1^° Prove that the scalar product satisfies the commutative…

$1^\circ$ Prove that the scalar product satisfies the commutative law: $\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{a}$. $2^\circ$ Prove that the scalar product satisfies the distributive law: $\vec{a}\cdot(\vec{b}+\vec{c})=\vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}$.

Explanation of Method

Commutative and distributive laws for the scalar (dot) product.

Step 1: Definition

\[ \vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta, \qquad \theta = \angle(\vec{a},\vec{b}) \in [0^\circ, 180^\circ]. \]

Equivalently, with the (signed) projection $\mathrm{Pr}_{\vec a}\vec b=|\vec b|\cos\theta$ of $\vec b$ onto the direction of $\vec a$: $\vec a\cdot\vec b = |\vec a|\,\mathrm{Pr}_{\vec a}\vec b$.

Step 2: Commutative law

The angle between two vectors does not depend on their order: $\angle(\vec{a},\vec{b}) = \angle(\vec{b},\vec{a}) = \theta$. Since multiplication of real numbers is commutative,

\[ \vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta = |\vec{b}||\vec{a}|\cos\theta = \vec{b}\cdot\vec{a}. \]

\[\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}\]

Geometrically: $|\vec a|$ times the projection of $\vec b$ onto $\vec a$ equals $|\vec b|$ times the projection of $\vec a$ onto $\vec b$ — both are $|\vec a||\vec b|\cos\theta$.

$\theta$ $O$ $\vec a$ $\vec b$ $\mathrm{Pr}_{\vec a}\vec b=|\vec b|\cos\theta$

$\vec a\cdot\vec b=|\vec a|\,\mathrm{Pr}_{\vec a}\vec b$

$\theta$ $O$ $\vec b$ $\vec a$ $\mathrm{Pr}_{\vec b}\vec a=|\vec a|\cos\theta$

$\vec b\cdot\vec a=|\vec b|\,\mathrm{Pr}_{\vec b}\vec a$ — the same number

Step 3: Key lemma: projection of a sum

Lemma. $\mathrm{Pr}_{\vec{a}}(\vec{b}+\vec{c}) = \mathrm{Pr}_{\vec{a}}\vec{b} + \mathrm{Pr}_{\vec{a}}\vec{c}$.

Proof. Put an axis $\ell$ through $O$ in the direction of $\vec a$ and give every point $X$ the signed coordinate $x_X$ of its orthogonal projection (foot of the perpendicular) on $\ell$. For any vector $\overrightarrow{XY}$, its projection on $\ell$ is $x_Y-x_X$ (the signed length from the foot of $X$ to the foot of $Y$). Draw $\vec b=\overrightarrow{OB}$ and attach $\vec c=\overrightarrow{BC}$ at $B$, so that $\vec b+\vec c=\overrightarrow{OC}$. Then

\[ \mathrm{Pr}_{\vec a}(\vec b+\vec c) = x_C - x_O = (x_B - x_O) + (x_C - x_B) = \mathrm{Pr}_{\vec a}\vec b + \mathrm{Pr}_{\vec a}\vec c. \qquad\blacksquare \]

$\text{axis of }\vec a$ $O$ $B$ $C$ $\vec b$ $\vec c$ $\vec b+\vec c$ $\mathrm{Pr}\,\vec b$ $\mathrm{Pr}\,\vec c$ $\mathrm{Pr}(\vec b+\vec c)$

With $\vec c$ attached at the tip of $\vec b$, the foot of $C$ is reached from $O$ by the two projections in turn: $\mathrm{Pr}(\vec b+\vec c)=\mathrm{Pr}\,\vec b+\mathrm{Pr}\,\vec c$.

Step 4: Distributive law

\[ \vec{a}\cdot(\vec{b}+\vec{c}) = |\vec{a}|\,\mathrm{Pr}_{\vec{a}}(\vec{b}+\vec{c}) = |\vec{a}|\left(\mathrm{Pr}_{\vec{a}}\vec{b} + \mathrm{Pr}_{\vec{a}}\vec{c}\right) = |\vec{a}|\,\mathrm{Pr}_{\vec{a}}\vec{b} + |\vec{a}|\,\mathrm{Pr}_{\vec{a}}\vec{c} = \vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} \]

\[\vec{a}\cdot(\vec{b}+\vec{c}) = \vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}\]

(If $\vec a=\vec 0$, both sides are $0$.) The proof uses only the geometric lemma above, not the distributive law itself.

Step 5: Corollaries

Property	Reason
$(\vec{a}+\vec{b})\cdot\vec{c} = \vec{a}\cdot\vec{c}+\vec{b}\cdot\vec{c}$	distributive + commutative
$(\lambda\vec{a})\cdot\vec{b} = \lambda(\vec{a}\cdot\vec{b})$	$\|\lambda\vec a\|=\|\lambda\|\|\vec a\|$; for $\lambda\lt0$ the angle becomes $180^\circ-\theta$ and $\cos$ changes sign
$\vec{a}\cdot\vec{a} = \|\vec{a}\|^2 \geq 0$	$\theta=0^\circ$
$\vec{a}\perp\vec{b} \iff \vec{a}\cdot\vec{b}=0$ ($\vec a,\vec b\neq\vec 0$)	$\cos 90^\circ=0$

Result

\[\boxed{\,\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}\,}\]

\[\boxed{\,\vec{a}\cdot(\vec{b}+\vec{c}) = \vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}\,}\]