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• a vector $\vec v$ moves $\vec x$ to $\vec x+\vec v$
• e.g. if $\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\def\vv#1#2{\mat{#1\\#2}}A=(-1,3)$, $B=(5,-4)$, which vector moves $A$ to $B$?
• Answer: $\vec{AB}=\vv 6{-7}$.
• Reason: $A+\vec {AB}=B$, so $\vec{AB}=B-A=\vv 5{-4}-\vv{-1}3=\vv6{-7}$.

Definition of $\vec{AB}$

If $A$ and $B$ are any points in $\mathbb{R}^n$, then the vector $\vec{AB}$ is defined by \[ \vec{AB}=B-A\] (on the right, we interpret points as column vectors so we can subtract them to get a column vector).

• $\vec{AB}$ is the vector which moves $A$ to $B$.

Example

In $\mathbb{R}^3$,

• if $A=(3,-4,5)$
• and $B=(11,6,-2)$
• then $\vec{AB}=\def\m#1{\mat{#1}}\m{11\\6\\-2}-\m{3\\-4\\5}=\m{8\\10\\-7}$.

The uses of vectors

Vectors are used in geometry and science to represent quantities with both a magnitude (size/length) and a direction. For example:

• displacements (in geometry)
• velocities
• forces

Recall that a column vector moves points. Its magnitude, or length, is how far it moves points.

Definition: the length of a vector

If $\vec v=\m{v_1\\v_2\\\vdots\\v_n}$ is a column vector in $\mathbb{R}^n$, then its magnitude, or length, or norm, is the number \[ \|\vec v\|=\sqrt{v_1^2+v_2^2+\dots+v_n^2}.\]

Examples

• $\left\|\m{4\\3}\right\|=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$

• \begin{align*}\left\|\m{1\\0\\-2\\3}\right\|&=\sqrt{1^2+0^2+(-2)^2+3^2}\\&=\sqrt{1+0+4+9}=\sqrt{14}\end{align*}

Exercise

Prove that if $c\in \mathbb{R}$ is a scalar and $\vec v$ is a vector in $\mathbb{R}^n$, then \[ \|c\vec v\|=|c|\,\|\vec v\|.\] That is, multiplying a vector by a scalar $c$ scales its length by $|c|$, the absolute value of $c$.

• Hints: $\sqrt{xy}=\sqrt{x}\sqrt{y}$, and $\sqrt{c^2}=|c|$.

Distance between two points

$\|\vec{AB}\|$ is the distance from point $A$ to point $B$

• since this is the length of vector which takes point $A$ to point $B$.

• e.g. how far from from $A=(1,2)$ to $B=(-3,4)$?
• $\small\|\def\m#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\vec{AB}\|=\left\|\m{-3\\4}-\m{1\\2}\right\|=\left\|\m{-4\\2}\right\|=\sqrt{(-4)^2+2^2}=\sqrt{20}$.

• e.g. what's the length of the main diagonal of the unit cube in $\mathbb{R}^3$?
• = distance from $0=(0,0,0)$ to $A=(1,1,1)$
• $\|\vec{0A}\|=\left\|\m{1\\1\\1}\right\|=\sqrt{1^2+1^2+1^2}=\sqrt3$.

Scalar multiplication and direction

Multiplying a vector by a scalar changes its length, but doesn't change its direction.

Definition: unit vectors

Proposition: finding a unit vector in the same direction as a given vector

If $\vec v$ is a non-zero vector, then $\vec w=\frac1{\|\vec v\|}\vec v$ is a unit vector (in the same direction as $\vec v$).

Proof

• Use the formula $\|c\vec v\|=|c|\,\|\vec v\|$ and the fact that $\|\vec v\|>0$:
• $\|\vec w\|=\left\|\frac1{\|\vec v\|}\vec v\right\|=\left|\frac1{\|\vec v\|}\right|\,\|\vec v\|=\frac1{\|\vec v\|}\,\|\vec v\| = 1$.
• So $\vec w$ is a unit vector.
• It's a scalar multiple of $\vec v$, so is in the same direction as $\vec v$. ■

Example

What is unit vector in the same direction as $\vec v=\m{1\\2}$?

• By the Proposition, $\frac{1}{\|\vec v\|}$ is a unit vector in the same direction as $\vec v$.
• $\|\vec v\|=\sqrt{1^2+2^2}=\sqrt5$
• So the unit vector in the same direction as $\vec v$ is:
  • $\frac1{\|\vec v\|}\vec v = \frac1{\sqrt 5}\vec v=\frac1{\sqrt5}\m{1\\2}=\m{1/\sqrt{5}\\2/\sqrt5}$.

Addition of vectors

If $\vec v=\vec{AB}$, then $\vec v$ moves $A$ to $B$, so $A+\vec v=B$.

If $\vec w=\vec {BC}$, then $\vec w$ moves $B$ to $C$, so $B+\vec w=C$.

What about $\vec v+\vec w$? We have $A+\vec v+\vec w=B+\vec w=C$. So $\vec v+\vec w=\vec{AC}$.

The triangle law for vector addition

The parallelogram law for vector addition

The dot product

Definition of the dot product

• 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 $\vec w=\m{4\\-7}$.

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

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: the relationship between angle and the dot product

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 $\vec w$.

• The proof will be given soon, but for now let's work out an example.

Example

• Let $\vec v=\m{1\\2}$ and $\vec w=\m{-2\\1}$
• Then $\dp vw=1(-2)+2(1)=-2+2=0$.
• On the other hand, $\|\vec v\|=\sqrt5=\|\vec w\|$
• So the angle $\theta$ between $\vec v$ and $\vec w$ has $ 0=\dp vw=\sqrt 5\times \sqrt 5 \times \cos\theta$
• So $5\cos\theta=0$, 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:

Proof of the Theorem

We wish to show that $\def\vv{\vec v} \def\ww{\vec w}\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$ where $\theta$ is the angle between $\vv$ and $\ww$.

• Recall the cosine rule:

Proof of the Theorem

We wish to show that $\def\vv{\vec v} \def\ww{\vec w}\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$ where $\theta$ is the angle between $\vv$ and $\ww$.

• Consider a triangle with two sides $\vv$ and $\ww$.
• By the triangle rule for vector addition, the third side $\vec x$ has $\ww+\vec x=\vv$, so $\vec x=\vv-\ww$:
• Apply the cosine rule: $ \|\vv-\ww\|^2=\|\vv\|\,\|\ww\|-2\vv\cdot\ww\cos\theta.$

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

Corollary

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

Corollary

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

Examples

1. The angle $\theta$ between $\def\c#1#2{\begin{bmatrix}#1\\#2\end{bmatrix}}\c12$ and $\c3{-4}$ has \[ \cos\theta=\frac{\c12\cdot\c3{-4}}{\left\|\c12\right\|\,\left\|\c3{-4}\right\|} =\frac{3-8}{\sqrt5\sqrt{25}}=-\frac1{\sqrt5},\] so $\theta=\cos^{-1}(-1/\sqrt5) \approx 2.03\,\text{radians}\approx 116.57^\circ$.
2. The points $A=(2,3)$, $B=(3,6)$ and $C=(-4,5)$ are the vertices of a right-angled triangle. Indeed, we have $\vec{AB}=\c36-\c23=\c13$ and $\vec{AC}=\c{-4}5-\c23=\c{-6}2$, 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.
3. To find a unit vector orthogonal to the vector $\vv=\c12$, we can first observe that $\ww=\c{-2}1$ has $\vv\cdot\ww=0$, so $\vv$ and $\ww$ are orthogonal; and then consider the vector $\vec u=\frac1{\|\ww\|}\ww$, which is a unit vector in the same direction as $\ww$, so is also orthogonal to $\vv$. Hence $\vec u=\frac1{\sqrt5}\c{-2}1=\c{-2/\sqrt5}{1/\sqrt5}$ is a unit vector orthogonal to $\vv=\c12$.