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    dot product

    What is the equation of a plane which is through (-2,3,2) and parallel to 3x + y + z = 4
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    Quote Originally Posted by EquinoX View Post
    What is the equation of a plane which is through (-2,3,2) and parallel to 3x + y + z = 4
    The new plane has the same normal

    \vec{n} = < 3,1,1>

    The plane containing your point is

    3(x+2) + 1(x-3) + 1(x-2) = 0

    or

    3x+y+z = -1
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    isn't that perpendicular? instead of parralel? what would I have to change if the question asks for parallel
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    Quote Originally Posted by EquinoX View Post
    isn't that perpendicular? instead of parralel? what would I have to change if the question asks for parallel
    No. The two parallel planes will have parallel normals.
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    so how would I work this problem out if it's perpendicular?
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    Quote Originally Posted by EquinoX View Post
    so how would I work this problem out if it's perpendicular?
    Planes are perpendicular is their normals are perpendicular. Find a vector lying on the plane and use this for the normal of the second plane (BTW - there is an infinite number of perpendicular planes since there are an infinite number of vectors lying on a given plane.)
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    I got confused by your answer as in the book there's a question with this answer:

    Find the equation of the plane perpendicular to -1 + 3j + 2k and passing through the point (1,0,4):

    -(x-1) + 3(y-0) + 2(z-4) = 0

    which is the same way you approach the parallel problem... can you explain this to me?
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    Quote Originally Posted by EquinoX View Post
    I got confused by your answer as in the book there's a question with this answer:

    Find the equation of the plane perpendicular to -1 + 3j + 2k and passing through the point (1,0,4):

    -(x-1) + 3(y-0) + 2(z-4) = 0

    which is the same way you approach the parallel problem... can you explain this to me?
    Ah, but in this question, you want a plane perpendicualr to

    - i + 3j + 2k or <-1,3,2> - this is a vector!

    So you want the line following this vector to be parallel to your normal. Hence, the bold numbers

    <br />
\bold{-1}(x-1) + \bold{3}(y-0) +\bold{2}(z-4) = 0.<br />
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    the answer to the question is not the bold numbers instead it is:

    -x + 3y + 2 = 7
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    Quote Originally Posted by danny arrigo View Post
    Ah, but in this question, you want a plane perpendicualr to

    - i + 3j + 2k or <-1,3,2> - this is a vector!

    So you want the line following this vector to be parallel to your normal. Hence, the bold numbers

    <br />
***\;\;\; \bold{-1}(x-1) + \bold{3}(y-0) +\bold{2}(z-4) = 0.<br />
    If you expand this (**), you get

    <br />
-x + 3 y + 2z = 7.<br />
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    Quote Originally Posted by EquinoX View Post
    the answer to the question is not the bold numbers instead it is:

    -x + 3y + 2 = 7
    I don't want to pick at you but I think it's high time that you read this here: http://www.mathhelpforum.com/math-he...l-posters.html
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    okay let me get this straighten out: my original question asks to solve a parallel problem and the example from the book that I present ask for perpendicular, and you solve both the same way.. why is that?
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  13. #13
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    Quote Originally Posted by EquinoX View Post
    ... my original question asks to solve a parallel problem
    Two planes p_1 and p_2 are parallel if the normal vectors \overrightarrow{n_1} and \overrightarrow{n_2} are collinear. That means:

    \overrightarrow{n_2} = k\cdot \overrightarrow{n_1}\ ,\ k\in \mathbb{R}

    see attachment.

    If you are looking for a plane perpendicular to a given plane, then there are an infinite number of those planes. Example: Open a door. Normally the door is perpendicular to the floor, but there isn't only one position the door can have.

    see attachment.

    and the example from the book that I present ask for perpendicular, and you solve both the same way.. why is that?
    If you have a vector ( \vec n)and a point Q with its stationary vector \vec q you can derive the equation of a plane which is perpendicular to the vector and contains the given point: Let R denote an arbitrary point in the plane with its stationary vector \vec r. Then the vector \vec r - \vec q must be perpendicular to the vector \vec n. That means: For all points in the plane the equation

    \vec n \cdot (\vec r - \vec q)=0

    is true and therefore this equation describes the plane completely (Point-normal-form of the equation of a plane)

    see attachment
    Attached Thumbnails Attached Thumbnails dot product-parallel_planes.png   dot product-perpend_planes.png   dot product-abstd_pktgerade3d.gif  
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