Vectors from Sum and Difference
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Theorem
Let $\mathbf a$ and $\mathbf b$ be vector quantities.
Let $\mathbf c = \mathbf a + \mathbf b$ and $\mathbf d = \mathbf a - \mathbf b$ be given.
Then:
\(\ds \mathbf a\) | \(=\) | \(\ds \dfrac 1 2 \paren {\mathbf c + \mathbf d}\) | ||||||||||||
\(\ds \mathbf b\) | \(=\) | \(\ds \dfrac 1 2 \paren {\mathbf c - \mathbf d}\) |
Proof
\(\ds \dfrac 1 2 \paren {\mathbf c + \mathbf d}\) | \(=\) | \(\ds \dfrac 1 2 \paren {\paren {\mathbf a + \mathbf b} + \paren {\mathbf a - \mathbf b} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {\mathbf a + \mathbf b + \mathbf a - \mathbf b}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {2 \mathbf a}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \mathbf a\) |
\(\ds \dfrac 1 2 \paren {\mathbf c - \mathbf d}\) | \(=\) | \(\ds \dfrac 1 2 \paren {\paren {\mathbf a + \mathbf b} - \paren {\mathbf a - \mathbf b} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {\mathbf a + \mathbf b - \mathbf a + \mathbf b}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {2 \mathbf b}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \mathbf b\) |
$\blacksquare$
Sources
- 1970: George Arfken: Mathematical Methods for Physicists (2nd ed.) ... (previous) ... (next): Chapter $1$ Vector Analysis $1.1$ Definitions, Elementary Approach: Exercise $1.1.1$