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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