Find the equation of the plane that passes through the points












0














Find the equation of the plane that passes through the points:
$$P_1=(1,1,2)\P_2=(2,3,3) \P_3=(3,-3,3)$$



The answer writes:
Let $x=vec{P_1P_2}=begin{bmatrix}1\2\1end{bmatrix}$ and $y=vec{P_1P_3}=begin{bmatrix}2\-4\1end{bmatrix}$
The normal vector $N$ must be orthogonal to both $X$ and $Y$. If we set $N=Xtimes Y=begin{bmatrix}6\1\-8end{bmatrix}$
Then $N$ will be a normal vector to the plane that passes through the given points. Using point $P_1$, we see that the equation of the plane is
$$6(x-1)+(y-1)-8(z-1)=0$$



But I don't know what the answer means, especially how it calculates X×Y and get $begin{bmatrix}6\1\-8end{bmatrix}$, I really have no idea.






linear-algebra orthogonality







share|cite|improve this question
























  • Have you computed the vectors $x$ and $y$ and their cross product $xtimes y$?
    – Dave
    yesterday










  • @Dave yes, but I get (2 -8 1)^T instead of (6 1 -8)^T
    – Shadow Z
    yesterday










  • I get $[6~1~-8]^T$. What steps are you taking to compute this (include these steps in the body of your question for us to see)?
    – Dave
    yesterday












  • I just simply times the the first, second, third number of the vectors....I have no idea how to compute..
    – Shadow Z
    yesterday






  • 2




    Found the leak! That is not what $xtimes y$ means. Rather, $xtimes y$ is the cross product of $x$ and $y$, which is computed as $$xtimes y=detbegin{bmatrix}e_1&e_2&e_3\x_1&x_2&x_3\y_1&y_2&y_3end{bmatrix}$$ where $e_i$ are the standard basis vectors in $mathbb R^3$ and $x_i$ and $y_i$ are the components of $x$ and $y$, respectively. Have you not covered cross product yet in your course, because it seems that this question is assuming you have.
    – Dave
    yesterday
















0














Find the equation of the plane that passes through the points:
$$P_1=(1,1,2)\P_2=(2,3,3) \P_3=(3,-3,3)$$



The answer writes:
Let $x=vec{P_1P_2}=begin{bmatrix}1\2\1end{bmatrix}$ and $y=vec{P_1P_3}=begin{bmatrix}2\-4\1end{bmatrix}$
The normal vector $N$ must be orthogonal to both $X$ and $Y$. If we set $N=Xtimes Y=begin{bmatrix}6\1\-8end{bmatrix}$
Then $N$ will be a normal vector to the plane that passes through the given points. Using point $P_1$, we see that the equation of the plane is
$$6(x-1)+(y-1)-8(z-1)=0$$



But I don't know what the answer means, especially how it calculates X×Y and get $begin{bmatrix}6\1\-8end{bmatrix}$, I really have no idea.






linear-algebra orthogonality







share|cite|improve this question
























  • Have you computed the vectors $x$ and $y$ and their cross product $xtimes y$?
    – Dave
    yesterday










  • @Dave yes, but I get (2 -8 1)^T instead of (6 1 -8)^T
    – Shadow Z
    yesterday










  • I get $[6~1~-8]^T$. What steps are you taking to compute this (include these steps in the body of your question for us to see)?
    – Dave
    yesterday












  • I just simply times the the first, second, third number of the vectors....I have no idea how to compute..
    – Shadow Z
    yesterday






  • 2




    Found the leak! That is not what $xtimes y$ means. Rather, $xtimes y$ is the cross product of $x$ and $y$, which is computed as $$xtimes y=detbegin{bmatrix}e_1&e_2&e_3\x_1&x_2&x_3\y_1&y_2&y_3end{bmatrix}$$ where $e_i$ are the standard basis vectors in $mathbb R^3$ and $x_i$ and $y_i$ are the components of $x$ and $y$, respectively. Have you not covered cross product yet in your course, because it seems that this question is assuming you have.
    – Dave
    yesterday














0












0








0







Find the equation of the plane that passes through the points:
$$P_1=(1,1,2)\P_2=(2,3,3) \P_3=(3,-3,3)$$



The answer writes:
Let $x=vec{P_1P_2}=begin{bmatrix}1\2\1end{bmatrix}$ and $y=vec{P_1P_3}=begin{bmatrix}2\-4\1end{bmatrix}$
The normal vector $N$ must be orthogonal to both $X$ and $Y$. If we set $N=Xtimes Y=begin{bmatrix}6\1\-8end{bmatrix}$
Then $N$ will be a normal vector to the plane that passes through the given points. Using point $P_1$, we see that the equation of the plane is
$$6(x-1)+(y-1)-8(z-1)=0$$



But I don't know what the answer means, especially how it calculates X×Y and get $begin{bmatrix}6\1\-8end{bmatrix}$, I really have no idea.






linear-algebra orthogonality







share|cite|improve this question















Find the equation of the plane that passes through the points:
$$P_1=(1,1,2)\P_2=(2,3,3) \P_3=(3,-3,3)$$



The answer writes:
Let $x=vec{P_1P_2}=begin{bmatrix}1\2\1end{bmatrix}$ and $y=vec{P_1P_3}=begin{bmatrix}2\-4\1end{bmatrix}$
The normal vector $N$ must be orthogonal to both $X$ and $Y$. If we set $N=Xtimes Y=begin{bmatrix}6\1\-8end{bmatrix}$
Then $N$ will be a normal vector to the plane that passes through the given points. Using point $P_1$, we see that the equation of the plane is
$$6(x-1)+(y-1)-8(z-1)=0$$



But I don't know what the answer means, especially how it calculates X×Y and get $begin{bmatrix}6\1\-8end{bmatrix}$, I really have no idea.






linear-algebra orthogonality




linear-algebra orthogonality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









Mostafa Ayaz

14.1k3937




14.1k3937










asked yesterday









Shadow Z

276




276












  • Have you computed the vectors $x$ and $y$ and their cross product $xtimes y$?
    – Dave
    yesterday










  • @Dave yes, but I get (2 -8 1)^T instead of (6 1 -8)^T
    – Shadow Z
    yesterday










  • I get $[6~1~-8]^T$. What steps are you taking to compute this (include these steps in the body of your question for us to see)?
    – Dave
    yesterday












  • I just simply times the the first, second, third number of the vectors....I have no idea how to compute..
    – Shadow Z
    yesterday






  • 2




    Found the leak! That is not what $xtimes y$ means. Rather, $xtimes y$ is the cross product of $x$ and $y$, which is computed as $$xtimes y=detbegin{bmatrix}e_1&e_2&e_3\x_1&x_2&x_3\y_1&y_2&y_3end{bmatrix}$$ where $e_i$ are the standard basis vectors in $mathbb R^3$ and $x_i$ and $y_i$ are the components of $x$ and $y$, respectively. Have you not covered cross product yet in your course, because it seems that this question is assuming you have.
    – Dave
    yesterday


















  • Have you computed the vectors $x$ and $y$ and their cross product $xtimes y$?
    – Dave
    yesterday










  • @Dave yes, but I get (2 -8 1)^T instead of (6 1 -8)^T
    – Shadow Z
    yesterday










  • I get $[6~1~-8]^T$. What steps are you taking to compute this (include these steps in the body of your question for us to see)?
    – Dave
    yesterday












  • I just simply times the the first, second, third number of the vectors....I have no idea how to compute..
    – Shadow Z
    yesterday






  • 2




    Found the leak! That is not what $xtimes y$ means. Rather, $xtimes y$ is the cross product of $x$ and $y$, which is computed as $$xtimes y=detbegin{bmatrix}e_1&e_2&e_3\x_1&x_2&x_3\y_1&y_2&y_3end{bmatrix}$$ where $e_i$ are the standard basis vectors in $mathbb R^3$ and $x_i$ and $y_i$ are the components of $x$ and $y$, respectively. Have you not covered cross product yet in your course, because it seems that this question is assuming you have.
    – Dave
    yesterday
















Have you computed the vectors $x$ and $y$ and their cross product $xtimes y$?
– Dave
yesterday




Have you computed the vectors $x$ and $y$ and their cross product $xtimes y$?
– Dave
yesterday












@Dave yes, but I get (2 -8 1)^T instead of (6 1 -8)^T
– Shadow Z
yesterday




@Dave yes, but I get (2 -8 1)^T instead of (6 1 -8)^T
– Shadow Z
yesterday












I get $[6~1~-8]^T$. What steps are you taking to compute this (include these steps in the body of your question for us to see)?
– Dave
yesterday






I get $[6~1~-8]^T$. What steps are you taking to compute this (include these steps in the body of your question for us to see)?
– Dave
yesterday














I just simply times the the first, second, third number of the vectors....I have no idea how to compute..
– Shadow Z
yesterday




I just simply times the the first, second, third number of the vectors....I have no idea how to compute..
– Shadow Z
yesterday




2




2




Found the leak! That is not what $xtimes y$ means. Rather, $xtimes y$ is the cross product of $x$ and $y$, which is computed as $$xtimes y=detbegin{bmatrix}e_1&e_2&e_3\x_1&x_2&x_3\y_1&y_2&y_3end{bmatrix}$$ where $e_i$ are the standard basis vectors in $mathbb R^3$ and $x_i$ and $y_i$ are the components of $x$ and $y$, respectively. Have you not covered cross product yet in your course, because it seems that this question is assuming you have.
– Dave
yesterday




Found the leak! That is not what $xtimes y$ means. Rather, $xtimes y$ is the cross product of $x$ and $y$, which is computed as $$xtimes y=detbegin{bmatrix}e_1&e_2&e_3\x_1&x_2&x_3\y_1&y_2&y_3end{bmatrix}$$ where $e_i$ are the standard basis vectors in $mathbb R^3$ and $x_i$ and $y_i$ are the components of $x$ and $y$, respectively. Have you not covered cross product yet in your course, because it seems that this question is assuming you have.
– Dave
yesterday










1 Answer
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According to geometry, a plane can be uniquely determined given the normal vector of it and a point lying on it (the reason is intuitively simple). To solve this problem, a number of arguments are employed:




$A_1$: The cross product of two linearly independent vectors, is always orthogonal on them (linearly independent in the case of two vectors means that they are not any multiple of each other).



$A_2$: If two points belong to a plane, then so is the corresponding vector whose endpoints are the given points of the plane.



$A_3$: The cross product of two vectors $v_1=ahat i+bhat j+chat k$ and $v_2=dhat i+ehat j+fhat k$ can be calculated as $$v_1times v_2=(bf-ce)hat i+(cd-af)hat j+(ae-bd)hat k$$




The normal vector then would be$$v=v_1times v_2=begin{bmatrix}1\2\1end{bmatrix}times begin{bmatrix}2\-4\1end{bmatrix}=begin{bmatrix}2times 1-1times(-4)\2times 1-1times 1\1times(-4)-2times 2end{bmatrix}=begin{bmatrix}6\1\-8end{bmatrix}$$therefore $$(x,y,z)cdot v=k$$for some constant $k$. By substitution we finally obtain$$6x+y-8z=-1$$






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    According to geometry, a plane can be uniquely determined given the normal vector of it and a point lying on it (the reason is intuitively simple). To solve this problem, a number of arguments are employed:




    $A_1$: The cross product of two linearly independent vectors, is always orthogonal on them (linearly independent in the case of two vectors means that they are not any multiple of each other).



    $A_2$: If two points belong to a plane, then so is the corresponding vector whose endpoints are the given points of the plane.



    $A_3$: The cross product of two vectors $v_1=ahat i+bhat j+chat k$ and $v_2=dhat i+ehat j+fhat k$ can be calculated as $$v_1times v_2=(bf-ce)hat i+(cd-af)hat j+(ae-bd)hat k$$




    The normal vector then would be$$v=v_1times v_2=begin{bmatrix}1\2\1end{bmatrix}times begin{bmatrix}2\-4\1end{bmatrix}=begin{bmatrix}2times 1-1times(-4)\2times 1-1times 1\1times(-4)-2times 2end{bmatrix}=begin{bmatrix}6\1\-8end{bmatrix}$$therefore $$(x,y,z)cdot v=k$$for some constant $k$. By substitution we finally obtain$$6x+y-8z=-1$$






    share|cite|improve this answer


























      0














      According to geometry, a plane can be uniquely determined given the normal vector of it and a point lying on it (the reason is intuitively simple). To solve this problem, a number of arguments are employed:




      $A_1$: The cross product of two linearly independent vectors, is always orthogonal on them (linearly independent in the case of two vectors means that they are not any multiple of each other).



      $A_2$: If two points belong to a plane, then so is the corresponding vector whose endpoints are the given points of the plane.



      $A_3$: The cross product of two vectors $v_1=ahat i+bhat j+chat k$ and $v_2=dhat i+ehat j+fhat k$ can be calculated as $$v_1times v_2=(bf-ce)hat i+(cd-af)hat j+(ae-bd)hat k$$




      The normal vector then would be$$v=v_1times v_2=begin{bmatrix}1\2\1end{bmatrix}times begin{bmatrix}2\-4\1end{bmatrix}=begin{bmatrix}2times 1-1times(-4)\2times 1-1times 1\1times(-4)-2times 2end{bmatrix}=begin{bmatrix}6\1\-8end{bmatrix}$$therefore $$(x,y,z)cdot v=k$$for some constant $k$. By substitution we finally obtain$$6x+y-8z=-1$$






      share|cite|improve this answer
























        0












        0








        0






        According to geometry, a plane can be uniquely determined given the normal vector of it and a point lying on it (the reason is intuitively simple). To solve this problem, a number of arguments are employed:




        $A_1$: The cross product of two linearly independent vectors, is always orthogonal on them (linearly independent in the case of two vectors means that they are not any multiple of each other).



        $A_2$: If two points belong to a plane, then so is the corresponding vector whose endpoints are the given points of the plane.



        $A_3$: The cross product of two vectors $v_1=ahat i+bhat j+chat k$ and $v_2=dhat i+ehat j+fhat k$ can be calculated as $$v_1times v_2=(bf-ce)hat i+(cd-af)hat j+(ae-bd)hat k$$




        The normal vector then would be$$v=v_1times v_2=begin{bmatrix}1\2\1end{bmatrix}times begin{bmatrix}2\-4\1end{bmatrix}=begin{bmatrix}2times 1-1times(-4)\2times 1-1times 1\1times(-4)-2times 2end{bmatrix}=begin{bmatrix}6\1\-8end{bmatrix}$$therefore $$(x,y,z)cdot v=k$$for some constant $k$. By substitution we finally obtain$$6x+y-8z=-1$$






        share|cite|improve this answer












        According to geometry, a plane can be uniquely determined given the normal vector of it and a point lying on it (the reason is intuitively simple). To solve this problem, a number of arguments are employed:




        $A_1$: The cross product of two linearly independent vectors, is always orthogonal on them (linearly independent in the case of two vectors means that they are not any multiple of each other).



        $A_2$: If two points belong to a plane, then so is the corresponding vector whose endpoints are the given points of the plane.



        $A_3$: The cross product of two vectors $v_1=ahat i+bhat j+chat k$ and $v_2=dhat i+ehat j+fhat k$ can be calculated as $$v_1times v_2=(bf-ce)hat i+(cd-af)hat j+(ae-bd)hat k$$




        The normal vector then would be$$v=v_1times v_2=begin{bmatrix}1\2\1end{bmatrix}times begin{bmatrix}2\-4\1end{bmatrix}=begin{bmatrix}2times 1-1times(-4)\2times 1-1times 1\1times(-4)-2times 2end{bmatrix}=begin{bmatrix}6\1\-8end{bmatrix}$$therefore $$(x,y,z)cdot v=k$$for some constant $k$. By substitution we finally obtain$$6x+y-8z=-1$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Mostafa Ayaz

        14.1k3937




        14.1k3937






























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