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• the vector $\vec{AB}$ joining $A$ to $B$
• length (norm) of a vector
• unit vectors
• triangle law and parallelogram law for vector addition

The dot product

Definition of the dot product

• In other words, $\vec v\cdot \vec w$ is the row-column product $(\vec v)^T\vec w$ of the transpose of $\vec v$ with $\vec w$.
• Note that while $\vec v$ and $\vec w$ are vectors, their dot product $\vec v\cdot \vec w$ is a scalar.

Example

Let $\vec v=\m{3\\5}$ and $\vec w=\m{4\\-7}$.

• $\vec v\cdot \vec w=\m{3\\5}\cdot \m{4\\-7} = 3(4)+5(-7)=12-35=-23$.

Properties of the dot product

For any vectors $\vec v$, $\vec w$ and $\vec u$ in $\mathbb{R}^n$, and any scalar $c\in \mathbb{R}$:

1. $\def\dp#1#2{\vec #1\cdot \vec #2}\dp vw=\dp wv$ (the dot product is commutative)
2. $\vec u\cdot(\vec v+\vec w)=\dp uv+\dp uw$
3. $(c\vec v)\cdot \vec w=c(\dp vw)$
4. $\dp vv=\|\vec v\|^2\ge 0$, and $\dp vv=0 \iff \vec v=0_{n\times 1}$

The proofs of these properties are exercises.

Angles and the dot product

Theorem: geometric dot product formula

If $\vec v$ and $\vec w$ are non-zero vectors in $\mathbb{R}^n$, then \[ \dp vw=\|\vec v\|\,\|\vec w\|\,\cos\theta\] where $\theta$ is the angle between $\vec v$ and $\vec w$.

• Why's it called a geometric formula?
  • It expresses the dot product using only geometric quantities: lengths and angle
• The proof will be given soon, but for now let's work out an example.

Example

What's the angle $\theta$ between $\vec v=\m{1\\2}$ and $\vec w=\m{-2\\1}$?

• $\|\vec v\|\,\|\vec w\|\cos\theta=\dp vw$
• $=1(-2)+2(1)=-2+2=0$.
• So $\|\vec v\|\,\|\vec w\|\cos\theta=0$
• We have $\|\vec v\|\,\|\vec w\|\ne 0$ (why?) so we can cancel this from both sides
• So $\cos\theta=0$
• So $\theta=\pi/2$ or $\theta=3\pi/2$ (measuring angles in radians).
• The angle between $\vec v$ and $\vec w$ is a right angle.
• We say $\vec v$ and $\vec w$ are orthogonal.

Picture of $\vec v=\m{1\\2}$ and $\vec w=\m{-2\\1}$

We can draw a convincing picture which indicates that these vectors are indeed at right angles:

Reminder: the cosine rule

• This is the key to proving the geometric dot product formula.

Proof of the geometric dot product formula

Aim: $\def\vv{\vec v}\def\ww{\vec w}\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$.

• Consider a triangle with two sides $\vv$ and $\ww$. By the triangle rule for vector addition, the third side is $\vv-\ww$:
• Use the cosine rule with $A=\theta$, $a=\|\vv-\ww\|$, $b=\|\vv\|$, $c=\|\ww\|$
• $ \|\vv-\ww\|^2=\|\vv\|\,\|\ww\|-2\|\vv\|\,\|\ww\|\,\cos\theta$

Proof of the geometric dot product formula, continued

Aim: $\def\vv{\vec v} \def\ww{\vec w}\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$.

• $\|\vv-\ww\|^2=\|\vv\|^2+\|\ww\|^2-2\|\vv\|\,\|\ww\|\,\cos\theta$
• So $2\|\vv\|\,\|\ww\|\cos\theta=\|\vv\|^2+\|\ww\|^2-\|\vv-\ww\|^2$
  • $=\vv\cdot\vv+\ww\cdot\ww-(\vv-\ww)\cdot(\vv-\ww)$
  • $=\vv\cdot\vv+\ww\cdot\ww-(\vv\cdot\vv+\ww\cdot\ww-\ww\cdot\vv-\vv\cdot\ww)$
  • $=2\vv\cdot\ww$.
• So $2\vv\cdot\ww=2\|\vv\|\,\|\ww\|\,\cos\theta$
• So $\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$. ■

Corollary 1: angle formula

If $\vv$ and $\ww$ are non-zero vectors and $\theta$ is the angle between them, then $\cos\theta=\displaystyle\frac{\vv\cdot\ww}{\|\vv\|\,\|\ww\|}$.

• Proof: since $\vv$ and $\ww$ are non-zero, we have $\|\vv\|\,\|\ww\|\ne0$ so we can divide both sides of the geometric dot product formula by $\|\vv\|\,\|\ww\|$ to get the angle formula.

Example 1

What is the angle $\theta$ between $\def\c#1#2{\left[\begin{smallmatrix}{#1}\\{#2}\end{smallmatrix}\right]}\c12$ and $\c3{-4}$?

• $ \cos\theta=\frac{\c12\cdot\c3{-4}}{\left\|\c12\right\|\,\left\|\c3{-4}\right\|} =\frac{3-8}{\sqrt5\sqrt{25}}=\frac{-5}{5\sqrt 5}=-\frac1{\sqrt5}$
• So $\theta=\cos^{-1}(-1/\sqrt5) \approx 2.03\,\text{radians}\approx 116.57^\circ$.

Corollary 2: orthogonal vectors

If $\vv$ and $\ww$ are non-zero vectors with $\vv\cdot\ww=0$, then $\vv$ and $\ww$ are orthogonal: they are at right-angles.

Example 2

Prove that $A=(2,3)$, $B=(3,6)$ and $C=(-4,5)$ are the vertices of a right-angled triangle.

• $\vec{AB}=\c36-\c23=\c13$
• $\vec{AC}=\c{-4}5-\c23=\c{-6}2$
• So $\vec{AB}\cdot\vec{AC}=\c13\cdot\c{-6}2=1(-6)+3(2)=0$
• So the sides $AB$ and $AC$ are at right-angles.
• So $ABC$ is a right-angled triangle.

Example 3

Find a unit vector orthogonal to the vector $\vv=\c12$.

• Observe that $\ww=\c{-2}1$ has $\vv\cdot\ww=0$
• So $\vv$ and $\ww$ are orthogonal
• $\vec u=\frac1{\|\ww\|}\ww$ is a unit vector in the same direction as $\ww$, (so is also orthogonal to $\vv$).
• So $\vec u=\frac1{\sqrt5}\c{-2}1=\c{-2/\sqrt5}{1/\sqrt5}$ is a unit vector orthogonal to $\vv=\c12$.

Orthogonal projection

Let $\def\pp{\vec p}\def\ww{\vec w}\def\vv{\vec v}\def\nn{\vec n}\ww$ non-zero, and $\vv$ any vector.

$\pp$ is the orthogonal projection of $\vv$ onto $\ww$ if:

1. $\pp$ is in the same direction as $\ww$; and
2. the vector $\nn=\vv-\pp$ is orthogonal to $\ww$.

• We write $\pp=\def\ppp{\text{proj}_{\ww}\vv}\ppp$.
• $\nn=\vv-\pp$ is the component of $\vv$ orthogonal to $\ww$.

Formula for $\pp=\ppp$

• $\pp$ same direction as $\ww$, so $\pp=c\ww$, some scalar $c$
• $\nn=\vv-\pp$ is orthogonal to $\ww$, so $\nn\cdot \ww=0$.
  • so $(\vv-\pp)\cdot \ww=0$
  • so $\vv\cdot\ww-\pp\cdot\ww=0$
  • so $\pp\cdot\ww=\vv\cdot\ww$
  • so $c\ww\cdot \ww=\vv\cdot\ww$
  • so $c\|\ww\|^2=\vv\cdot\ww$
  • so $c=\frac{\vv\cdot\ww}{\|\ww\|^2}$.
• So $\color{blue}{\pp=\ppp=\frac{\vv\cdot\ww}{\|\ww\|^2}\ww}.$

Example

$\vv=\def\c#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}\c12{-1}$ and $\ww=\c2{-1}4$

• $\ppp=\frac{\vv\cdot\ww}{\|\ww\|^2}\ww$
  • $=\frac{2-2-4}{2^2+(-1)^2+4^2}\c2{-1}4$
  • $=-\frac4{21}\c2{-1}4$
• component of $\vv$ orthogonal to $\ww$ is $\nn=\vv-\ppp$
  • $\nn=\c12{-1}-\left(-\frac4{21}\right)\c2{-1}4=\frac1{21}\c{29}{38}{-5}$.