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Math Help - Vectors. Did not know where to post this one.

  1. #1
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    Vectors. Did not know where to post this one.

    Okay so I read about the equation of a plane that is parallel to 2 vectors and passes through a point(say 'P').

    And i am thinking, if we have 2 vectors to which the plane is parallel, why do we need another point? Or should that point be a parameter in determining the equation of a plane.

    So I argue that there must be two cases:-
    1) The plane that contains the 2 vectors to which it is parallel does not contain the point P - possible and absurd. Say the plane is parallel to  \frac{x}{1} = \frac{y}{0} = \frac{z}{0} and  \frac{x}{0} = \frac{y}{1} = \frac{z}{0} and I want it to pass through say (0,0,4). I cannot have such a plane.

    (2) So the point has to lie on that plane. Say the vectors are ~a and ~b . So the k1a + k2b gives us the entire plane for different values og k1 and k2.

    My question is that if it also has to pass through P (with ~OP=P) isn't k1a + k2b a sufficient equation ? I think so.

    But my book says differently. It says that the equation will become P+k1a + k2b .

    Please help !
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    Re: Vectors. Did not know where to post this one.

    There are infinitely many planes that are parallel to any two given vectors. Knowing a point on the plane allows us to find a unique solution.

    I don't really understand your argument. In (1), you talk about a plane parallel to the lines given by

    x = t, y = 0, z = 0

    and

    x = 0, y = t, z = 0

    (at least I think that's what you mean - I changed to a parametric form to avoid the division by zero). Then you say that there is no such plane that passes through (0,0,4), but this is clearly untrue: The plane z = 4 meets both conditions.
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    Re: Vectors. Did not know where to post this one.

    I get it. Really dumb of me. I guess i misunderstood 'parallel'. I thought it meant the plane also had to contain those vectors. Ahh! Silly of me. Thanks.
    Yes, thats what i meant with those lines but the argument is now kaput.
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  4. #4
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    Re: Vectors. Did not know where to post this one.

    Quote Originally Posted by pratique21 View Post
    Okay so I read about the equation of a plane that is parallel to 2 vectors and passes through a point(say 'P').

    And i am thinking, if we have 2 vectors to which the plane is parallel, why do we need another point? Or should that point be a parameter in determining the equation of a plane.

    So I argue that there must be two cases:-
    1) The plane that contains the 2 vectors to which it is parallel does not contain the point P - possible and absurd. Say the plane is parallel to  \frac{x}{1} = \frac{y}{0} = \frac{z}{0} and  \frac{x}{0} = \frac{y}{1} = \frac{z}{0} and I want it to pass through say (0,0,4). I cannot have such a plane.

    (2) So the point has to lie on that plane. Say the vectors are ~a and ~b . So the k1a + k2b gives us the entire plane for different values og k1 and k2.

    My question is that if it also has to pass through P (with ~OP=P) isn't k1a + k2b a sufficient equation ? I think so.

    But my book says differently. It says that the equation will become P+k1a + k2b .

    Please help !
    1. Two non-collinear vectors define the direction of a plane in 3D-space. Imagine a (closed) book: All pages are parallel that means they have the same direction. If you want to pick one specific page you must know at least one point of this page.

    2. I've attached a sketch to illustrate what I meant: Only the greyed plane contains the point P. To reach any point on this plane you start at the origin, go to P and then a multiple of \vec u and a multiple of \vec v.
    Attached Thumbnails Attached Thumbnails Vectors. Did not know where to post this one.-ebenenstuetzvektor.png  
    Thanks from pratique21
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    Re: Vectors. Did not know where to post this one.

    Thanks a lot. I get it now.
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