Dot Product Distributes over Addition

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Theorem

Let $\mathbf u, \mathbf v, \mathbf w$ be vectors in the real Euclidean space $\R^n$.


Then:

$\paren {\mathbf u + \mathbf v} \cdot \mathbf w = \mathbf u \cdot \mathbf w + \mathbf v \cdot \mathbf w$


Proof 1

\(\ds \left({\mathbf u + \mathbf v}\right) \cdot \mathbf w\) \(=\) \(\ds \sum_{i \mathop = 1}^n \left({u_i + v_i}\right) w_i\) Definition of Vector Sum and Definition of Dot Product
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \left({u_i w_i + v_i w_i}\right)\) Real Multiplication Distributes over Real Addition
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n u_i w_i + \sum_{i \mathop = 1}^n v_i w_i\)
\(\ds \) \(=\) \(\ds \mathbf u \cdot \mathbf w + \mathbf v \cdot \mathbf w\) Definition of Dot Product

$\blacksquare$


Proof 2

The following proof is valid in the context of a Cartesian $3$-space:


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.

Dot-product-distributes-over-addition.png

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.

$\blacksquare$


Proof 3

From Dot Product Operator is Bilinear:

$\paren {c \mathbf u + \mathbf v} \cdot \mathbf w = c \paren {\mathbf u \cdot \mathbf w} + \paren {\mathbf v \cdot \mathbf w}$

Setting $c = 1$ yields the result.

$\blacksquare$


Sources

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