Dot/scalar product

We continue with a definition of the dot product, sometimes also referred to as the scalar product.

Given two vectors $\boldsymbol{a}=\left(a_{1}, a_{2}, a_{3}\right)$ and $\boldsymbol{b}=\left(b_{1}, b_{2}, b_{3}\right)$, expressed in terms of the standard basis vectors $\{\mathbf{i}, \mathbf{j}, \mathbf{k}\}$, the dot product is defined as,

We note the following properties:

• $u \cdot(v+w)=u \cdot v+u \cdot w$
• $u \cdot v=v \cdot u$
• $\boldsymbol{v} \cdot \mathbf{0}=0$
• $(c \boldsymbol{u}) \cdot \boldsymbol{v}=\boldsymbol{v} \cdot(c \boldsymbol{u})=c(\boldsymbol{u} \cdot \boldsymbol{v})$
• $\boldsymbol{u} \cdot \boldsymbol{u}=|\boldsymbol{u}|^{2}$
• If $\boldsymbol{u} \cdot \boldsymbol{u}=0$, then $\boldsymbol{u}=\mathbf{0}$

We move to a geometric interpretation of the dot product. Figure 9.4 depicts two vectors, $\boldsymbol{a}$ and $\boldsymbol{b}$, as well as the vector $\boldsymbol{a}-\boldsymbol{b}$. The angle between $\boldsymbol{a}$ and $\boldsymbol{b}$ is given by $\theta$.

Figure 9.4: Geometric interpretation of the dot product.

We have the following theorem: given $\boldsymbol{a}$ and $\boldsymbol{b}$, the dot product is defined by

Starting from the Law of Cosines, we have

Expanding the LHS of Eq. (9.17) and using the properties listed above, gives

Plugging back to Eq. (9.17)

which yields Eq. (9.16)

If the angle $\theta=\pi / 2$, i.e. $\boldsymbol{a}$ and $\boldsymbol{b}$ are perpendicular, then the dot product is zero. In an orthonormal basis such as $\{\mathbf{i}, \mathbf{j}, \mathbf{k}\}$, we have the following identities:

Consider again the two vectors $\boldsymbol{a}=\left(a_{1}, a_{2}, a_{3}\right)$ and $\boldsymbol{b}=\left(b_{1}, b_{2}, b_{3}\right)$. If we are asked to find the angle $\theta$ between the vectors we use the dot product as follows,

Moreover, using Eq. (9.18), we can find the length of the missing edge in a triangle. Given $\boldsymbol{a}$ and $\boldsymbol{b}$, the length of the missing edge is given by $|\boldsymbol{a}-\boldsymbol{b}|$ (see Fig. 9.4).

Remarks

1. The triangle inequalities

Substituting $-1 \leq \cos \theta \leq 1$ in Eq. (9.17) yields

since Eq. (9.17) takes its minimum value at $a^{2}+b^{2}-2 a b=(a-b)^{2}$ and its maximum at $a^{2}+b^{2}+2 a b=(a+b)^{2}$. From Eq. (9.20), taking the positive square root of each term gives the triangle inequalities,

2. Suppose we have the identity

for $\boldsymbol{a} \neq \mathbf{0}$. Note that this does not necessarily mean that $\boldsymbol{b}=\boldsymbol{c}$. Rather,

which leads to,

It follows that the above is true if, either $\boldsymbol{b}=\boldsymbol{c}$ or if $\boldsymbol{a}$ is perpendicular to $\boldsymbol{b}-\boldsymbol{c}$.

Exercise - What is the angle between:

1. $\boldsymbol{a}=(1,2,3)$ and $\boldsymbol{b}=(-2,2,4)$ ?
2. $\boldsymbol{a}=(1,3,-1)$ and $\boldsymbol{b}=(-1,-1,-5)$ ?
3. $\boldsymbol{a}=(1,0,2)$ and $\boldsymbol{b}=(-1,2,3)$ ?