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Thread: Help with vectors

  1. #1
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    Help with vectors

    Hello I need help with the following:

    u = 2i +3j v= -i + 2j

    Find a cartesian equation for the plane P containing the point Q(1,2,3) and parallel to u and w.

    Find the perpendicular distance between the plane P and the origin
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  2. #2
    Senior Member Twig's Avatar
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    hi

    Do you know what a cross product is?

    You could use it to get a vector that is orthogonal to both u and w.

    However in this case, itīs clear that both u and v are in the xy plane so to speak. So you will get a normal pointing "straight up" from the xy-plane.

    You will need your normal to satisfy $\displaystyle (0,0,1) \cdot (1-x,2-y,3-z)=0$ . So you get $\displaystyle z = 3 $ . Note here that the $\displaystyle \cdot $ means dot-product.

    So you have a plane parallell to the xy-plane, with perpendicular distance of 3 to the origin.
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  3. #3
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    Hello, gva0324!

    Are you sure of the wording of this problem?
    As given, it is too simple . . .


    Given: .$\displaystyle \begin{array}{ccc}\vec u &=& 2i +3j \\ \vec v&=& \text{-}i + 2j \end{array}$

    (a) Find a cartesian equation for the plane $\displaystyle P$ containing the point $\displaystyle Q(1,2,3)$
    . . . and parallel to $\displaystyle \vec u$ and $\displaystyle \vec v$

    Since $\displaystyle \vec u$ and $\displaystyle \vec v$ are in the $\displaystyle xy$-plane, then plane $\displaystyle P$ is parallel to the $\displaystyle xy$-plane.

    Since plane $\displaystyle P$ contains $\displaystyle Q(1,2,3)$, its equation is: .$\displaystyle z \:=\:3$




    (b) Find the perpendicular distance between the plane $\displaystyle P$ and the origin

    The point on plane $\displaystyle P$ nearest the origin is $\displaystyle (0,0,3).$

    Its distance from the origin is $\displaystyle 3.$

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