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Find two distinct points in the intersection of two given planes

by RoRi

Let

    \[ P = (1,1,1), \quad A = (2,-1,3), \quad B = (-1,0,2) \quad Q = (2,3,1), \quad C = (1,2,3), \quad D = (3,2,1). \]

Then let M, M' be the two planes

    \[ M = \{ P + sA + tB \}, \qquad M' = \{ Q + sC + tD \}. \]

Find two distinct points on the intersection of M and M'.

First, we have

    \begin{align*} M &= \{ (1,1,1) + s(2,-1,3) + t(-1,0,2) \} = \{ (1+2s-t, 1-s, 1+3s+2t) \} \\ M' &= \{ (2,3,1) + s'(1,2,3) + t'(3,2,1) \} = \{ (2+s'+3t', 3 + 2s' + 2t', 1+3s' + t') \}. \end{align*}

So, the points (x,y,z) in the intersection are those such that

    \begin{align*} x &= 1+2s-t & \text{and} && x &= 2 + s' + 3t' \\ y &= 1-s & \text{and} && y &= 3 + 2s' + 2t' \\ z &= 1+3s + 2t &\text{and} && z&= 1+3s' + t'. \end{align*}

From the first set of equations we have the equations

    \[ x = 1+2s -t, \qquad y = 1-s, \qquad z = 1+3s + 2t. \]

The second equation implies s = y+1, and so the first equation implies t = 3 - x - 2y. Then,

    \begin{align*} && z &= 1 + 3(1-y) + 2(3-x-2y) \\ && &= 10 - 7y - 2x \\ \implies && 2x + 7y + z &= 10. \end{align*}

From the second set of equations we have

    \[ x = 2 + s' + 3t', \qquad y = 3 + 2s' + 2t', \qquad z = 1 + 3s' + t'. \]

The first equation implies s' = x - 2 - 3t'. The second equation then implies

    \[ y = 3 + 2(x- 2 - 3t') + 2t' = -1 + 2x - 4t' \quad \implies \quad t' = \frac{2x-y-1}{4}. \]

Therefore, s' = x - 2 - 3 \left( \frac{2x-y-1}{4} \right). So, finally we have

    \begin{align*} z &= 1 + 3 \left( x - 2 - 3 \left( \frac{2x-y-1}{4} \right) \right) + \frac{2x-y-1}{4} \\ &= -5 + 3x - 2 (2x-y-1) \\ &= -3 -x + 2y. \end{align*}

Therefore, we obtain the Cartesian equation

    \[ x - 2y + z = -3. \]

Thus, the set of points in the intersection of these two planes are those points (x,y,z) such that

    \begin{align*} 2x+7y+z &= 10 \\ x - 2y + z &= -3. \end{align*}

In this case z is arbitrary. First, taking z = 6 we have

    \begin{align*} 2x+7y + 6 &= 10 & \implies && 2x + 7y &= 4 \\ x - 2y + 6 &= -3 & \implies && x - 2y &= -9. \end{align*}

These implies x = -5 and y = 2. On the other hand, taking z = 17 we have

    \begin{align*} 2x+7y + 17 &= 10 & \implies && 2x+7y &= -7 \\ x - 2y + 17 &= -3 & \implies && x - 2y &= -20. \end{align*}

These imply x = -14 and y = 3. Hence, we have

    \[ (-5,2,6), \ (-14,3,17) \in M \cap M'. \]

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