Find the values of A and b for which the planes are parallel

Question

I was doing a problem in Linear Algebra where I have to find all values of let's say A and B for which the planes are parallel.

\begin{align} 2x - 4Ay + 2Bz &= 2 + 2B \tag1\\ 3Bx - 3y + 3z &= 3 + 3B \tag2 \end{align}

First I divided the (1) equation by 2 and (2) equation by 3, then I extracted the plane vector from both planes and two planes are parallel if their vector planes are the same (linear or co-linear) and I got:

1 / B = 2A / 1 = B / 1 = 1+B / 1+B

From this we can clearly see that B = 1 and A = 1/2, But the professor said that there are more A and B for which these planes are parallel but I don't see how can I find them, can anybody give me a hint how to find the other solutions ?!

Answer 1

Note that

\begin{align} 2x - 4Ay + 2Bz &= 2 + 2B \tag1\\ 3Bx - 3y + 3z &= 3 + 3B \tag2 \end{align}

are parallel if and only if normal vectors

$$\vec n_1=(2,-4A,2B) \quad \vec n_2=(3B,-3,3)$$

are multiple that is $\vec n_2=k\cdot \vec n_1$ for some $k\in \mathbb{R}$

• $3B=2k$
• $-3=-4Ak$
• $3=2kB$

Solving the system we find $k=\frac32$ and then

• $A=\frac12$
• $B=1$

Answer 2

Two planes $$Px+Qy+Rz=S$$ $$Tx+Uy+Vz=W$$ are parallel if and only if there exists some real number $k$ such that $$k(P,Q,R)=(T,U,V)$$

If, furthermore, $kS=W$, they are the same plane.

Answer 3

The planes will be parallel when their normals are parallel, i.e., that $(2,-4A,2B) = k(3B,-3,3)$ for some scalar $k$. You can avoid introducing the extra variable $k$ by taking advantage of working in $\mathbb R^3$: the parallel condition can be expressed as $(2,-4A,2B) \times (3B,-3,3)=0$, which will give you three equations in $A$ and $B$ only, namely, $$\begin{align} -12A+6B &= 0 \\ 6B^2-6 &= 0 \\ 12AB-6 &=0. \end{align}$$