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

by RoRi

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

    \[ 3x - z = 0. \]

One comment

  1. William C says:

    I think we can skip the very last part if we just use t(0,1,0) (since the plane is parallel to (0,1,0) so we know those points will be on the plane, and (0,1,0) and (1,-5,3) are linearly independent)

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