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Please explain volume of parallelepiped

  1. Aug 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Ok so my book says that the volume of a parallelepiped is:

    V= |b cross c||a||cos(theta)| = |a dot (b cross c)|, where a, b, and c are vectors

    I get the "|b cross c||a||cos(theta)|" part because I can see the geometry but I don't get how they get from that to |a dot (b cross c)|.





    2. Relevant equations



    3. The attempt at a solution

    I see how |a||cos(theta)| looks kinda familiar so I tried to figure out using that.

    So I know a dot b = |a||b|cos(theta), so

    |a|cos(theta) = (a dot b)/|b| but then if I plug that quantity in it doesn't seem to work.

    So can anyone help me see why |b cross c||a||cos(theta)| = |a dot (b cross c)| ?

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 27, 2011 #2

    LCKurtz

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    Think of a, b, and c with their tails together. b x c is a vector perpendicular to the plane of b and c and |b x c| is numerically equal to the area of the parallelogram formed by b and c. To get the volume of the parallelepiped formed with this parallelogram as a base and a as the third side you need to multiply the area of the base by the height of the parallelepiped. But the height of the parallelepiped is the component of a perpendicular to the plane of b and c. If theta is the angle between b x c and a, then this height is h = |a|cos(theta). So the volume is V = h|bxc| = |a|cos(theta)|bxc| = a dot (bxc). Does that help?
     
  4. Aug 27, 2011 #3
    I already understood how the volume is |bxc||a||cos(theta)| but I don't understand how that equals |a dot(bxc)|. Can you explain how |bxc||a||cos(theta)|= |a dot(bxc)|? Thanks.
     
  5. Aug 27, 2011 #4

    LCKurtz

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    Oh, that's all you are asking. That is just the basic property of dot product:

    [tex]\vec A \cdot \vec B = |\vec A||\vec B|\cos\theta[/tex]

    applied to a and bxc. Here θ is the angle between a and bxc.
     
  6. Aug 30, 2011 #5
    Ok so I'm still not seeing it.

    I want to show that |bxc||a||cos(theta)| = |a dot (bxc)|

    ok my first step: I know that a dot b = |a||b|cos(theta) so I can solve for |a|cos(theta)

    So now I have |bxc|(a dot b)/||b| = |a dot (bxc)|

    So I plug it in and get:

    |bxc|[(a dot b)/|b|] and now I am stuck.
     
  7. Aug 30, 2011 #6

    LCKurtz

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    Just do what I suggested above. Replace B by BxC in the identity.
     
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