Dot Product Distributes over Addition/Proof 2
Theorem
- $\paren {\mathbf u + \mathbf v} \cdot \mathbf w = \mathbf u \cdot \mathbf w + \mathbf v \cdot \mathbf w$
Proof
Let the vectors $\mathbf u$, $\mathbf v$ and $\mathbf w$ be embedded in a Cartesian $3$-space.
It is noted that $\mathbf u$, $\mathbf v$ and $\mathbf w$ are not necessarily coplanar.
Let instances of $\mathbf u$ and $\mathbf w$ be selected so their initial points are at some point $O$.
Let an instance of $\mathbf v$ be selected so its initial point is positioned at the terminal point $U$ of $\mathbf u$.
Let the terminal point $\mathbf v$ be $V$.
Let $UA$ be dropped perpendicular to $\mathbf w$.
Let $VB$ be dropped perpendicular to $\mathbf w$.
Let another instance of $\mathbf w$ be selected so that its initial point is at $U$.
Let $VC$ be dropped perpendicular to this second instance of $\mathbf w$.
Let $CB$ be dropped from $C$ to the first instance of $\mathbf w$.
We have that:
- $UA \parallel CB$
and:
- $UC \parallel AB$
Thus $\Box ABCU$ is a parallelogram.
Hence:
- $AB = UC$
Then we have that:
\(\ds OA\) | \(=\) | \(\ds OU \cos \angle \mathbf u, \mathbf w\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\mathbf u} \cos \angle \mathbf u, \mathbf w\) | ||||||||||||
\(\ds AB\) | \(=\) | \(\ds UV \cos \angle \mathbf v, \mathbf w\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\mathbf v} \cos \angle \mathbf v, \mathbf w\) | ||||||||||||
\(\ds OB\) | \(=\) | \(\ds OV \cos \angle \paren {\mathbf u + \mathbf v}, \mathbf w\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\paren {\mathbf u + \mathbf v} } \cos \angle \paren {\mathbf u + \mathbf v}, \mathbf w\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\mathbf u} \cos \angle \mathbf u, \mathbf w + \norm {\mathbf v} \cos \angle \mathbf v, \mathbf w\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \norm {\paren {\mathbf u + \mathbf v} } \norm {\mathbf w} \cos \angle \paren {\mathbf u + \mathbf v}, \mathbf w\) | \(=\) | \(\ds \norm {\mathbf u} \norm {\mathbf w} \cos \angle \mathbf u, \mathbf w + \norm {\mathbf v} \norm {\mathbf w} \cos \angle \mathbf v, \mathbf w\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {\mathbf u + \mathbf v} \cdot \mathbf w\) | \(=\) | \(\ds \mathbf u \cdot \mathbf w + \mathbf v \cdot \mathbf w\) | Definition of Dot Product |
Hence the result.
$\blacksquare$
Sources
- 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product: $(2.7)$
- 1957: D.E. Rutherford: Vector Methods (9th ed.) ... (previous) ... (next): Chapter $\text I$: Vector Algebra: $\S 2$.
- 1992: Frederick W. Byron, Jr. and Robert W. Fuller: Mathematics of Classical and Quantum Physics ... (previous) ... (next): Volume One: Chapter $1$ Vectors in Classical Physics: $1.3$ The Scalar Product