Matéria: Geometria Analítica Professor: Carlos Roberto Vianna

Aluno: Vitor Ferreira do Nascimento nº 68 GRR: 20157165

Exercício 2(a-h), pg.147 Livro: Álgebra Vetorial e Geometria Analítica, Jacir J.

Venturi

2-Sejam os pontos A=(68,0,0), B=(2,2,1) e C=(1,1,-1), determinar:

a) A medida do lado (a)

𝑎 = |𝐵𝐶| 𝐵𝐶 = (𝐵 − 𝐶) = (2,2,1) − (1,1, −1) logo 𝐵𝐶 = (1,1,2) então

|𝐵𝐶| = √𝐵𝐶2 ∴ |𝐵𝐶| = √12 + 12 + 22 = √6 a=√6 u.c

b) A medida do ângulo Â

𝐴𝐵 = (𝐵 − 𝐴)

(𝐵 − 𝐴) = (2,2,1) − (68,0,0)

𝐴𝐵 = (−66,2,1) |𝐴𝐵| = √−662 + 22 + 12 = √4361

𝐴𝐶 = (𝐶 − 𝐴) (𝐶 − 𝐴) = (1,1, −1) − (68,0,0) = (−67,1, −1)

|𝐴𝐶| = √−672 + 12 + (−1)2 = √4491

cos 𝜃 =𝐴𝐶. 𝐴𝐵

|𝐴𝐶|. |𝐴𝐵|=

(−67,1, −1). (−66,2,1)

√4491. √4361=

4423

√19585251

𝑎𝑟𝑐 𝑐𝑜𝑠 ≅ 2,26°

c) A área do triângulo ABC

(𝐵 − 𝐴)𝑥(𝐶 − 𝐴) =𝑖 𝑗 𝑘

−66 2 1−67 1 1

= (−3, −133,68)

|(𝐵 − 𝐴)𝑥(𝐶 − 𝐴)| = √−32 + (−1332) + 682 = √22322

𝑛 =(𝐵 − 𝐴)𝑥(𝐶 − 𝐴)

|(𝐵 − 𝐴)𝑥(𝐶 − 𝐴)|=

(−3, −133,68)

√22322 ∴ 𝑛 = (

−3

√22322,

−133

√22322,

68

√22322)

𝑆𝐴𝐵𝐶 = 12⁄ (−3, −133,68). (

−3

√22322,

−133

√22322,

68

√22322)

= 12⁄ (

9

√22322,

17689

√22322,

4624

√22322)

𝑆𝐴𝐵𝐶 = 12⁄ (

22322

√22322) ∴ 𝑆𝐴𝐵𝐶 = 70,7 𝑢. 𝑎.

d) A altura relativa do vértice do triângulo ABC

ℎ𝐴 =|(𝐴 − 𝐵)𝑥(𝐶 − 𝐵)|

|(𝐶 − 𝐵)|

(𝐴 − 𝐵) = (66, −2, −1) (𝐶 − 𝐵) = (−1, −1, −2) |(𝐶 − 𝐵)| = √6

(𝐴 − 𝐵)𝑥(𝐶 − 𝐵) =𝑖 𝑗 𝑘

66 −2 −1−1 −1 −2

= (3,133, −68)

|(𝐴 − 𝐵)𝑥(𝐶 − 𝐵)| = √32 + 1332 + (−68)2

ℎ𝐴 =√22322

√6 ∴ ℎ𝐴 = 70 𝑢. 𝑐.

e) O pé da normal baixada de A sobre a reta BC

𝑁 = 𝐵 + [(𝐴 − 𝐵). 𝑛]𝑛 (𝐴 − 𝐵) = (66, −2, −1)

𝑛 =(𝐵 − 𝐶)

|𝐵 − 𝐶|= (

1

√6,

1

√6,

2

√6)

𝑁 = (2,2,1) + [(66, −2, −1) (1

√6,

1

√6,

2

√6)] (

1

√6,

1

√6,

2

√6)

𝑁 = (2,2,1) + (62

√6) (

1

√6,

1

√6,

2

√6) ∴ 𝑁 = (2,2,1) + (

62

6,62

6,124

6)

𝑁 = (37

3,37

3,65

3) 𝑜𝑢 𝑁 = (12.333,12.333,21.667)

f) A altura relativa a O do tetraedro OABC

ℎ𝑶 = 𝑑(𝑂,𝛼) =(𝐵 − 𝐴)𝑥(𝐶 − 𝐴). (𝑂 − 𝐴)

|(𝐵 − 𝐴)𝑥(𝐶 − 𝐴)|

𝑑(𝑂,𝛼) =(−3, −133,68). (−68,0,0)

√22322 ∴ 𝑑(𝑂,𝛼) =

204

√22322

𝑑(𝑂,𝛼) ≅ 1,365 𝑢. 𝑐.

g) O pé da normal baixada de O sobre o plano ABC

𝑁 = 𝑂 + [(𝐴 − 𝑂). 𝑛]𝑛

𝑁 = [(68,0,0). (−3

√22322,

−133

√22322,

68

√22322)] (

−3

√22322,

−133

√22322,

68

√22322)

𝑁 = (−204

√22322) (

−3

√22322,

−133

√22322,

68

√22322)

𝑁 = (612

22322,

27132

22322,

−13872

22322) 𝑁 = (0.0274,1.2155, −0.6214)

h) O volume do tetraedro OABC

𝑉𝑶𝑨𝑩𝑪 =𝑉𝑝𝑟𝑖𝑠𝑚𝑎

𝟑

𝑉𝑂𝐴𝐵𝐶 =𝑆𝐴𝐵𝐶 . 𝑑(𝑂,𝛼)

3

𝑉𝑂𝐴𝐵𝐶 =70,7 ∗ 1,365

3 ∴ 𝑉𝑂𝐴𝐵𝐶 = 32,17 𝑢. 𝑣.