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Thread: Finding the area of a triangle on the space given three points

  1. #1
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    Finding the area of a triangle on the space given three points

    Hi there, I must solve the next problem. The sentence says:
    Find the area of a triangle wich vertex are the points: $\displaystyle V_1(1,5,-2)$ $\displaystyle V_2(0,0,0)$ and $\displaystyle V_3(3,5,1)$.

    What I did was define the vectors: $\displaystyle \vec{v_1}=(1,5,-2)$ y $\displaystyle \vec{v_3}=(3,5,1)$, considering the point $\displaystyle v_2$ is the origin, these vectors are $\displaystyle \vec{v_2v_1}$ y $\displaystyle \vec{v_2v_3}$.

    So I took the vectors $\displaystyle \vec{v_1}$ and $\displaystyle \vec{v_3}$ and then what I did was getting the module of the vectorial product, which is the area of the parallelogram conformed by these two vectors, and then I've divide it on two:

    $\displaystyle A=\displaystyle\frac{||v_1\wedge{v_3}||}{2}$

    $\displaystyle v_1\wedge{v_3}=(15,-7,-10)\Rightarrow{||(15,-7,-10)||=\sqrt[ ]{347}}$

    Then: $\displaystyle A=\displaystyle\frac{||v_1\wedge{v_3}||}{2}=\displ aystyle\frac{\sqrt[ ]{347}}{2}$

    Is it right?

    Bye, and thanks for commenting.
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  2. #2
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    Quote Originally Posted by Ulysses View Post
    Hi there, I must solve the next problem. The sentence says:
    Find the area of a triangle wich vertex are the points: $\displaystyle V_1(1,5,-2)$ $\displaystyle V_2(0,0,0)$ and $\displaystyle V_3(3,5,1)$.

    What I did was define the vectors: $\displaystyle \vec{v_1}=(1,5,-2)$ y $\displaystyle \vec{v_3}=(3,5,1)$, considering the point $\displaystyle v_2$ is the origin, these vectors are $\displaystyle \vec{v_2v_1}$ y $\displaystyle \vec{v_2v_3}$.

    So I took the vectors $\displaystyle \vec{v_1}$ and $\displaystyle \vec{v_3}$ and then what I did was getting the module of the vectorial product, which is the area of the parallelogram conformed by these two vectors, and then I've divide it on two:

    $\displaystyle A=\displaystyle\frac{||v_1\wedge{v_3}||}{2}$

    $\displaystyle v_1\wedge{v_3}=(15,-7,-10)\Rightarrow{||(15,-7,-10)||=\sqrt[ ]{347}}$

    Then: $\displaystyle A=\displaystyle\frac{||v_1\wedge{v_3}||}{2}=\displ aystyle\frac{\sqrt[ ]{347}}{2}$

    Is it right?

    Bye, and thanks for commenting.
    $\displaystyle \frac{\sqrt{374}}{2}$
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  3. #3
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    Thanks, didn't note that :P
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