I have four planes in an ordinary orthogonal (x, y, z)-coordinate system – given by the equations:
$$ \begin{cases} α_1 :& x &+ 2ay + az &= a^2 \\ α_2 :& x &+ ay + az &= a \\ α_3 :& x &+ a^2 z &= a^3 \\ α_4 :& ax &+ ay + a^2 z &= a \\ \end{cases} $$ I'm asked to find a value for (a), for which the four planes do not have any point in common. What steps do you recommend taking towards a solution?
Consider following system os equations:
$$\begin {cases} Ax+By+Cz=D \\A_1x+B_1y+C_1z=D_1\end{cases}$$
If:
$$\frac{A_1}A=\frac{B_1}B=\frac{C_1}C=\frac{D_1}D$$
then the system is consistent and it represent a line. Now we find thses for for equations:
1&2: $1/1=1,2a/a=2, 1/1=1, a^2/a=a$ or $(1, 2, 1, a)$
2&3: $(1, 0, a, a^2)$
3&4: $(1/a, 0, a, a^2)$
1&4: $(a, 1/2, a, 1/a)$
1&3: $(1, 0, a, a)$
2&4: $(a, 1, a, 1)$
As can be seen all systems of couple of plains are not consistent for any $a$ except the system of the couple 2&4 which can be consistent if $a=1$. Hence the condition is $a\neq 1$