Definition

Given vectors $\vec{v}$ and $\vec{w}$, the dot product $\vec{v}\cdot\vec{w}$ is defined by

  1. $\vec{v}\cdot\vec{w}=\Vert\vec{v}\Vert\cdot\Vert\vec{w}\Vert\cos\theta$ where $\theta$ is the angle between $\vec{v}$ and $\vec{w}$ such that $0\le\theta\le\pi$.
  2. $\vec{v}\cdot\vec{w}=v_1w_1+v_2w_2+\cdots+v_nw_n$

Proof of the Dot Product Equality

Your book mentions that one uses the law of cosines in order to prove that the algebraic and geometric formulas for the dot product are equal. We’ll expand that a bit here. First, we’ll state the law of cosines.

Law of Cosines: Given the triangle above, the following identity holds.

\[c^2=a^2+b^2-2ab\cos\gamma\]

Of course, you can form the law of cosines for $a^2$ or $b^2$ as well. This is really just a generalization of the Pythagorean Theorem. If the angle at $\gamma$ is $\pi/2$ radians, then the cosine of $\gamma$ will be zero, giving the usual formula. We can use this formula in our current situation by considering the following diagram.

The law of cosines then states that

\[\Vert\vec{v}-\vec{w}\Vert^2=\Vert\vec{v}\Vert^2+\Vert\vec{w}\Vert^2-2\Vert\vec{v}\Vert\Vert\vec{w}\Vert\cos\theta\]

Expanding the left side gives

\[\Vert\vec{v}-\vec{w}\Vert^2=\left(v_1^2-2v_1w_1+w_1^2\right)+\cdots+\left(v_n^2-2v_nw_n+w_n^2\right)\]

Since $\Vert\vec{v}\Vert^2=v_1^2+v_2^2+\cdots+v_n^2$, we can cancel terms on each side and find ourselves left with

\[-2v_1w_1-2v_2w_2-\cdots-2v_nw_n=-2\Vert\vec{v}\Vert\Vert\vec{w}\Vert\cos\theta\]

Dividing by $-2$ then gives us the desired equality:

\[v_1w_1+v_2w_2+\cdots+v_nw_n=\Vert\vec{v}\Vert\Vert\vec{w}\Vert\cos\theta\]

Properties of the Dot Product

The most important geometric use of the dot product deals with “orthogonal” vectors. The notion of orthogonality is a generalization of perpendicularity. We have the following useful theorem:

Theorem: Two vectors $\vec{v}$ and $\vec{w}$ in $n$-dimensional space are orthogonal if and only if $\vec{v}\cdot\vec{w}=0$.

Proof: Two vectors are orthogonal when the angle between them is $\pi/2$ radians. Since $\cos(\pi/2)=0$, this proves the theorem.

There are also several nice algebraic properties of the dot product. Namely,

  • $\vec{v}\cdot\vec{w}=\vec{w}\cdot\vec{v}$
  • For any $\lambda\in\mathbb{R}$ we have $\vec{v}\cdot(\lambda\vec{w})=(\lambda\vec{v})\cdot\vec{w}=\lambda(\vec{v}\cdot\vec{w})$
  • $(\vec{v}+\vec{w})\cdot\vec{u}=\vec{v}\cdot\vec{u}+\vec{w}\cdot\vec{u}$

Example 1

Determine conditions for the vector $\vec{b}=b_1\vec{i}+b_2\vec{j}+b_3\vec{k}$ to be orthogonal to the vector $\vec{a}=\vec{i}+\vec{j}-\vec{k}$.

Solution: It isn’t enough here to find only one vector $\vec{b}$ orthogonal to $\vec{a}$. We search for a formula to tell use which vectors work for $\vec{b}$. Notice that

\[\vec{a}\cdot\vec{b}=a_1b_1+a_2b_2+a_3b_3=b_1+b_2-b_3.\]

If $\vec{a}$ and $\vec{b}$ are orthogonal, then $\vec{a}\cdot\vec{b}=0$, implying that $\vec{b}$ is orthogonal to $\vec{a}$ if and only if $b_1+b_2=b_3$.

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