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Determine which points are on the plane determined by (1,1,-1), (3,3,2), (3,-1,-2)

by RoRi

Let

    \[ P = (1,1,-1), \qquad Q = (3,3,2), \qquad R = (3,-1,-2). \]

There is a unique plane M containing these three points. Determine which of the following points are also on M.

  1. \left( 2,2, \frac{1}{2} \right);
  2. \left( 4,0,-\frac{1}{2} \right);
  3. (-3,1,-3);
  4. (3,1,3);
  5. (0,0,0).
First, we have

    \begin{align*} M &= \{ P + s(Q-P) + t(R-P) \} \\ &= \{ (1,1,-1) + s(2,2,3) + t(2,-2,-1) \} \\ &= \{ (1+2s+2t, 1+2s-2t, -1+3s-t) \}. \end{align*}

  1. The point \left( 2,2, \frac{1}{2} \right) is on M since

        \[ 1+2s+2t = 2 \quad \implies \quad s = \frac{1-2t}{2}. \]

    Then,

        \[ 1+2s-2t = 2 \quad \implies \quad 1+1-2t-2t = 2 \quad \implies \quad t = 0 \quad \implies \quad s = \frac{1}{2}. \]

    And then,

        \[ -1 + 3s -t = \frac{1}{2}. \]

    So, \left(2,2,\frac{1}{2}\right) = (1+2s+2t, 1+2s-2t, -1+3s-t) for s = \frac{1}{2} and t= 0.

  2. The point \left( 4,0, -\frac{1}{2} \right) is on M since

        \[ 1+2s+2t = 4 \quad \implies \quad s = \frac{3-2t}{2}. \]

    Then,

        \[ 1+2s-2t = 0 \quad \implies \quad 1+3-2t-2t = 0 \quad \implies \quad t = 1 \quad \implies \quad s = \frac{1}{2}. \]

    And then,

        \[ -1 + 3s -t = -\frac{1}{2}. \]

    So, \left(4,0,-\frac{1}{2}\right) = (1+2s+2t, 1+2s-2t, -1+3s-t) for s = \frac{1}{2} and t= 1.

  3. The point (-3,1,-3) is on M since

        \[ 1+2s+2t = -3 \quad \implies \quad s = \frac{-4-2t}{2}. \]

    Then,

        \[ 1+2s-2t = 1 \quad \implies \quad 1-4-2t-2t = 1 \quad \implies \quad t = -1 \quad \implies \quad s = -1. \]

    And then,

        \[ -1 + 3s -t = -3. \]

    So, (-3,1,-3) = (1+2s+2t, 1+2s-2t, -1+3s-t) for s = t = -1.

  4. The point (3,1,3) is not on M since

        \[ 1 + 2s + 2t = 3 \quad \implies \quad s = \frac{2-2t}{2}. \]

    Then,

        \[ 1 + 2s - 2t = 1 \quad \implies \quad 1 + 2 - 2t - 2t = 1 \quad \implies \quad t = \frac{1}{2} \quad \implies \quad s= \frac{1}{2}. \]

    And then,

        \[ -1 + 3s - t = 0 \neq 3. \]

    Hence, (3,1,3) is not on the plane.

  5. The point (0,0,0) is not on M since

        \[ 1 + 2s + 2t = 0 \quad \implies \quad s = \frac{-1-2t}{2}. \]

    Then,

        \[ 1 + 2s - 2t = 0 \quad \implies \quad 1 -1 - 2t - 2t = 0 \quad \implies \quad t = 0 \quad \implies \quad s= -\frac{1}{2}. \]

    And then,

        \[ -1 + 3s - t = -\frac{5}{2} \neq 0. \]

    Hence, (0,0,0) is not on the plane.

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