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What is the volume and area determined by three points?

  1. Jan 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Let P = (2, 2, 0), Q = (0, 4, 1) and R = (-1, 2, 3) in the space [tex] \Re^{3}.
    [/tex]
    a) Determine the area of the rectangle determined by vectors [tex]\overrightarrow{PQ}[/tex] and [tex]\overrightarrow{PR}[/tex].
    b) Determine the volume of the tetrahedral determined by vectors [tex]\overrightarrow{PQ}[/tex] and [tex]\overrightarrow{PR}[/tex], and the origin O, OPQR.

    3. The attempt at a solution
    a)
    [tex]
    |\overrightarrow{PQ}| = \sqrt {4 + 4 +1} = 3
    [/tex]

    [tex]
    |\overrightarrow{PR}| = \sqrt {9 + 0 + 9} = 3 \sqrt {2}
    [/tex]

    [tex]
    Area = |\overrightarrow{PQ}| * |\overrightarrow{PR}|
    = 9 \sqrt {2}
    [/tex]

    b)
    [tex]
    Volume = Area * |\overrightarrow{RO}|
    |\overrightarrow{RO}| = \sqrt {1 + 4 + 9}
    = \sqrt {14}
    [/tex]

    The volume is
    [tex]
    Volume = 9 \sqrt{2} * \sqrt {14}
    = 18 \sqrt {7}
    [/tex]

    Please, comment any mistakes.
     
    Last edited: Jan 19, 2009
  2. jcsd
  3. Jan 19, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    PR and PQ are not orthogonal, they don't determine a rectangle. Ditto for the second problem. You should be using the cross product and the dot product to solve these problems. Do you know any relations between them and the area and volume?
     
  4. Jan 19, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If you mean "parallelogram" and "rectangular prism", then the area of the parallelogram determined by the two vectors [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex] is given by [itex]\vec{u}\times\vec{v}= |u||v|sin(\theta)[/itex] while the rectangular prism determined by the three vectors [itex]\vec{u}[/itex], [itex]\vec{v}[/itex], and [itex]\vec{w}[/itex] has volume [itex]\vec{u}\cdot(\vec{v}\times\vec{w})[/itex].
     
  5. Jan 19, 2009 #4
    You are right. We need to use Sarrus.
    The b -part changes to:

    b)
    [tex]
    Volume = Area * |\overrightarrow{PQ} x \overrightarrow{PR}|
    [/tex]

    [tex]
    \overrightarrow{PQ} x \overrightarrow{PR} = (6, 3, 6) // by Sarrus
    [/tex]

    [tex]
    |\overrightarrow{PQ} x \overrightarrow{PR}| = 9
    [/tex]

    The volume is
    [tex]
    Volume = 9 \sqrt{2} * 9
    = 81 \sqrt {2}
    [/tex]

    Please, suggest any improvements.
     
  6. Jan 19, 2009 #5
    I mean parallelogram.
    Thank you for the correction!
     
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