Finding the Equations for Planes of Different Faces of a Box

Question

The question is as follows:

The illustration at bottom shows a rectangular box, three of whose edges coincide with the coordinate axes, and one of whose vertices is $(3, 6, 4)$. Give an equation for the plane containing the front face of the box. Do the same for the top and right faces.

For the front face, I found the slope by calculating $\frac{\Delta z}{\Delta y}$ which led me to $\frac{4}{6} = \frac{2}{3}$. However, I don't know whether it should be $\frac{2}{3}x$, $\frac{2}{3}y$, or $\frac{2}{3}z$ nor do I know what the equation will result in a $x$, $y$, or $z$ value.

I hope that by receiving assistance in this part of the problem, then I may be able to calculate the equations for the other planes of the two other faces. Any help will be greatly appreciated.

Answer 1

Planes don't have "slope", they have normal vectors. But this problem is easier than that. The front face is parallel to the $yz$ plane and has $x$-intercept $3$. Every point in that plane has $x$-coordinate $3$ and there is no restriction on the $y$ and $z$ coordinates. So the equation of the plane is $x=3.$

Answer 2

You can write the equation of a plane by knowing a point that belongs to the plane ${\bf x}_0 = (x_0, y_0, z_0)$ and a vector perpendicular to the plane ${\bf n}$. The equation then becomes

$$ {\bf n}\cdot({\bf x} - {\bf x}_0) = 0 \tag{1} $$

For example, for the front face

• ${\bf x}_0 = (3, 0, 0)$ and
• ${\bf n} = (1, 0, 0)$

The equation of the plane is then

$$ (1, 0, 0)\cdot[(x, y, z) - (3, 0, 0)] = 0 \\ \Rightarrow ~~~x - 3 = 0 $$

I will leave the other two for you to work out