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Math Help - 3D Vectors Parallelogram Area

  1. #1
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    3D Vectors Parallelogram Area

    Hi can anybody help me I have to determine if PQRS is a parallelogram and then find the area
    1- P (1,1,1) Q (-2,3,-2) R (-1,0,1) S (2,-2,2)
    2- P (-1,1,1) Q(3,2,-2) R(5,3,-5) S (1,2,-2)
    What is the procedure to do this?
    Also i have to determine the area of the triangle P(4,-1,3) Q(1,-5,2) R(1,1,-6)
    One last thing what is the formula to calculate the angle between 2 vectors when is a 3d vector?
    any help will be appreciated it
    thanks
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  2. #2
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    For the area of the triangle, you can use the cross product.

    \overline{P_{1}P_{2}}=(-3,-4,-1)....[1]

    \overline{P_{1}P_{3}}=(-3,2,-9)...[2]

    \overline{P_{1}P_{2}}\times\overline{P_{1}P_{3}}=(  38,-24,-18)....[3]

    Consequently, from the area of a triangle, use the norm of [3], \sqrt{(38)^{2}+(-24)^{2}+(-18)^{2}}=2\sqrt{586}

    A=\frac{1}{2}||\overline{P_{1}P_{2}}\times\overlin  e{P_{1}P_{3}}||=\sqrt{586}\approx{24.21}
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  3. #3
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    Quote Originally Posted by Flash View Post
    Hi can anybody help me I have to determine if PQRS is a parallelogram and then find the area
    1- P (1,1,1) Q (-2,3,-2) R (-1,0,1) S (2,-2,2)
    2- P (-1,1,1) Q(3,2,-2) R(5,3,-5) S (1,2,-2)
    What is the procedure to do this?
    Also i have to determine the area of the triangle P(4,-1,3) Q(1,-5,2) R(1,1,-6)
    One last thing what is the formula to calculate the angle between 2 vectors when is a 3d vector?
    any help will be appreciated it
    thanks
    A parallelogram
    - has 2 pairs of parallel sides
    - the parallel sides have the same length
    - the mean value of the coordinates of opposite vertices is the midpoint of both diagonals.

    to #1: Calculate the midpoint of the diagonals (because you don't know which vertices are opposite you have to check in 2 steps):

    M(P,Q)=\left(-\frac12~,~ 1~,~-\frac12\right) and M(R,S)=\left(\frac12~,~ -1~,~\frac32\right)...... \implies~ false

    M(P,R)=\left(0~,~ \frac12~,~1\right) and M(Q,S)=\left(0~,~ \frac12~,~0\right)...... \implies~ false

    Therefore PQRS is not a parallelogram

    to #2:

    Same procedure: Check twice. Finally you get

    M(P,R)=\left(2~,~ 2~,~-2\right) and M(Q,S)=\left(2~,~2~,~-2\right)...... \implies~ right

    Check if |\overrightarrow{PQ}|=|\overrightarrow{RS}|

    \overrightarrow{PQ}=(4, 1, -3) and

    \overrightarrow{RS}=(-4, -1, 3)

    Obviously both vectors have the same length.

    To calculate the area you are supposed to know that the absolute value of the cross-product yields the value of the area:

    \overrightarrow{PQ} \times \overrightarrow{RS} = (4, 1, -3) \times (-4, -1, 3) = (6, -24, -8)~\implies~|(6, -24, -8)|=\sqrt{676} = 26
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  4. #4
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    Quote Originally Posted by earboth View Post
    To calculate the area you are supposed to know that the absolute value of the cross-product yields the value of the area:

    \overrightarrow{PQ} \times \overrightarrow{RS} = (4, 1, -3) \times (-4, -1, 3) = (6, -24, -8)~\implies~|(6, -24, -8)|=\sqrt{676} = 26
    thanks for the answer but I have a question, I tried doing the cross product and it gives (0,0,0) what am I doing wrong? this is what I did (I dont know how to use matrices to calculate the Cross Product)

    1 -3 4 1
    -1 3 -4 -1

    (1)(3) - (-1)(-3) =0
    (-3)(-4) - (4)(3) =0
    (4)(-1) - (1)(-4) =0
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  5. #5
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    Hello,

    I think this is because there is a typo... It should be the cross product \overrightarrow{PQ}\times \overrightarrow{QR}.

    Your method is correct

    To get it in the form of a matrix :


    \begin{pmatrix} a \\ b \\ c \end{pmatrix} \times \begin{pmatrix} c \\ d \\e \end{pmatrix}

    Consider \begin{pmatrix} a & c \\ b & d \\ c & e \end{pmatrix}

    The cross product will be :
    + determinant of the remaining after removing the first line
    - determinant of the remaining after removing the second line
    + determinant of the remaining after removing the third line

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  6. #6
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    Quote Originally Posted by Flash View Post
    thanks for the answer but I have a question, I tried doing the cross product and it gives (0,0,0) what am I doing wrong? this is what I did (I dont know how to use matrices to calculate the Cross Product)

    1 -3 4 1
    -1 3 -4 -1

    (1)(3) - (-1)(-3) =0
    (-3)(-4) - (4)(3) =0
    (4)(-1) - (1)(-4) =0
    Sorry for the confusion: Since \overline{PQ} \parallel \overline{RS} the cross-product must yield the null vector.

    Probably you have done the calculations suggested by Moo and have got the value of the area as A=\sqrt{40} \approx 6.32

    But check my results. As you have seen I'm the MoD (= master of desaster) if it comes to calculations.
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