Conditions for Homogeneity/Plane
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Theorem
The plane $P = \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$ is homogeneous if and only if $\gamma = 0$.
Proof
Sufficient Condition
Let the plane $P = \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$ be homogeneous.
Then the origin $\tuple {0, 0, 0}$ lies on $P$.
Hence:
\(\ds \alpha_1 0 + \alpha_2 0 + \alpha_3 0\) | \(=\) | \(\ds \gamma\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \gamma\) | \(=\) | \(\ds 0\) |
$\Box$
Necessary Condition
Let the equation of $P$ be $\alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = 0$.
Then:
- $0 = \alpha_1 0 + \alpha_2 0 + \alpha_3 0 \in P$
and so:
- $\tuple {0, 0, 0}$ lies on $P$
Hence $P$ is homogeneous.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations