Condition for Planes being 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 $P$ iff there is a $\gamma' \in \R$ such that:
- $P' = \left\{{ \left({x_1, x_2, x_3}\right) \in \R^3 : \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma' }\right\}$
Proof
Necessary Condition
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Sufficient Condition
Let $P' \ne P$ be a plane given by the equation:
- $\alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma'$
Suppose we have a point $\vec x = \left({x_1, x_2, x_3}\right) \in P \cap P'$.
Then, as $\vec x \in P$, 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 $P \cap P' = \varnothing$, i.e., $P$ 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
- Seth Warner: Modern Algebra (1965): $\S 28$
