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Math Help - Vectors question/confusion

  1. #1
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    Vectors question/confusion

    Determine the value for x such that the point A(-1,3,4), B(-2,3,-1), and C(-5,6, x) all lie on a plane that contains the origin.

    I'm not sure what they're asking for. The answer in the book is -7.
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  2. #2
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    Try using this:
    If 3 vectors lie on the same plane their Triple product is 0.
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  3. #3
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    You have four points: A, B, C and O (origin). The question is asking for a plane, so try to find an equation of a plane that involves all of these four points. Start by constructing two vectors from these points, say \overrightarrow{OA} and \overrightarrow{OB}. Then find a Cartesian equation of this plane (i.e., in the form Ax+By+Cz=D). Since the plane intersects the origin, D=0. See if you can then solve the equation for x.

    Edit: I suppose VincentP's method is more sophisticated, but both work
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  4. #4
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    Quote Originally Posted by IanCarney View Post
    Determine the value for x such that the point A(-1,3,4), B(-2,3,-1), and C(-5,6, x) all lie on a plane that contains the origin.
    I'm not sure what they're asking for. The answer in the book is -7.
    Here is another way. Think of A,~B,~\&~C as vectors.
    The solve C\cdot(A\times B)=0.
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  5. #5
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    Hello, IanCarney!

    Here is an elementary solution . . .


    Determine the value for \,a such that the points A(\text{-}1,3,4),\;B(\text{-}2,3,\text{-}1),\;C(\text{-}5,6,a)
    all lie on a plane that contains the origin. . (Answer: -7)

    A plane has the general equation: . Ax + By + Cz + D \:=\:0
    . . If the plane contains the Origin, then D \,=\,0.

    \text{Let }\vec u \:=\:\overrightarrow{BA} \:=\:\langle 1,0,5\rangle
    \text{Let }\vec v \:=\:\overrightarrow{BC} \:=\:\langle \text{-}3,3,a+1\rangle

    \text{The normal to the plane is: }\:\vec n \;=\;\vec u \times \vec v \;=\;\begin{vmatrix}i & j & k \\ 1 & 0 & 5 \\ \text{-}3 & 3 & a\!+\!1 \end{vmatrix}

    . . \vec n \;=\;i(0\!-\!15) - j(a\!+\!1\!+\!15) + k(3\!-\!0) \;=\;\text{-}15i - (a\!+\!16)j + 3k

    . . . . =\;\langle \text{-}15,\,\text{-}(a\!+\!16),\:3\rangle \;=\;\langle 15,\:a\!+\!16,\:\text{-}3\rangle


    The equation of the plane through A(\text{-}1,3,4) with \vec n \,=\,\langle 15, a\,+\,\!1,\text{-}3\rangle is:

    . . . . . . 15(x+1) + (a+16)(y-3) - 3(z - 4) \;=\;0

    . . . 15x + 15 + (a+16)y - 3(a+16) - 3z + 12 \;=\;0

    . . . . . . . . . . . 15x + (a+16)y - 3z \underbrace{-\:3a - 21}_{\text{This is }D} \;=\;0


    Therefore: . -3a - 21 \:=\:0 \quad\Rightarrow\quad \boxed{a \:=\:-7}

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