3ª PROVA DE VETORES E GEOMETRIA ANALÍTICA

Prévia do material em texto

VERIFIQUE SE O TRIÂNGULO DE VÉRTICES A(-3,-2,3), B(1,1,2) E 
C(-3,1,2) É RETÂNGULO. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
B 
C A 
𝐴𝐵 = B – A 
𝐴𝐵 = (1,1,2) – (-3,-2,2) 
𝐴𝐵 = (4,3,0) 
𝐴𝐶 = C – A 
𝐴𝐶 = (-3,1,2) – (-3,-2,2) 
𝐴𝐶 = (0,3,0) 
𝐵𝐴 = - 𝐴𝐵 
𝐵𝐴 = (-4,-3,0) 
 
𝐵𝐶 = C - B 
𝐵𝐶 = (-3,1,2) – (1,1,2) 
𝐵𝐶 = (-4,0,0) 
𝐶𝐴 = - 𝐴𝐶 
𝐶𝐴 = (0,-3,0) 
 
𝐶𝐵 = - 𝐵𝐶 
𝐶𝐵 = (4,0,0) 
 
𝐴𝐵 . 𝐴𝐶 = 0 
𝐴𝐵 . 𝐴𝐶 = (4,3,0) . (0,3,0) 
𝐴𝐵 . 𝐴𝐶 = (0 + 9 + 0) 
𝐴𝐵 . 𝐴𝐶 = 9 
𝐵𝐴 . 𝐵𝐶 = 0 
𝐵𝐴 . 𝐵𝐶 = (-4,-3,0) . (-4,0,0) 
𝐵𝐴 . 𝐵𝐶 = (16 + 0 + 0) 
𝐵𝐴 . 𝐵𝐶 = 16 
𝐶𝐴 . 𝐶𝐵 = 0 
𝐶𝐴 . 𝐶𝐵 = (0,-3,0) . (4,0,0) 
𝐶𝐴 . 𝐶𝐵 = (0 + 0 + 0) 
𝐶𝐴 . 𝐶𝐵 = 0 
Portanto o triângulo é retângulo em C. 
 
DADOS A(-3,-2, √ ) E B(-2,-3,-2 √ ), CALCULAR OS ÂNGULOS 
DIRETORES DO VETOR . 
 
 
 
 
 
 
 
 
DETERMINAR A PROJEÇÃO ORTOGONAL DE = (1,-3,-2) SOBRE 
 = (-1,1,1) E DECOMPOR COMO SOMA COM , SENDO 
 // E  . 
 
 
 
 
 
 
 
 
 
 
 
Cos  = 
𝑥
|𝐴𝐵 |
 
Cos  = 
1
2
 
Cos  = 60º 
 
Cos β = 
𝑦
|𝐴𝐵 |
 
Cos β = 
−1
2
 
Cos β = 120º 
 
Cos ϒ = 
𝑧
|𝐴𝐵 |
 
Cos ϒ = 
−√2
2
 
Cos ϒ = 135º 
AB = B – A 
AB = (-2,-3,-2√2) – (-3,-2, −√2) 
AB = (1,-1, −√2) 
|AB | = 12 + (−1)2 + (− 2)
2
 
AB = √1 + 1 + 2 
AB = √4 
AB = 2 
Proju 𝑣 = 
 𝑣 . 𝑢 
 𝑢 . 𝑢 
 . 𝑢 
Proju 𝑣 = 
−6
 3
 . (-1,1,1) 
Proju 𝑣 = (−2) . (-1,1,1) 
Proju 𝑣 = (2,-2,-2) 
 
V1 = Proju 𝑣 
V1 = (2,-2,-2) 
V = V1 + V2 
(1,-3,-2) = (2,-2,-2) + V2 
V2 = (1,-3,-2) – (2,-2,-2) 
V2 = (-1,-1,0) 
v . u = (1,-3,-2) . (-1,1,1) 
v . u = 1 . (-1) + (-3) . 1 + (-2) . 1 
v . u = -1 - 3 – 2 
v . u = -6 
u . u = (-1,1,1) . (-1,1,1) 
u . u = (-1) . (-1) + 1 . (1) + 1 . (1) 
u . u = 1 + 1 + 1 
u . u = 3 
CALCULAR A ÁREA DO TRIÂNGULO ABC, SABENDO QUE O PONTO 
C(-2,0,X) É EQUIDISTANTE DE A(0,-1,-2) E B(1,-1,-1). 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
|CA | = |CB | 
(√𝑋2 + 4𝑋 + 9 )2 = (√𝑋2 + 2𝑋 + 11 )2 
X2 + 4X + 9 = X2 + 2X +11 
X2 – X2 + 4X + 2X + 9 – 11 = 0 
2X – 2 = 0 
2X = 2 
X = 1 
CA = A – C 
CA = (0,-1,-2) – (-2,0,X) 
CA = (2,-1,-2-X) 
C𝐵 = B – C 
C𝐵 = (1,-1,-1) – (-2,0,X) 
C𝐵 = (3,-1,-1-X) 
|C𝐴 | = 22 + (−1)2 + (−𝑋 − 2 )2 
|C𝐴 |= 4 + 1 + (𝑋)2 + 2(−𝑋)(−2 ) + (−2)2 
|C𝐴 | = √5 + 𝑋2 + 4𝑋 + 4 
|C𝐴 | = √𝑋2 + 4𝑋 + 9 
|C𝐵 | = 32 + (−1)2 + (−𝑋 − 1 )2 
|C𝐵 |= 9 + 1 + (−𝑋)2 + 2(−𝑋)(−1 ) + (−1)2 
|C𝐵 | = √10 + 𝑋2 + 2𝑋 + 1 
|C𝐵 | = √𝑋2 + 2𝑋 + 11 
𝐴𝐵 = B – A 
𝐴𝐵 = (1,-1,-1) – (0,-1,-2) 
𝐴𝐵 = (1,0,1) 
𝐴𝐶 = C – A 
𝐴𝐶 = (-2,0,1) – (0,-1,-2) 
𝐴𝐶 = (-2,1,3) 
 
s∆ = 
|𝐴𝐵 𝑋 𝐴𝐶 |
2
 
|𝐴𝐵 𝑋 𝐴𝐶 | = 
𝑖 𝑗 𝑘 
1 0 1
−2 1 3
 
= 0 1
1 3
 𝑖 - 1 1
−2 3
 𝑗 + 1 0
−2 1
 𝑘 
= (0-1) 𝑖 - (3-(-2)) 𝑗 + (1-0) 
= (-1,-5,1) 
|𝐴𝐵 𝑋 𝐴𝐶 | = (−1)2 + (−5)2 + 12 
 𝐴𝐵 𝑋 𝐴𝐶 = √1 + 25 + 1 
 𝐴𝐵 𝑋 𝐴𝐶 = √27 
 𝐴𝐵 𝑋 𝐴𝐶 = √32. 3 
|𝐴𝐵 𝑋 𝐴𝐶 | = 3√3 
s∆ = 
|𝐴𝐵 𝑋 𝐴𝐶 |
2
 
s∆ = 
3√3
2
 u.a