Determine whether a line is parallel to given planes

by RoRi

April 29, 2016

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.

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

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.
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 $M$ .
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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