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Find the Cartesian equation for a plane parallel to a given vector and passing through the intersection of given planes

Let M be the plane which is parallel to \mathbf{j} and which passes through the intersection of the planes

    \[ x + 2y + 3z = 4 \qquad \text{and} \qquad 2x + y + z = 2. \]

Find a Cartesian equation for M.


First, we find the intersection of the two planes. This is the set of points which simultaneously satisfy

    \begin{align*}  x + 2y + 3z &= 4 \\  2x + y + z &= 2. \end{align*}

From the first equation we have

    \[ y = \frac{1}{2} (4 - x - 3z). \]

Plugging this into the second equation we have

    \[ 2x + 2 - \frac{1}{2} x - \frac{3}{2} z + z = 2 \quad \implies \quad z = 3x. \]

This then gives us y = 2-5x. So the line is the set of points (x, 2-5x, 3x) where x is arbitrary. Thus,

    \[ L = \{ (0,2,0) + t(1,-3,3) \}. \]

The plane must then contain this line and be parallel to the vector (0,1,0). Since it is parallel to (0,1,0) its normal must be perpendicular to (0,1,0), thus,

    \[ M = \{ (0,2,0) + s(1,-5,3) + t(b_1, b_2, b_3) \} \]

and we must have

    \[ (0,1,0) \cdot (1,-5,3) \times (b_1, b_2, b_3) = 0 \quad \implies \quad b_3 = 3b_1. \]

Therefore,

    \[ M = \{ (0,2,0) + s(1,-5,3) + t(1,0,3) \}. \]

This has the Cartesian equation

    \[ 3x - z = 0. \]

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