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Copythat Group Title

Draw a parallelogram whose adjacent sides are determined by the vectors [4,6] and [3,5]. Find the area of the parallelogram.

  • one year ago
  • one year ago

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  1. CliffSedge Group Title
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    Isn't it dot product which finds the area?

    • one year ago
  2. CliffSedge Group Title
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    You can use the origin (0,0) as the common start point for those two vectors; that would make it easiest to draw.

    • one year ago
  3. henpen Group Title
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    No, I don't think it's dot product, as dot product is a measure of how much the vectors are in each other's direction, so it would have high dottiness if they were like this|dw:1350760137445:dw|but very low area (dottiness=cos(angle between vectors))

    • one year ago
  4. henpen Group Title
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    Cross product most likely, but forget about the result being a vector.

    • one year ago
  5. CliffSedge Group Title
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    Oh, that's right, cross product because it is a*b*sin(Θ).

    • one year ago
  6. CliffSedge Group Title
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    b*sin(Θ) is the height of the parallelogram. Duh, should have remembered the basic trig on that one.

    • one year ago
  7. Copythat Group Title
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    Is there any way to solve for the area without trig?

    • one year ago
  8. CliffSedge Group Title
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    Not really...

    • one year ago
  9. CliffSedge Group Title
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    Well, doing the cross-product has the trig built-in, so you wouldn't actually be using a trig function explicitly.

    • one year ago
  10. henpen Group Title
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    I suppose you could try to use pythagoras

    • one year ago
  11. CliffSedge Group Title
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    ^ what do you mean? By stacking up right triangles and using subtraction?

    • one year ago
  12. henpen Group Title
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    |dw:1350764732234:dw| find B perpendicular with A by finding the 'gradient' of A... you can do it, but it's really fiddly

    • one year ago
  13. henpen Group Title
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    No- there must be an elegant way of finding B perp A

    • one year ago
  14. henpen Group Title
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    as a function of Ax, Bx, Ay and By

    • one year ago
  15. CliffSedge Group Title
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    I'd rather just do the cross-product.

    • one year ago
  16. CliffSedge Group Title
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    \[\left[\begin{matrix}i & j & k \\ 4 & 6 & 0 \\ 3 & 5 & 0\end{matrix}\right] = 4\times5 - 6\times3=2\]

    • one year ago
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