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Find the parametric equation for a line through a point and parallel to two planes

by RoRi

We say that a line is parallel to a plane if the direction vector of the line is parallel to the plane. Let L be the line containing the point (1,2,3) and parallel to the planes

    \[ x + 2y + 3z = 4, \qquad 2x + 3y + 4z = 5. \]

Find a vector parametric equation for L.

The normal vectors of the planes are N_1 = (1,2,3) and N_2 = (2,3,4). So, the direction vector A = (a_1, a_2, a_3) of L will be perpendicular to both of these,

    \begin{align*} N_1 \cdot A &= 0 & \implies && a_1 + 2a_2 + 3a_3 &= 0 \\ N_2 \cdot A &= 0 & \implies && 2a_1 + 3a_2 + 4a_3 &= 0. \end{align*}

From the first equation we have x = -2y - 3z. Plugging this into the second equation we obtain y = -2z, which then gives us x = z. Since z is arbitrary, we take z = 1 to obtain a direction vector A = (1,-2,1). Therefore, the vector parametric equation for the line is

    \[ X(t) = (1,2,3) +t(1,-2,1). \]

Determine which points are on a plane given by a parametric equation

by RoRi

Let M be a plane defined by the scalar parametric equations

    \[ x = 1+s-2t, \qquad y = 2+s+4t, \qquad z = 2s+t. \]

  1. Determine which of the following points are on M: \ \ (0,0,0), \ (1,2,0), \ (2,-3,-3).
  2. Find vectors P,A,B such that M = \{ P + sA + tB \}.
  1. The point (0,0,0) is not on M since

        \[ 2s+t = 0 \quad \implies \quad t = -2s. \]

    Then,

        \[ 2+s+4t = 0 \quad \implies \quad 2+s+4(-2s) = 0 \quad \implies \quad s = \frac{2}{7}, \quad t = -\frac{4}{7}. \]

    But then the first coordinate fails since

        \[ 1+s-2t = 1 + \frac{2}{7} + \frac{8}{7} = \frac{17}{7} \neq 0. \]

    The point (1,2,0) is on M since the system of equations

        \begin{align*} 1+s-2t &= 1 \\ 2 + s + 4t &= 2 \\ 2s + t &= 0 \end{align*}

    has a solution s = t = 0. Therefore, (1,2,0) = (1+s-2t, 2+s+4t,2s+t) for s = t = 0.

    The point (2,-3,-3) is on M since the system of equations

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

    has a solution s = t = -1.

  2. Since (1,2,0) is on the plane we take P = (1,2,0). Then from the parametric equations we get the vectors A and B as the coordinates of s and t. So, A = (1,1,2) and B = (-2,4,1). Then we have

        \[ M = \{ (1,2,0) + s(1,1,2) + t(-2,4,1) \}. \]