Points on a line p equidistant from two planes

Question

Let $\pi_1:x+y-z=1$ and $\pi_2: x-y+z=5$, and line $p: (1,0,0)+\lambda(1,1,1)$. Find all points on $p$ equidistant from both planes.

I forgot how to solve these problems but my idea is that we substitute line $p$ in both of the planes? Can someone help me?

Answer 1

HINT: the Hessian Normalform of your two planes are given by $$\frac{x+y-z-1}{\pm \sqrt{3}}=0$$ and $$\frac{x-y+z-5}{\pm\sqrt{3}}=0$$ then we get $$\left|\frac{x_g+y_g-z_g-1}{\sqrt{3}}\right|=d$$ and $$\left|\frac{x_g-y_g+z_g-5}{\sqrt{3}}\right|=d$$ where $$x_g=1+\lambda,y_g=\lambda,z_g=\lambda$$

Answer 2

Hints:

Step 1: Find formulas $d(P,\pi_1)$ and $d(P,\pi_2)$ for the distance between a point $P = (a,b,c)$ and the planes $\pi_1$ and $\pi_2$.

If you need a hint for this, one way to do it would be to take any point $(x, y, z) \in \pi_1$ and consider the unit normal vector $\frac{1}{\sqrt{3}}(1, 1, -1)$ in relation to the difference vector $(a,b,c) - (x,y,z)$... can you take it from here?

Step 2: Now, write out the equation $d(P, \pi_1) = d(P, \pi_2)$ and substitute the formula for the line $p$.

You should get a quadratic equation on $\lambda$.