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

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

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 here is an example.

Example

If $\vec v=\m{1\\2}$ and $\vec w=\m{-2\\1}$, then $\dp vw=1(-2)+2(1)=-2+2=0$. On the other hand, we have $\|\vec v\|=\sqrt5=\|\vec w\|$, so the angle $\theta$ between $\vec v$ and $\vec w$ satisfies \[ 0=\dp vw=\sqrt 5\times \sqrt 5 \times \cos\theta\] so $5\cos\theta=0$, so $\cos\theta=0$, so $\theta=\pi/2$ or $\theta=3\pi/2$ (measuring angles in radians). This tells us that the angle between $\vec v$ and $\vec w$ is a right angle. We say that these vectors are orthogonal. 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:

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 $\vec x=\vv-\ww$:

Applying the cosine rule gives \[ \|\vv-\ww\|^2=\|\vv\|^2+\|\ww\|^2-2\|\vv\|\,\|\ww\|\,\cos\theta.\] On the other hand, we know that $\|\vec x\|^2=\vec x\cdot\vec x$, so \begin{align*}\|\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.\end{align*} So \[\|\vv\|^2+\|\ww\|^2-2\|\vv\|\,\|\ww\|\,\cos\theta=\|\vv\|^2+\|\ww\|^2-2\vv\cdot\ww\cos\theta.\] Subtracting $\|\vv\|^2+\|\ww\|^2$ from both sides and dividing by $-2$ gives $\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$. ■

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 $\vv=\c12$.

The orthogonal projection of one vector onto another

Let $\def\ww{\vec{w}}\def\vv{\vec{v}}\def\uu{\vec{u}}\ww$ be a non-zero vector, and let $\vv$ be any vector. We call a vector $\def\pp{\vec p}\def\nn{\vec{n}}\pp$ the orthogonal projection of $\vv$ onto $\ww$, and write $\pp=\def\ppp{\text{proj}_{\ww}\vv}\ppp$, if

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

We can use these properties of $\pp$ to find a formula for $\pp$ in terms of $\vv$ and $\ww$.

1. Since $\pp$ is in the same direction as $\ww$, we have $\pp=c\ww$ for some scalar $c\in \mathbb{R}$.
2. Since $\nn=\vv-\pp$ is orthogonal to $\ww$, we have $\nn\cdot \ww=0$. Hence \begin{align*}&&(\vv-\pp)\cdot \ww&=0\\&\implies& \vv\cdot\ww-\pp\cdot\ww&=0\\&\implies& \pp\cdot\ww&=\vv\cdot\ww\\&\implies& c\ww\cdot \ww&=\vv\cdot\ww\\&\implies& c\|\ww\|^2&=\vv\cdot\ww\\&\implies& c&=\frac{\vv\cdot\ww}{\|\ww\|^2}.\end{align*}

So we obtain the orthogonal projection formula: \[ \pp=\ppp=\frac{\vv\cdot\ww}{\|\ww\|^2}\ww.\]

We call $\nn=\vv-\ppp$ the component of $\vv$ orthogonal to $\ww$.

Example

If $\def\vv{\vec v}\def\pp{\vec p}\def\ppp{\text{proj}_{\ww}\vv}\def\ww{\vec w}\def\nn{\vec n}\vv=\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c12{-1}$ and $\ww=\c2{-1}4$, then \begin{align*}\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\end{align*} and the component of $\vv$ orthogonal to $\ww$ is \begin{align*}\nn&=\vv-\ppp\\&=\c12{-1}-\left(-\frac4{21}\right)\c2{-1}4\\&=\c12{-1}+\c{8/21}{-4/21}{16/21}\\&=\c{29/21}{38/21}{-5/21}.\end{align*}

The cross product of vectors in $\mathbb{R}^3$

Definition: the standard basis vectors

We define $\def\i{\vec \imath}\i=\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c100$, $\def\j{\vec \jmath}\j=\c010$ and $\def\k{\vec k}\k=\c001$. These are the standard basis vectors of $\mathbb{R}^3$.

Note that any vector $\vec v=\c{v_1}{v_2}{v_3}$ may be written as a linear combination of these vectors (that is, a sum of scalar multiplies of $\i$, $\j$ and $\k$), since \[ \def\vc#1{\c{#1_1}{#1_2}{#1_3}}\vec v=\vc v=\c{v_1}{0}{0}+\c{0}{v_2}{0}+\c{0}{0}{v_3} = v_1\i+v_2\j+v_3\k.\]

Definition: the cross product

If $\vec v=\vc v$ and $\vec w=\vc w$ are vectors in $\mathbb{R}^3$, then we define $\def\vv{\vec v}\vv\times\def\ww{\vec w}\ww$ to be the vector given by the determinant \[ \vv\times\ww=\def\cp#1#2#3#4#5#6{\begin{vmatrix}\i&\j&\k\\#1&#2&#3\\#4&#5&#6\end{vmatrix}}\def\cpc#1#2{\cp{#1_1}{#1_2}{#1_3}{#2_1}{#2_2}{#2_3}}\cpc vw.\] We interpret this determinant by expanding along the first row: \[\vv\times\ww=\cpc vw=\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\vm{v_2&v_3\\w_2&w_3}\i-\vm{v_1&v_3\\w_1&w_3}\j+\vm{v_1&v_2\\w_1&w_2}\k=\c{v_2w_3-v_3w_2}{-(v_1w_3-v_3w_1)}{v_1w_2-v_2w_1}\]

Example

Let $\vv=\c13{-1}$ and $\ww=\c21{-2}$. We have \[ \vv\times\ww=\cp13{-1}21{-2}=\c{3(-2)-1(-1)}{-(1(-2)-(-1)2)}{1(1)-3(2)}=\c{-5}0{-5}\] and \[ \ww\times\vv=\cp21{-2}13{-1}=\c{1(-1)-(-2)3}{-(2(-1)-(-2)1)}{2(3)-1(1)}=\c{5}0{5}.\] Observe that $\vv\times\ww=-\ww\times \vv$. Moreover, \[ \vv\times \vv=\cp13{-1}13{-1}=\c000=\vec0\] and \[ \ww\times\ww=\cp21{-2}21{-2}=\c000=\vec0.\]

Example: cross products of standard basis vectors

We have \[ \i\times\j=\cp100010=\c001=\k,\] \[ \j\times\k=\cp010001=\c100=\i\] \[ \k\times\i=\cp001100=\c010=\j\]

Proposition: properties of the cross product

For any vectors $\def\uu{\vec u}\uu$, $\vv$ and $\ww$ in $\mathbb{R}^3$ and any scalar $c\in\mathbb{R}$, we have:

1. $\uu\times(\vv+\ww)=\uu\times\vv+\uu\times\ww$
2. $\vv\times\ww=-\ww\times\vv$
3. $(c\vv)\times \ww=c(\vv\times\ww)=\vv\times(c\ww)$
4. $\vv\times\vv=\vec0$
5. $\vv\times \vec0=\vec0$
6. $\vv\times \ww$ is orthogonal to both $\vv$ and $\ww$

Proof

1. This is a tedious (but easy) bit of algebra.
2. Swapping two rows in a determinant changes the sign, so \[ \vv\times\ww=\cpc vw=-\cpc wv=-\ww\times\vv.\]
3. Scaling one row in a determinant scales the determinant in the same way, so \[ (c\vv)\times\ww=\cpc {cv}w=c\cpc vw=c\vv\times\ww.\]
4. The determinant of a matrix with a repeated row is zero.
5. The determinant of a matrix with a zero row is zero.
6. Observe that $\uu\cdot (\vv\times \ww)=\vm{u_1&u_2&u_3\\v_1&v_2&v_3\\w_1&w_2&w_3}$. The determinant of a matrix with a repeated row is zero, so \[\vv\cdot (\vv\times \ww)=\vm{v_1&v_2&v_3\\v_1&v_2&v_3\\w_1&w_2&w_3}=0\] so $\vv$ is orthogonal to $\vv\times\ww$; and similarly, \[\ww\cdot(\vv\times \ww)=\vm{w_1&w_2&w_3\\v_1&v_2&v_3\\w_1&w_2&w_3}=0\] so $\ww$ is orthogonal to $\vv\times\ww$. ■

Theorem: the dot product/cross product length formula

For any vectors $\vv$ and $\ww$ in $\mathbb{R}^3$, we have \[ \|\vv\times\ww\|^2+(\vv\cdot\ww)^2=\|\vv\|^2\,\|\ww\|^2.\]

Proof

Let $D$ be the sum of $v_i^2w_j^2$ over all $i,j\in\{1,2,3\}$ with $i=j$. (So $D=v_1^2w_1^2+v_2^2w_2^2+v_3^2w_3^2$.)

Let $F$ be the sum of $v_i^2w_j^2$ over all $i,j\in\{1,2,3\}$ with $i\ne j$. (So $F=v_1^2w_2^2+v_2^2w_1^2+\dots+v_3^2w_2^2$, with 6 terms on the right hand side.)

Let $C$ be the sum of $v_iw_iv_jw_j$ over all $i,j\in\{1,2,3\}$ with $i<j$. (So $C=v_1w_1v_2w_2+v_1w_1v_3w_3+v_2w_2v_3w_3$.)

Then $D+F$ is the sum of $v_i^2w_j^2$ over all $i,j\in \{1,2,3\}$.

Expanding the formulae for $\|\vv\|^2$ and $\|\ww\|^2$, we get $\|\vv\|^2\|\ww\|^2=D+F$.

Expanding the formula for the cross product, we get $\|\vv\times\ww\|^2=F-2C$.

Expanding the formula for the dot product, we get $(v\cdot w)^2=D+2C$.

So $\|\vv\times\ww\|^2+(v\cdot w)^2=F-2C+D+2C=F+D=\|\vv\|^2\|\ww\|^2.■$

Corollary: the length of $\vec v\times\vec w$

For any vectors $\vv$ and $\ww$ in $\mathbb{R}^3$, we have \[ \|\vv\times\ww\|=\|\vv\|\,\|\ww\|\,\sin\theta\] where $\theta$ is the angle between $\vv$ and $\ww$ (with $0\le\theta\le\pi$).

Proof

Recall that $\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$. Now \begin{align*}\|\vv\times\ww\|^2&= \|\vv\|^2\,\|\ww\|^2-(\vv\cdot\ww)^2\\ &=\|\vv\|^2\,\|\ww\|^2-\|\vv\|^2\|\ww\|^2\cos^2\theta\\ &=\|\vv\|^2\,\|\ww\|^2(1-\cos^2\theta)\\ &=\|\vv\|^2\,\|\ww\|^2\sin^2\theta.\end{align*} Since $\sqrt {a^2}=a$ if $a\ge0$ and $\sin\theta\ge0$ for $0\le\theta\le\pi$, taking square roots of both sides gives \[ \|\vv\times\ww\|=\|\vv\|\,\|\ww\|\,\sin\theta. ■ \]

Geometry of the cross product

Let $\def\vv{\vec v}\vv$ and $\def\ww{\vec w}\ww$ be vectors in $\def\bR{\mathbb{R}}\bR^3$.

The area of a triangle

Consider a triangle with sides $\vv$ and $\ww$ (and a third vector, namely $\vv-\ww$). Thinking of $\vv$ as the base, the length of the base is $b=\|\vv\|$ and the height of this triangle (measured at right angles to the base) is $h=\|\ww\|\sin \theta$ where $\theta$ is the angle between $\vv$ and $\ww$.

Hence the area of this triangle is $\tfrac12 bh=\tfrac12\|\vv\|\,\|\ww\|\sin\theta$, which is equal to $\tfrac12\|\vv\times\ww\|$ (by the formula for $\|\vv\times\ww\|$ which appears above).