The point A (4, 3, c) is equidistant from the planes P1 and P2. Calculate the two possible values of c

Question

The point $A (4, 3, c)$ is equidistant from the planes $P_1$ and $P_2$. Calculate the two possible values of $c$.

• Plane $P_1$ has equation $r\cdot (2,-2,1)=1$
• Plane $P_2$ has equation $r\cdot (-6,3,2)=-1$

I have tried to find the plane that is equidistant from $P_1$ and $P_2$ but I can't find the normal to that plane.

Answer 1

let $M_1, M_2$ are the feet of perpendiculars from the points $A$ to the respective planes $p_1, p_2.$ then we have $$M_1 = (4 + 2t, 3-2t, c+t), M_2= ( 4 -6s, 3+3s, c + 2s) \text{ for real }s, t.$$

three constrains are: $$2(4+2t)-2(3-2t)+(c+t) = 1\\ -6(4-6s)+3(3+3s) +2(c+2s) = -1\\t^2(2^2 + (-2)^2 + 1^2) = s^2((-6)^2 + 3^2 + 2^2) $$

you can find $c$ from these.

Answer 2

Before we do anything, let's convert the planes into normal form, such that the vectors used are length 1.

$P_1$ has equation $\vec r\cdot\left(\frac{2}{3},-\frac{2}{3},\frac{1}{3}\right)=\frac{1}{3}$.

$P_2$ has equation $\vec r\cdot\left(-\frac{6}{7},\frac{3}{7},\frac{2}{7}\right)=-\frac{1}{7}$.

This is convenient, because it means for a general plane $\vec r\cdot \vec v=c$, the distance from a point $\vec x$ from the plane is $D=|c - \vec x\cdot \vec v|$. It's especially convenient in this case, because the vectors were already integer length.

So, given this, we find the distances from the point $A$ to the two planes:

$$\begin{align}D_1 &= \left|\frac{1}{3}-(4,3,c)\cdot\left(\frac{2}{3},-\frac{2}{3},\frac{1}{3}\right)\right|\\ &=\left|\frac{1}{3}-\left(\frac{8}{3}-2+\frac{c}{3}\right)\right|\\ &=\left|-\frac{1}{3}-\frac{c}{3}\right|\\ &=\left|\frac{c+1}{3}\right|\end{align}$$

$$\begin{align}D_2 &= \left|-\frac{1}{7}-(4,3,c)\cdot\left(-\frac{6}{7},\frac{3}{7},\frac{2}{7}\right)\right|\\ &=\left|-\frac{1}{7}-\left(\frac{-24}{7}+\frac{9}{7}+\frac{2c}{7}\right)\right|\\ &=\left|\frac{2c}{7}-2\right|\end{align}$$

Now we have to equate these. This is somewhat annoying with absolute value and fractions, so I'm going to get rid of both, by multiplying both by a common number and then squaring.

$$21D_1=|7c+7|$$ $$21D_2=|6c-42|$$

$$49c^2+98c+49=36c^2-504c+1764$$ $$13c^2+602c-1715=0$$

Applying the quadratic formula now gives us the answers.

$$\begin{align}c&=\frac{-602\pm\sqrt{602^2-4\cdot13\cdot1715}}{2\cdot13}\\ &=\frac{-602\pm56\sqrt{114}}{26}\\ &=\frac{-301\pm28\sqrt{114}}{13}\\ &\approx\{-49, \frac{35}{13}\} \end{align}$$