Find volume of cube with the help of eqn of plane

Question

The volume of cube whose two faces lie on the plane 6x-3y+2z+1=0 and 6x-3y+2z+4=0?

Answer 1

Since we have parallel planes we can find the distance between the planes.

Let y = x = 0.

So 6(0) - 3(0) + 2z + 4 = 0 --> 2z = -4 --> z = -2.

Thus a point in that plane is (0,0,2). Now use formula to find distance between the other plane and this point. That is,

$D = \frac{|ax+by+cz+d|}{\sqrt{a^{2}+b^{2}+c^{2}}}$

So $D = \frac{|6(0)+ -3(0)+2(-2)+1|}{\sqrt{6^{2}+(-3)^{2}+2^{2}}} = \frac{|-4 + 1 |}{\sqrt{36+9+4}} = \frac{3}{7}$.

Thus a cube with faces of these two planes would have a distance/length of $\frac{3}{7}$.

So $(\frac{3}{7})^{3} = \frac{27}{343}$