Finding the value for $a$ for which three planes don't intersect

Question

Find the value of $a$ for which the three planes \begin{align*} \Pi_1 \colon \phantom{2} x - 2y + z &= 7 \\ \Pi_2 \colon 2x + y - 3z &= 9 \\ \Pi_3 \colon \phantom{2} x + y - az &= 3 \end{align*} do not intersect.

(Original image here.)

I'm attempting to solve this problem. I tried solving these equations simultaneously and arrived at $a= 1/z + 2$. I don't know how to continue or if what I've been doing is even correct.

Some help would be greatly appreciated.

Working:

$x-2y+z=7$

$2x+y-3z=9$

$x+y-az=3$

$4x+2y-6z=18$

$5x-5z=25$

$2x+2y-2az=6$

$3x-2az+z=13$

$15x-15z=75$

$15x-10az+5z=65$

$-15z+10az-5z=10$

$-20z+10az=10$

$2-a=-1/z$

$a=1/z+2$

Answer 1

Guide:

• We do not want the columns to span $\mathbb{R}^3$ or it will be consistent.

• Let $C_i$ be the $i$-th column.

• If the third column is a linear combination of the first two, determine $a_1$ and $a_2$ where $C_3=a_1C_1+a_2C_2$ using the first two rows.

• Using $a_1$ and $a_2$ that you found in the previous step, you should be able to compute $a$ and verify that it satisfy the condition.

Remark about your attempt:

• After you obtain $-20z+10az=10$, we have $z(2-a)=-1$. Now, we note that if $a=2$, then we get a contradiction, which is the value that you are looking for.

• Suppose $a \ne 2$, then you can solve for $z$ in terms of $a$ and in turn solve for $x$ and then $y$.

• General writing remark for improving readability, you might like to label your equations and include how are each line obtain, which operation was performed.

• Things will be neater after you learn about Gaussian eliminations.

Answer 2

The planes do not intersect iff the system of equations has no solution. A necessary condition for that is that the determinant of the coefficient matrix vanishes. We have $$\det\begin{bmatrix}1&-2&1\\2&1&-3\\1&1&-a\end{bmatrix} = 10-5a.$$ There’s only one possible value of $a$ for which this determinant is equal to zero. Check this against the original system of equations to eliminate the possibility that you have an infinite number of solutions instead.