Find the distance between the parallel planes x + y - z + 4 = 0 and x + y - z + 5 = 0.

Hint: In this question it is given that we have to find the distance between the parallel planes $$x+y-z+4=0$$ and $$x+y-z+5=0$$. So to find the distance between two parallel planes we need to the distance formula which states that, if $$ax+by+cz+d_{1}=0$$ and $$ax+by+cz+d_{2}=0$$ be two parallel planes then the perpendicular distance between,
$$\mathrm{D} =\dfrac{|d_{2}-d_{1}|}{\sqrt{a^{2}+b^{2}+c^{2}} }$$..........(1)

Complete step-by-step solution:
Here the given equations are
$$x+y-z+4=0$$...........(2)
$$x+y-z+5=0$$...........(3)
Now comparing equation (2) with $$ax+by+cz+d_{1}=0$$ and equation (3) with $$ax+by+cz+d_{2}=0$$ , we get,
$$a=1,\ b=1,\ c=-1,\ d_{1}=4\ \text{and} \ d_{2}=5$$
Therefore, by formula (1) we can say that the distance between the planes (2) and (3) is,
$$\mathrm{D} =\dfrac{|d_{2}-d_{1}|}{\sqrt{a^{2}+b^{2}+c^{2}} }$$
$$=\dfrac{\left\vert 5-4\right\vert }{\sqrt{1^{2}+1^{2}+\left( -1\right)^{2} } }$$
$$=\dfrac{\left\vert 1\right\vert }{\sqrt{1+1+1} }$$
$$=\dfrac{1}{\sqrt{3} }$$
So the distance is $$\dfrac{1}{\sqrt{3} }$$.

Note: While solving this type of question you need to know that the distance between two planes implies the perpendicular distance, and the perpendicular distance is also called the shortest distance. If the perpendicular distance is given between two planes then the plane must be parallel to each other because perpendicular distance always measures in between two parallel planes.