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Math Help - Vector-help!

  1. #1
    Junior Member
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    Vector-help!

    Find the cartesian equation of the plane which passes through the point
    (3,-4, 1) and which is parrallel to the plane containing the point (1,2,-1) and the line r=t( i+j+k)
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  2. #2
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    Hello, kingkaisai2!

    Find the equation of the plane which passes through the point P(3,-4,1) and which is
    parallel to the plane containing the point A(1,2,-1) and the line \vec{r} \:=\:t(\vec{i} + \vec{j} + \vec{k})

    To write the equation of a plane, we need a point (x_1,y_1,z_1) and its normal direction: \vec{n} = \langle a,b,c\rangle
    . . Then the equation is: . a(x-x_1) + b(y-y_1) + c(z-z_1)\:=\:0

    We have a point P(3,-4,1), so we need the normal vector: \vec{n}

    Since this plane is parallel to the plane containing point A and line \vec{r},
    . . it has the same normal vector.

    The line \vec{r}\:=\:t(\vec{i} + \vec{j} + \vec{k}) has parametric equations: . \begin{Bmatrix}x = t \\ y = t \\ z = t\end{Bmatrix}
    For t = 1, we have the point: B(1,1,1)
    For t = 2, we have the point: C(2,2,2)

    Hence, the plane contains points: A(3,-4,1),\;B(1,1,1),\;C(2,2,2)

    The normal vector is perpendicular to \overrightarrow{AB} and \overrightarrow{AC}.

    Since \overrightarrow{AB} = \langle -2,5,0\rangle and \overrightarrow{AC} = \langle -1,6,1\rangle
    . . \vec{n}\;=\;\overrightarrow{AB} \times \overrightarrow{AC} \;= \;\begin{vmatrix}\vec{i} & \vec{j} & \vec{k} \\ -2 & 5 & 0 \\ -1 & 6 & 1\end{vmatrix} \;=\;\langle 5,2,-7\rangle

    So the desired plane has point P(3,-4,1) and normal \vec{n} = \langle 4,2,-7\rangle

    Can you finish it now?

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