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Math Help - vector equation of line through point perpendicular to plane

  1. #1
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    vector equation of line through point perpendicular to plane

    Hi,

    How do I find the vector equation of this.
    given the line travels through point (1,0,6) and perpendicular to the plane x+3y+z=5

    I'm guessing here by do I do something like this
    (x-1)+(y-0)+(z-6) = x+3y+z-5 ? ??
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  2. #2
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    Quote Originally Posted by Craka View Post
    Hi,

    How do I find the vector equation of this.
    given the line travels through point (1,0,6) and perpendicular to the plane x+3y+z=5

    I'm guessing here by do I do something like this
    (x-1)+(y-0)+(z-6) = x+3y+z-5 ? ??
    The Normal Vector to the plane is \vec n =1 \vec i + 3 \vec j + 1 \vec k You can get this by taking the gradient of the plane.

    To find the vector equation of the plane

    \vec r(t)=P_0+t\cdot \vec v

    so we get

    \vec r(t)=(\vec i +6 \vec k) +t(1 \vec i + 3 \vec j + 1 \vec k)=(1+t)\vec i +(3t)\vec j +(6+t) \vec k
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Craka View Post
    Hi,

    How do I find the vector equation of this.
    given the line travels through point (1,0,6) and perpendicular to the plane x+3y+z=5

    I'm guessing here by do I do something like this
    (x-1)+(y-0)+(z-6) = x+3y+z-5 ? ??
    the direction vector of the line is the normal vector to the plane. so we have a line passing through (1,0,6) in the direction <1,3,1>. how would you write the equation for this line?

    You know why the normal vector of the plane is the direction vector of the line, right?

    EDIT: well, if i'm going to be beaten by anyone, it might as well be TES... you mind me calling you TES, TheEmptySet?
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    Quote Originally Posted by Craka View Post
    Hi,

    How do I find the vector equation of this.
    given the line travels through point (1,0,6) and perpendicular to the plane x+3y+z=5

    I'm guessing here by do I do something like this
    (x-1)+(y-0)+(z-6) = x+3y+z-5 ? ??
    From the equation of the plane you already know it's normal vector: \vec n = (1,3,1)

    Therefore the equation of the line is:

    l: \vec r=(1,0,6)+t \cdot (1,3,1)

    Beaten again ...
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  5. #5
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    Given a line of the form ax + by + cz + d = 0, the vector (a,b,c) is perpendicular to the line. This is the vector component that is parallel to your line, i.e. the \vec{v} of the vector equation: \vec{r} = \vec{r}_{0} + \lambda \vec{v}. You have a point on the line and thus you can find \vec{r}_{0} to complete your equation.

    EDIT: WOW BEaTEN BY 3 PEOPLE????
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    Quote Originally Posted by Jhevon View Post

    EDIT: well, if i'm going to be beaten by anyone, it might as well be TES... you mind me calling you TES, TheEmptySet?
    Fine by be I have been called ALOT worse
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  7. #7
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    Quote:
    Originally Posted by Craka
    Hi,

    How do I find the vector equation of this.
    given the line travels through point (1,0,6) and perpendicular to the plane x+3y+z=5

    I'm guessing here by do I do something like this
    (x-1)+(y-0)+(z-6) = x+3y+z-5 ? ??


    The Normal Vector to the plane is You can get this by taking the gradient of the plane.

    To find the vector equation of the plane



    so we get

    It is a normal vector as the plane equation is not equal to zero, right?

    sorry didn't realise needed to use the equation of that format. Obviously if there were a j component that would be included in (i+6k) ?
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    Given the plane ax + by + cz + d = 0, the vector (a,b,c) is orthogonal to the plane.

    The j component is present as 0j.
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  9. #9
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    Quote Originally Posted by Craka View Post
    It is a normal vector as the plane equation is not equal to zero, right?

    sorry didn't realise needed to use the equation of that format. Obviously if there were a j component that would be included in (i+6k) ?
    Yes it would need to be included. The notation I used is what I am used to. There are many other ways this could have been written for example

    n=<1,3,1> \\\ P=<1,0,6>

    Then the equation of the line would be

    r(t)=<1,0,6>+t<1,3,1>=<1+t,3t,6+t>

    They also could have been written in parametric form.

    So I would use whatever form your professor or book uses.
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