Algebra Lineal. Producto cruz de dos vectores

Algebra Lineal.

Producto cruz de dos vectores.

Responda las siguientes preguntas de evaluación.

[pic 1]

1.- i x k – k x i[pic 2]

1. 0
2. J
3. 2j
4. -2j

[pic 3]

2.- i x (j x K) [pic 4]

1. 0
2. -1
3. 1
4. i – j + k

Encuentra el producto cruz de U X V.

[pic 5]

1. - U= 2 i, - 3 j, + 5 k      V= 3 i, - j, - k.

[pic 6]

U x V==  i - j+  k=[pic 11][pic 7][pic 8][pic 9][pic 10]

[pic 12]

[(-3) (-1) – (5) (-1)] I - [(2) (-1) - (5) (3)] j + [(2) (-1) – (-3) (3)] k

[3 - (-5)] i - [-2 -15] j+ [-2 - (-9)] k

[3 + 5] i - [-17] j+ [-2 + 9)] k

8 i + 17 j +7 k    =     < 8, 17, 7 >

[pic 13]

2. - U= 10 i, + 7 j, - 3 k      V= - 3 i, + 4 j, - 3 k. [pic 14]

[pic 15]

U x V==  i - j+  k=[pic 20][pic 16][pic 17][pic 18][pic 19]

[pic 21]

[(7) (-3) – (-3) (4)] I - [(10) (-3) - (-3) (-3)] j + [(10) (4) – (7) (-3)] k

[-21 - (-12)] i - [-30 -9] j+ [40 - (-21)] k

[-21 + 12] i - [-39] j+ [40 + 21)] k

-9 i + 39 j + 61 k    =     < -9, 39, 61 >

[pic 22]

Encuentre el área del paralelogramo con los vértices consecutivos.

(1, - 2, 3); (2, 0, 1); (0, 4, 0)[pic 23]

        B                                 C        [pic 24][pic 25][pic 26][pic 27][pic 28]

                   A                             D

[pic 29][pic 30][pic 31]

D= A + BC   = A + AD[pic 32]

BC= C – B= (0, 4, 0) – (2, 0, 1) = (-2, 4, -1)

D= (1, -2, 3) + (-2, 4, -1) = (-1, 2, 2)[pic 33]

AB= B – A= (2, 0, 1) – (1, -2, 3) = (1, 2, -2)[pic 34]

AD= D – A= (-1, 2, 2) - (1, -2, 3) = (-2, 4, -1)

AB x AD==  i - j+  k=[pic 39][pic 40][pic 35][pic 36][pic 37][pic 38]

[pic 41]

[(2) (-1) – (-2) (4)] I - [(1) (-1) - (-2) (-2)] j + [(1) (4) – (2) (-2)] k

[-2 - (-8)] i - [-1 -4] j+ [4 - (-4)] k

[-2 + 8] i - [-5] j+ [4 + 4)] k

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