Taller de algebra_compressed
Enviado por andrea03972021 • 11 de Julio de 2023 • Trabajos • 3.114 Palabras (13 Páginas) • 27 Visitas
[pic 1]
Determine 3u-2v si u= (-4,-3) y v= (5/2,2)
Al escribir su respuesta solo ingrese el vector resultado en la forma: (a, b)
5
3u − 2v = 3(−4, −3) − 2 ([pic 2]
2
, 2)
5
3u − 2v = (−4 × 3, −3 × 3) − ([pic 3]
2
× 2,2 × 2)
Punto 2:
Dado los vectores
3u − 2v = (−12, −9) − (5,4)
3u − 2v = (−12 − 5, −9 − 4)
3u − 2v = (−17, −13)
𝑢 = 〈2,3, −1〉
1
𝑣 = 〈−1,2, 〉[pic 4]
2
El ángulo alfa determinado por ellos está dado por la expresión:
𝑢⃗→ ∙ 𝑣→ = |𝑢⃗→| ∙ |𝑣→| cos 𝛼
[pic 5]
〈2,3, −1〉 ∙ 〈−1,2,
1〉 = |√(2)2 + (3)2 + (−1)2| ∙ |√(−1)2 + (2)2 + (1)
2 2[pic 6][pic 7][pic 8][pic 9]
| cos 𝛼
[pic 10]
1 21
(2 × −1) + (3 × 2) + (−1 ×
) = √14 ∙ √
2 4[pic 11][pic 12]
cos 𝛼
[pic 13]
7
2 = √14 ∙[pic 14]
7
√21
2 cos 𝛼[pic 15]
= cos 𝛼[pic 16]
√14 ∙ √21
[pic 17]
√6
6 = cos 𝛼[pic 18][pic 19]
[pic 20]
𝛼 = cos−1 (√6) =
6
[pic 21]
𝑢 = 〈5,0,0〉
𝑣 = 〈2.5 , 1.44 , 4.08〉
𝑤 = 〈2.5 , 4.33 , 0〉
𝑉𝑜𝑙𝑢𝑚𝑒𝑛 𝑡𝑒𝑡𝑟𝑎𝑒𝑑𝑟𝑜 = |𝑢⃗→ ∙ (𝑣→ × 𝑤⃗→)|
1
𝑉𝑜𝑙𝑢𝑚𝑒𝑛 𝑡𝑒𝑡𝑟𝑎𝑒𝑑𝑟𝑜 = 6 |[pic 22]
𝑉 𝑡𝑒𝑡𝑟𝑎𝑒𝑑𝑟𝑜 =
1
{|5(0 − 4.08𝑥4.33) − 0(0 − 4.08𝑥2.5) + 0(2.5𝑥4.33 − 1.44𝑥2.5)|}[pic 23]
6
1
𝑉𝑜𝑙𝑢𝑚𝑒𝑛 𝑡𝑒𝑡𝑟𝑎𝑒𝑑𝑟𝑜 =
|−88.332|
6[pic 24]
𝑉𝑜𝑙𝑢𝑚𝑒𝑛 𝑡𝑒𝑡𝑟𝑎𝑒𝑑𝑟𝑜 = 14.722 𝑢2
[pic 25]
1 | ||
6 | ||
5 | 0 | 0 |
2. | 5 1.44 | 4.08| |
2. | 5 4.33 | 0 |
𝐴 = 〈1,2,3〉
𝐵 = 〈3 , 2, 1〉
𝐴𝑟𝑒𝑎 = |𝐴𝑥𝐵|
𝑖 𝐴𝑟𝑒𝑎 = 𝑑𝑒𝑡 |1 | 𝑗 2 | 𝑘 3| |
3 | 2 | 1 |
𝐴𝑟𝑒𝑎 = |𝑖(2 − 6) − 𝑗(1 − 9) + 𝑘(2 − 6)|
𝐴𝑟𝑒𝑎 = |−4𝑖 + 8𝑗 − 4𝑘|
[pic 26]
𝐴𝑟𝑒𝑎 = √(−4)2 + (8)2 + (−4)2
[pic 27]
𝐴𝑟𝑒𝑎 = √96
[pic 28]
𝐴𝑟𝑒𝑎 = 4√6 ≈ 9.8 𝑐𝑚2
[pic 29]
〈𝑥 , 𝑦, 𝑧〉 = 𝑎→ 𝑥 𝑏⃗→
𝑖 〈𝑥 , 𝑦, 𝑧〉 = 𝑑𝑒𝑡 |3 | 𝑗 −1 | 𝑘 0| |
0 | −3 | 1 |
〈𝑥 , 𝑦, 𝑧〉 = 𝑖(−1 − 0) − 𝑗(3 − 0) + 𝑘(−9 − 0)
〈𝑥 , 𝑦, 𝑧〉 = −𝑖 − 3𝑗 − 9𝑘
𝑥 = −1 ; 𝑦 = −3 ; 𝑧 = −9
[pic 30]
[pic 31]
- Para determinar si el tetraedro es regular, debemos hallar magnitudes de todos los vectores determinados por las aristas del tetraedro y determinar que es el mismo valor.
𝐻1 = (1,1,1)
𝐻2 = (0,0,1)
𝐻3 = (0,1,0)
𝐻4 = (1,0,0)
⃗𝐻⃗⃗⃗1⃗⃗𝐻⃗⃗⃗⃗2→ = (0,0,1) − (1,1,1) = (−1, −1,0)[pic 32][pic 33]
𝑠𝑖 |⃗𝐻⃗⃗⃗1⃗⃗𝐻⃗⃗⃗⃗2→| = √(−1)2 + (−1)2 + (0)2 = √2 ≈ 1.41
⃗𝐻⃗⃗⃗1⃗⃗𝐻⃗⃗⃗⃗3→ = (0,1,0) − (1,1,1) = (−1,0, −1)[pic 34][pic 35]
𝑠𝑖 |⃗𝐻⃗⃗⃗1⃗⃗𝐻⃗⃗⃗⃗3→| = √(−1)2 + (0)2 + (−1)2 = √2 ≈ 1.41
⃗𝐻⃗⃗⃗1⃗⃗𝐻⃗⃗⃗⃗4→ = (1,0,0) − (1,1,1) = (0, −1, −1)
𝑠𝑖 |⃗𝐻⃗⃗⃗1⃗⃗𝐻⃗⃗⃗⃗4→| = √(0)2 + (−1)2 + (−1)2 = √2 ≈ 1.41[pic 36][pic 37]
⃗𝐻⃗⃗⃗2⃗⃗𝐻⃗⃗⃗⃗3→ = (0,1,0) − (0,0,1) = (0,1, −1)
𝑠𝑖 |⃗𝐻⃗⃗⃗2⃗⃗⃗𝐻⃗⃗⃗3→| = √(0)2 + (1)2 + (−1)2 = √2 ≈ 1.41[pic 38][pic 39]
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