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Math Help - vector..again

  1. #1
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    vector..again

    the 4 points A,B,C and D have the following co-ordinates in a Cartesian coordinate system A = (2,2,2), B = (5,2,4), C = (-2,4,-1), D = (0,0,6).
    i) find the volume of the parallelepiped whose edges are the lines AB,AC and AD.
    ii)calculate the angle BAC
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  2. #2
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    Do you know

    V=|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})| ?

    where

    \mathbf{a}=3\mathbf{i}+2\mathbf{k}

    \mathbf{b}=-4\mathbf{i}+2\mathbf{j}-3\mathbf{k}

    \mathbf{b}=-2\mathbf{i}-2\mathbf{j}+4\mathbf{k}
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  3. #3
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    Quote Originally Posted by ashes View Post
    the 4 points A,B,C and D have the following co-ordinates in a Cartesian coordinate system A = (2,2,2), B = (5,2,4), C = (-2,4,-1), D = (0,0,6)
    ....
    ii)calculate the angle BAC
    Use the transscription of the dot-product:

    \cos(\theta) = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{ | \overrightarrow{AB}| \cdot |\overrightarrow{AC}|}


    For confrimation only: I've got \cos(\theta) = -\frac{18}{\sqrt{13} \cdot \sqrt{29}} ~\implies~ \theta \approx 157.979^\circ

    But check my results!! When it comes to calculation I'm a master of desaster.
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  4. #4
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    reply

    k got the first part..but which formula to use to get angle BAC? i could not map it out on my drawing..
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  5. #5
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    Quote Originally Posted by ashes View Post
    k got the first part..but which formula to use to get angle BAC? i could not map it out on my drawing..
    Consider the stationary vectors pointing at the points A, B, C and D.

    Then the vector

    \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}

    According to the coordinates of the points you get:

    \overrightarrow{AB} = (5, 2, 4) - (2, 2, 2,) = (3, 0, 2)

    etc. ...

    I've attached a very rough sketch of the 4 points and I've marked the angle \angle(BAC)
    Attached Thumbnails Attached Thumbnails vector..again-raumwinkel.gif  
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  6. #6
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    Is your rough sketch a parallelpiped, cos i thought a parallelepiped is a polyhedron with all its faces being a parallelogram.. care to enlighten me cos im very bad at 3d visualisation..
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  7. #7
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    by formula i got the same answer as yours..its only the visualizing thingy which i cant seem to get everytime i do vectors..thanks btw..
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  8. #8
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    Quote Originally Posted by ashes View Post
    Is your rough sketch a parallelpiped, cos i thought a parallelepiped is a polyhedron with all its faces being a parallelogram.. care to enlighten me cos im very bad at 3d visualisation..
    No of course not. I only placed the 4 points in a 3-D-coordinate system to give an impression of the distances.

    I've attached a sketch of the complete parallelepiped.
    Attached Thumbnails Attached Thumbnails vector..again-parallepiped.gif  
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