3x = 0 → -3x = -2z → x = 2z/3 
-2x – 2y = 0 
𝑥 . (1, 2, -2) = 12 
x + 2y – 2z = 12 
2z/3 – 2z/3 – 2z = 12 
z = -6 
 
y = 6/3 
y = 2 
 
x = 2 . (-6)/3 
x = -12/3 
x = -4 
Logo temos que 𝑥 = (-4, 2 -6) 
6º) Dados os vetores �⃗⃗� = (3, 1, 1), �⃗⃗� = (-4, 1, 3) e �⃗⃗⃗� = (1, 2, 0), determinar �⃗⃗� de modo que �⃗⃗� ⊥ �⃗⃗⃗� e �⃗⃗� . 
�⃗⃗� = �⃗⃗� . 
Tomando 𝑥 = (x, y, z) 
𝑥 ⊥ �⃗⃗� → 𝑥 . �⃗⃗� = 0 
(x, y, z) . (1, 2, 0) = 0 
x + 2y = 0 
𝑥 x �⃗�  𝑣 
x . 𝑣 = 0 → (x, y, z) . (-4, 1, 3) → -4x + y + 3z = 0 
�⃗� . 𝑣  = → (3, 1, 1) . (-4, 1, 3) = -12 + 1 + 3 = -8 
Não existe 𝑥 pois 𝑣 teria que ser simultaneamente ortogonal a 𝑥 e �⃗� , o que não ocorre em �⃗� . 
7º) Levando em conta a figura 3.13, calcular: 
a) 𝑶𝑭⃗⃗⃗⃗⃗⃗ x 𝑶𝑫⃗⃗⃗⃗ ⃗⃗ 
 
 𝑖 𝑗 �⃗� 
𝑎 0 𝑎
0 𝑎 𝑎
 
 𝑖 𝑗 
𝑎 0
0 𝑎
 = a2 �⃗� – a2 𝑗 – a2 𝑖 = (-a2, -a2, a2), mas 
𝑂𝐹⃗⃗⃗⃗ ⃗ = F – O = (a, 0, a) – (0, 0, 0) = (a, 0, a) 
𝑂𝐷⃗⃗⃗⃗⃗⃗ = D – O → D = (0, a, a) 
b) 𝑨𝑪⃗⃗⃗⃗ ⃗ x 𝑭𝑨⃗⃗⃗⃗ ⃗ 
 
 𝑖 𝑗 �⃗� 
−𝑎 𝑎 0
0 0 −𝑎
 
 𝑖 𝑗 
−𝑎 𝑎
0 0
 = - a2 𝑖 – a2 𝑗 = (-a2, -a2, 0) mas, 
𝐴𝐶⃗⃗⃗⃗ ⃗ = C – A → (0, a, 0) – (a, 0, 0) = (-a, a, 0) 
𝐹𝐴⃗⃗⃗⃗ ⃗ = A – F = (a, 0, 0) – (a, 0, a) = (0, 0, -a) 
c) 𝑨𝑩⃗⃗⃗⃗⃗⃗ x 𝑨𝑪⃗⃗⃗⃗ ⃗ 
 
 𝑖 𝑗 �⃗� 
0 𝑎 0
−𝑎 𝑎 0
 
 𝑖 𝑗 
0 𝑎
−𝑎 𝑎
 = a2 �⃗� = (0, 0, a2), mas... 
𝐴𝐵⃗⃗⃗⃗ ⃗ = B – A → (a, a, 0) – (a, 0, 0) = (0, a, 0) 
𝐴𝐶⃗⃗⃗⃗ ⃗ = (-a, a, 0) 
d) 𝑬𝑪⃗⃗⃗⃗ ⃗ x 𝑬𝑨⃗⃗⃗⃗ ⃗ 
 
 𝑖 𝑗 �⃗� 
0 𝑎 −𝑎
𝑎 0 −𝑎
 
 𝑖 𝑗 
0 𝑎
𝑎 0
 = -a2 𝑖 – a2 𝑗 – a2�⃗� = (-a2, -a2, -a2), mas... 
𝐸𝐶⃗⃗⃗⃗ ⃗ = C – E → (0, a, 0) – (0, 0, a) = (0, a, -a) 
𝐸𝐴⃗⃗⃗⃗ ⃗ = A – E → (a, 0, 0) – (0, 0, a) = (a, 0, -a) 
e) 𝑶𝑨⃗⃗⃗⃗⃗⃗ . (𝑶𝑪⃗⃗⃗⃗⃗⃗ x 𝑶𝑬⃗⃗⃗⃗⃗⃗ ) 
 
 𝑖 𝑗 �⃗� 
0 𝑎 0
0 0 𝑎
 
 𝑖 𝑗 
0 𝑎
0 0
 = a2 𝑖 = (a2, 0, 0), mas... 
𝑂𝐴⃗⃗⃗⃗ ⃗ = A – O → (a, 0, 0) 
𝑂𝐶⃗⃗⃗⃗ ⃗ = C – O → (0, a, 0) 
𝑂𝐸⃗⃗⃗⃗ ⃗ = E – O → (0, 0, a) 
Logo 𝑂𝐴⃗⃗⃗⃗ ⃗ . (𝑂𝐶⃗⃗⃗⃗ ⃗ x 𝑂𝐸⃗⃗⃗⃗ ⃗) = (a, 0, 0) . (a2, 0, 0) = a3 
f) 𝑮𝑩⃗⃗⃗⃗⃗⃗ x 𝑨𝑭⃗⃗⃗⃗ ⃗ 
 
 𝑖 𝑗 �⃗� 
0 0 𝑎
0 0 𝑎
 = duas linhas iguais, logo, pela propriedade dos determinantes, o produto vetorial é 
 0⃗ . 
8º) Sejam os vetores �⃗⃗� = (1, -2, 1), �⃗⃗� = (1, 1, 1) e �⃗⃗⃗� = (1, 0, -1). 
a) Utilizar o produto escalar para mostrar que os vetores são, dois a dois, ortogonais. 
�⃗� x 𝑣 = 
 
 𝑖 𝑗 �⃗� 
1 −2 1
1 1 1
 
𝑖 𝑗 
1 −2
1 1
 = -2 𝑖 + 𝑗 + �⃗� - 𝑗 - 𝑖 + 2 �⃗� = (-3, 0, 3) 
(�⃗� x 𝑣 ) . �⃗� = (-3, 0, 3) . (1, -2, 1) = 0 
(�⃗� x 𝑣 ) . 𝑣 = (-3, 0, 3) . (1, 1, 1) = 0 
 
�⃗� x �⃗⃗� = 
 
 𝑖 𝑗 �⃗� 
1 −2 1
1 0 −1
 
𝑖 𝑗 
1 −2
1 0
 = 2 𝑖 + 𝑗 + 𝑗 + 2 �⃗� = (2, 2, 2) 
(�⃗� x �⃗⃗� ) . �⃗� = (2, 2, 2) . (1, -2, 1) = 0 
(�⃗� x �⃗⃗� ) . �⃗⃗� = (2, 2, 2) . (1, 0, -1) = 0 
 
𝑣 x �⃗⃗� = 
 
 𝑖 𝑗 �⃗� 
1 1 1
1 0 −1
 
𝑖 𝑗 
1 1
1 0
 = - 𝑖 + 𝑗 + 𝑗 - �⃗� = (-1, 2, -1) 
(𝑣 x �⃗⃗� ) . 𝑣 = (-1, 2, -1) . (1, 1, 1) = 0 
(𝑣 x �⃗⃗� ) . �⃗⃗� = (-1, 2, -1) . (1, 0, -1) = 0