0
$\begingroup$

I know this should be quite an easy exercise, but I've been having some problems in solving it. Here it goes:

I'm asked to determine the equations of the two planes of $\Bbb R^3$ that include the origin and are parallel to the line defined by: $x=0$ and $y=z+2$. Also the distance from each plane to the line is $1$.

If anyone could explain me the logic behind its resolution, it would be much appreciated.

  • 2
    If $x=0$ and $y=z=2$, you don't get a line but a single point. – 2017-02-19
  • 0
    My bad, I meant y=z+2 – 2017-02-19
  • 0
    There are infinite planes including the origin and parallel to a given line. – 2017-02-19

1 Answers 1

0

Let the line be $L_1$ and the planes be $P_1$ and $P_2$. Because both planes include the origin and are parallel to $L_1$, they must include the line defined by $x=0\land y=z$, hereafter referred to as $L_2$.

image 1

It is easy to see that the distance between $L_1$ and $L_2$ equals the distance between $(0,0,0)$ and $(0,1,-1)$, which is $\sqrt2$.

Consider the plane perpendicular to the $Y-Z$ plane and including the line $x=0\land y=-z$, hereafter referred to as $P_3$. It intersects $L_1$ and $L_2$ at $(0,0,0)$ and $(0,1,1)$ respectively. The projection of $P_3$ on the $X-Y$ plane is as follows (the projection of plane $P_A$ onto plane $P_B$ is obtained by projecting every point on $P_A$ onto $P_B$):

image 2

$z$-values included to ease actual distance calculation.

Note that the $\sqrt2$ is the actual distance while the distance in the projection (looking at $x$- and $y$-value only) is only $1$. This is because the vertical distances in this projection is $\frac1{\sqrt2}$ times the actual distances.