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Determine whether a line is parallel to given planes

We say that a line L in the direction of a vector X is parallel to a plane M if X is parallel to M. Consider the line L through the point (1,1,1) and parallel to the vector (2,-1,3). Determine whether L is parallel to the following planes.

  1. The plane through the point (1,1,-2) and spanned by (2,1,3) and (\frac{3}{4}, 1,1).
  2. The plane through the points (1,1,-2), \ (3,5,2), \ (2,4,-1).
  3. The plane determined by the Cartesian equation x + 2y + 3z = -3.

  1. This asks if X = (2,-1,3) is in the span of \{ (2,1,3), (\frac{3}{4}, 1,1) \}, i.e., does there exist s,t such that

        \begin{align*}  2s + \frac{3}{4}t &= 2 \\  s + t &= -1 \\  3s + t &= 3. \end{align*}

    From the second equation we have s = -1-t. Then, from the first, 2 = 2 - 2t + \frac{3}{4}t which implies

        \[ t = -\frac{16}{5} \qquad \implies \qquad s = \frac{11}{5}. \]

    But then,

        \[ 3s + t = 3 \left( \frac{11}{5} \right) - \frac{16}{5} = \frac{17}{5} \neq 3. \]

    Thus, there is no solution, so the line is not parallel to the plane.

  2. The plane through the points (1,1,-2), \ (3,5,2), \ (2,4,-1) is the set of points

        \[ M = \{ (1,1,-2) + s((3,5,2) - (1,1,-2)) + t((2,4,-1) - (1,1,-2)) \} = \{ (1,1,-2) + s(2,4,4) + t(1,3,1) \}. \]

    For X to be in the span of \{ (2,4,4), (1,3,1) \} we must have s,t such that

        \begin{align*}  2s + t &= 2 \\  4s + 3t &= -1 \\  4s + t &= 3. \end{align*}

    From the first equation we have t = 2 - 2s. Then from the second we have 4s + 6 - 6s = -1 which implies

        \[ s = \frac{7}{2} \qquad \implies \qquad t = -5. \]

    But then,

        \[ 4s + t = 14 -5 = 9 \neq 3. \]

    Hence, L is not parallel to M.

  3. The plane with Cartesian equation x + 2y + 3z = -3 is the set of points

        \[ M = \{ (x,y,z) \mid x+2y + 3z = -3 \}. \]

    The points (-3,0,0), \ (0, -\frac{3}{2}, 0), \ (0,0,-1) are all in M. So,

        \begin{align*}  M &= \{ (-3,0,0) + s((0, -\frac{3}{2}, 0) - (-3,0,0)) + t((0,0,-1) - (-3,0,0)) \} \\  &= \{ (-3,0,0) + s(3, -\frac{3}{2}, 0) + t(3,0,-1) \}. \end{align*}

    Thus, we ask if X= (2,-1,3) is in the span of \{ (3,-\frac{3}{2},0), (3,0,-1) \}. This requires that there exist s,t such that

        \begin{align*}  3s + 3t &= 2 \\  -\frac{3}{2}s &= -1 \\  -t &=3. \end{align*}

    But, this fails since the second and third equations implies s = \frac{2}{3} and t = -3. But then

        \[ 3s+3t = 2 - 9 = -7 \neq 2. \]

    Hence, this line is not parallel to the plane.

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