# Condition for Planes to be Parallel

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## Theorem

Let $P: \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$ be a plane in $\R^3$.

Then the plane $P'$ is parallel to $P$ if and only if there is a $\gamma' \in \R$ such that:

- $P' = \set {\tuple {x_1, x_2, x_3} \in \R^3 : \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma'}$

## Proof

### Necessary Condition

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### Sufficient Condition

Let $P' \ne P$ be a plane given by the equation:

- $\alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma'$

Aiming for a contradiction, suppose we have a point:

- $\mathbf x = \tuple {x_1, x_2, x_3} \in P \cap P'$

Then, as $\mathbf x \in P$, it also satisfies:

- $\alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$

It follows that $\gamma = \gamma'$, so $P = P'$.

This contradiction shows that $P \cap P' = \O$, that is, $P$ and $P'$ are parallel.

The remaining case is when $P' = P$.

By definition, $P$ is parallel to itself.

The result follows.

$\blacksquare$

## Also see

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text V$: Vector Spaces: $\S 28$: Linear Transformations