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Basic Vector Problem

  1. Sep 20, 2008 #1
    1. The problem statement, all variables and given/known data

    What is the area of a parallelogram that has points A(1,2,-3), B(2,0,4), & C(5,1,-2) as three of its four vertices?

    2. Relevant equations
    none, really

    3. The attempt at a solution

    First, I found two vectors, [tex]\vec{AB}[/tex] and [tex]\vec{BC}[/tex].

    [tex]\vec{AB}[/tex] is <1,-2,7> and [tex]\vec{BC}[/tex] is <-3,-1,6>

    Then the cross product between the two: -9i + 3j + 7k.

    And then the magnitude: [tex]\sqrt{81+9+49}[/tex] = [tex]\sqrt{139}[/tex]

    Is this the correct way of solving this problem? Thank you.
  2. jcsd
  3. Sep 20, 2008 #2
    2. Relevant equations
    Area of the thing = area of triangle * 2

    What's area of triangle?
    Write in terms of sin theta
  4. Sep 20, 2008 #3


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    Staff Emeritus
    Science Advisor

    Please, please, please don't just memorize formulas without learning what they mean! You have, in fact, done the calculation correctly but that is meaningless if you don't know it is correct and, better, understand why it is correct!

    One definition of cross product is "The cross product of two vectors, u and v is the vector at right angles to both (in the "right hand" sense) with length |u||v|sin[itex]\theta[/itex] where [/itex]\theta[/itex] is the angle between the two vectors".

    Now, remember that the area of a parallelogram is "base time height" where the height is measured from one vertex perpendicular to the base. That gives a right triangle where the length of one side is the hypotenuse and the height is the side opposite angle [itex]\theta[/itex]. If v forms the base and u is the other side, then the height is |u|sin(\theta) so the area of the parallelogram is |u||v|sin[itex]\theta[/itex]. Look familiar?
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