Find the volume of rectangular box determined by the following 3 planes with the equations:
$ax + by + cz = \pm A$
$dx + ey + fz = \pm B$
$gx + hy + iz = \pm C$
I tried to make the changes of variable:
$X=ax+by+cz$
$Y=dx+ey+fz$
$Z=gx+hy+iz$
And now with Jacobian matrix If determinant is $0$ the volume is null and why can I find the volume when:
$X=A, X=-A$ are the first pair of parallel planes.
$Y=B, Y=-B$ are the second pair of parallel planes.
$Z=C, Z=-C$ are the third pair of parallel planes.
The distance between the parallel planes $$ax+by+cz = (+ or -) A$$ is $$ D _1 = \frac {2|A|}{\sqrt {a^2+b^2+c^2}}$$
Similarly$$ D _2= \frac {2|B|}{\sqrt {d^2+e^2+f^2}}$$ and $$ D _3= \frac {2|C|}{\sqrt {g^2+h^2+i^2}}$$ Thus the volume of the box is $$ V = \frac {2|A|}{\sqrt {a^2+b^2+c^2}}\frac {2|B|}{\sqrt {d^2+e^2+f^2}}\frac {2|C|}{\sqrt{g^2+h^2+i^2}}.$$