vectors!!! (20 pts)

1) With respect to right-handed Cartesian coordinates, let a = [1, -2, 0],

b = [-2, 3, 0], c = [2, -4, -1], and d = [3, -1, 5]. Find (a x b) x (c x d).

2) Let a = [1, 1, 0], b = [3, 2, 1], and c = [1, 0, 2] . Find the angle between a+c

and b+c.

3) For what c are 3x + z = 5 and 8x - y + cz = 9 orthogonal?

1 Answer

Rating
  • 6 years ago
    Favorite Answer

    #3

    Consider that the normal vectors of the planes are orthogonal.

    (3, 0, 1).(8, -1, c) = 0

    24 + c = 0

    c = -24

    2014-01-26 23:08:38 補充:

    #2

    a + c = [2, 1, 2] and b + c = [4, 2, 3]

    Let the required angle be θ.

    (a + c).(b + c) = |a + c| × |b + c| × cosθ

    8 + 2 + 6 = √(2²+1²+2²) √(4²+2²+3²) cosθ

    16 = √9 √29 cosθ

    cosθ = 16/(3√29)

    θ = 7.955800083°

    2014-01-26 23:10:20 補充:

    #1

    I think you know how to do it.

    Just tedious.

    (a, b, c) × (x, y, z)

     |i j k|

    =|a b c|

     |x y z|

    2014-02-05 04:10:49 補充:

    [#1]

    a x b = (1, -2, 0) × (-2, 3, 0)

     | i j k|

    =| 1-2 0| = -k = (0, 0, -1)

     |-2 3 0|

    c x d = (2, -4, -1) × (3, -1, 5)

     | i j k|

    =| 2-4-1| = -21i-13j+10k = (-21, -13, 10)

     | 3-1 5|

    (a x b) x (c x d) = (0, 0, -1) × (-21, -13, 10)

     |  i  j  k|

    =|  0  0 -1| = -13i+21j = (-13, 21, 0)

     |-21-13 10|

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    [#2]

    a + c = [2, 1, 2] and b + c = [4, 2, 3]

    Let the required angle be θ.

    (a + c).(b + c) = |a + c| × |b + c| × cosθ

    8 + 2 + 6 = √(2²+1²+2²) √(4²+2²+3²) cosθ

    16 = √9 √29 cosθ

    cosθ = 16/(3√29)

    θ = 7.955800083°

    ♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬♪♫♬

    [#3]

    Consider that the normal vectors of the planes are orthogonal.

    (3, 0, 1).(8, -1, c) = 0

    24 + c = 0

    c = -24

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