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