Equation of a plane that crosses the axes at points equidistant from the origin.

Question

Give the equation of a plane that crosses the axes at points equidistant from the origin. Explain your reasoning.

I know the equation should be on a 45 degree angle looking towards the axis. I have the picture in my head, but I don't know.

Answer 1

The answer is $x+y+z=1$. Check that it is a plane and passes through $(1,0,0),(0,1,0),(0,0,1)$ which are all the same distance (one) from the origin.

Answer 2

Your plane does not pass thorough the origin , so itd equation is of the form: $$ ax+by+cz=1 $$ Now the point where this plane intercept the axis are:

for $(y,z)=(0,0) \rightarrow x= \dfrac{1}{a}$

for $(x,z)=(0,0) \rightarrow y= \dfrac{1}{b}$

for $(x,y)=(0,0) \rightarrow z= \dfrac{1}{c}$

since you want $\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}=k$ (the same distance)

the searched equation is: $x+y+z=k$