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).

The area of a parallelogram

Consider a parallelogram, two of whose sides are $\def\bR{\mathbb{R}}\def\vv{\vec v}\def\ww{\vec w}\vv$ and $\ww$.

This has double the area of the triangle considered above, so its area is $\|\vv\times\ww\|$.

Example

A triangle with two sides $\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\vv=\c13{-1}$ and $\ww=\c21{-2}$ has area $\tfrac12\|\vv\times\ww\|=\tfrac12\left\|\c13{-1}\times\c21{-2}\right\|=\tfrac12\left\|\c{-5}0{-5}\right\|=\tfrac52\left\|\c{-1}0{-1}\right\|=\tfrac52\sqrt2$, and the parallelogram with sides $\vv$ and $\ww$ has area $\|\vv\times\ww\|=5\sqrt2$.

The volume of a parallelepiped in $\mathbb R^3$

Let $\def\uu{\vec u}\uu$, $\vv$ and $\ww$ be vectors in $\bR^3$.

Consider a parallelepiped, with three sides given by $\uu$, $\vv$ and $\ww$.

Call the face with sides $\vv$ and $\ww$ the base of the parallelpiped. The area of the base is $A=\|\vv\times\ww\|$, and the volume of the parallelpiped is $Ah$ where $h$ is the height, measured at right-angles to the base.

One vector which is at right-angles to the base is $\vv\times\ww$. It follows that $h$ is the length of $\vec p=\text{proj}_{\vv\times\ww}\uu$, so \[ h=\|\text{proj}_{\vv\times\ww}\uu\|=\left\|\frac{\uu\cdot(\vv\times\ww)}{\|\vv\times\ww\|^2}\vv\times\ww\right\| = \frac{\uu\cdot(\vv\times\ww)}{\|\vv\times\ww\|^2}\|\vv\times\ww\| = \frac{|\uu\cdot(\vv\times\ww)|}{\|\vv\times\ww\|}\] so the volume is \[ V=Ah=\|\vv\times\ww\|\frac{|\uu\cdot(\vv\times\ww)|}{\|\vv\times\ww\|}\] or \[ V=|\uu\cdot(\vv\times\ww)|.\] Now $\uu\cdot(\vv\times \ww)=\det\left(\begin{bmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{bmatrix}\right)$, so $V$ is the absolute value of this determinant: \[ V=\left|\quad\det\left( \begin{bmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{bmatrix}\right)\quad \right|.\]

Example

Find volume of a parallelepiped whose vertices include $A=(1,1,1)$, $B=(2,1,3)$, $C=(0,2,2)$ and $D=(3,4,1)$, where $A$ is an adjacent vertex to $B$, $C$ and $D$.

Solution

The vectors $\vec{AB}=\c102$, $\vec{AC}=\c{-1}11$ and $\vec{AD}=\c230$ are all edges of this parallepiped, so the volume is \[ V=\left|\quad \begin{vmatrix}1&0&2\\-1&1&1\\2&3&0\end{vmatrix}\quad \right| = | 1(0-3)-0+2(-3-2)| = |-13| = 13.\]

Planes and lines in $\mathbb{R}^3$

Recall that a typical plane in $\bR^3$ has equation \[ ax+by+cz=d\] where $a,b,c,d$ are constants. If we write \[ \def\nn{\vec n}\nn=\c abc\] then we can rewrite the equation of this plane in the form \[ \nn\cdot \c xyz=d.\] If $A=\def\cc#1{(x_{#1},y_{#1},z_{#1})}\cc1$ and $B=\cc2$ are both points in this plane, then the vector $\vec{AB}$ is said to be in the plane, or to be parallel to the plane. Observe that \[ \vec n\cdot \vec{AB}=\nn\cdot\def\cp#1{\c{x_{#1}}{y_{#1}}{z_{#1}}}\left(\cp2-\cp1\right) = \nn\cdot\cp2-\nn\cdot\cp1=d-d=0,\] so \[\nn\cdot\vv=0\] for every vector $\vv$ in the plane. In other words: the vector $\nn$ is orthogonal to every vector in the plane.

We call a vector with this property a normal vector to the plane.

Examples

1. Find a unit normal vector to the plane $x+y-3z=4$.

Solution: The vector $\nn=\c11{-3}$ is a normal vector to this plane, so $\vv=\frac1{\|\nn\|}\nn=\frac1{\sqrt{11}}\c11{-3}$ is a unit normal vector to this plane. Indeed, $\vv$ is a unit vector and it's in the same direction as the normal vector $\nn$, so $\vv$ is also a normal vector.

2. Find the equation of the plane with normal vector $\def\nn{\vec n}\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c1{-3}2$ which contains the point $(1,-2,1)$. Then find three other points in this plane.

Solution: the equation is $x-3y+2z=d$, and we can find $d$ by subbing in $(x,y,z)=(1,-2,1)$: $1-3(-2)+2(1)=d$, so $d=9$ and the equation of the plane is \[ x-3y+2z=9.\] Some other points in this plane are $(9,0,0)$, $(0,1,6)$, $(1,1,\tfrac{11}2)$. (We can find these by inspection).

3. Find the equation of the plane parallel to the vectors $\c111$ and $\c1{-1}1$ containing the point $(3,0,1)$.

Solution: a normal vector is $\nn=\c111\times\c1{-1}1=\cp1111{-1}1=\c{2}0{-2}$, so the equation is $2x+0y-2z=2(3)-2(1)=4$, or $2x-2z=4$, or $x-z=2$.

4. Find the equation of the plane containing the points $A=(1,2,0)$, $B=(3,0,1)$ and $C=(4,3,-2)$.

Solution: $\def\nn{\vec n}\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\vec{AB}=\c2{-2}1$ and $\vec{AC}=\c31{-2}$ are both vectors in this plane. We want to find a normal vector $\nn$ which is be orthogonal to both of these. The cross product of two vectors is orthogonal to both, so we can take the cross product of $\vec{AB}$ and $\vec{AC}$: \[ \nn=\vec{AB}\times\vec{AC}=\def\cp#1#2#3#4#5#6{\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\#1&#2&#3\\#4&#5&#6\end{vmatrix}}\cp2{-2}131{-2}=\c378\] so the equation of the plane is $3x+7y+8z=d$, and we find $d$ by subbing in a point in the plane, say $A=(1,2,0)$, which gives $d=3(1)+7(2)+8(0)=17$. So the equation is \[ 3x+7y+8z=17.\]

Orthogonal planes and parallel planes

Let $\Pi_1$ be a plane with normal vector $\nn_1$, and let $\Pi_2$ be a plane with normal vector $\nn_2$.

1. $\Pi_1$ and $\Pi_2$ are orthogonal or perpendicular planes if they meet at right angles. The following conditions are equivalent:
   1. $\Pi_1$ and $\Pi_2$ are orthogonal planes;
   2. $\nn_1\cdot\nn_2=0$;
   3. $\nn_1$ is a vector in $\Pi_2$;
   4. $\nn_2$ is a vector in $\Pi_1$.
2. $\Pi_1$ and $\Pi_2$ are parallel planes if they have the same normal vectors. In other words, if $\Pi_1$ has equation $ax+by+cz=d_1$ then any parallel plane $\Pi_2$ has an equation with the same left hand side: $ax+by+cz=d_2$.

Examples

1. Find the equation of the plane $\Pi$ passing through $A=(1,3,-3)$ and $B=(4,-2,1)$ which is orthogonal to the plane $x-y+z=5$.

Solution: The plane $x-y+z=5$ has normal vector $\c1{-1}1$, so this is a vector in $\Pi$. Moreover, $\vec{AB}=\c3{-5}4$ is also a vector in $\Pi$, so it has normal vector \[ \nn=\c1{-1}1\times\c3{-5}4=\cp1{-1}13{-5}4=\c1{-1}{-2}.\] So the equation of $\Pi$ is $x-y-2z=d$ and subbing in $A=(1,3,-3)$ gives $d=1-3-2(-3)=4$, so the equation of $\Pi$ is \[ x-y-2z=4.\]

2. The plane parallel to $2x-4y+5z=8$ passing through $(1,2,3)$ is $2x-4y+5z=2(1)-4(2)+5(3) = 10$, or $2x-4y+5z=10$.

3. Find the equation of the plane $\Pi$ which contains the line of intersection of the planes \[ \Pi_1: x-y+2z=1\quad\text{and}\quad \Pi_2: 3x+2y-z=4,\] and is perpendicular to the plane $\Pi_3:2x+y+z=3$.

Solution: To find the line of intersection of $\Pi_1$ and $\Pi_2$, we must solve the system of linear equations \begin{gather}x-y+2z=1\\3x+2y-z=4.\end{gather} We can solve this linear system in the usual way, by applying EROs to the matrix $\begin{bmatrix}1&-1&2&1\\3&2&-1&4\end{bmatrix}$: \begin{align*} \def\go#1#2{\begin{bmatrix}#1\\#2\end{bmatrix}} \def\ar#1{\\[6pt]\xrightarrow{#1}&} &\go{1&-1&2&1}{3&2&-1&4} \ar{R2\to R2-3R1} \go{1&-1&2&1}{0&5&-7&1} \ar{R1\to 5R1+R2} \go{5&0&3&6}{0&5&-7&1} \ar{R1\to\tfrac15R1,\,R2\to\tfrac15R2} \go{1&0&3/5&6/5}{0&1&-7/5&1/5} \end{align*} So the line $L$ of intersection is given by \[ L: \c xyz=\c{\tfrac65}{\tfrac15}0+t\c{-\tfrac35}{\tfrac75}1,\quad t\in\mathbb{R}.\] So $\c{-\tfrac35}{\tfrac75}1$ is a direction vector along $L$, and also $5\c{-\tfrac35}{\tfrac75}1=\c{-3}75$ is a vector along $L$. So $\c{-3}75$ is a vector in the plane $\Pi$. Moreover, taking $t=2$ gives the point $(0,3,2)$ in the line $L$, so this is a point in $\Pi$.

Since $\Pi$ is perpendicular to $\Pi_3$, which has normal vector $\nn_3=\c211$, the vector $\c211$ is in $\Pi$.

So a normal vector for $\Pi$ is \[ \nn=\c211\times\c{-3}75 = \cp211{-3}75=\c{-2}{-13}{17}\] hence $\Pi$ has equation $-2x-13y+17z=d$, and subbing in the point $(0,3,2)$ gives $d=0-13(3)+17(2)=-39+34=-5$, so $\Pi$ has equation $-2x-13y+17z=-5$, or \[ 2x+13y-17z=5.\]

The distance to a plane

The distance from a point to a plane

Let $\Pi$ be a plane in $\def\rt{\mathbb{R}^3}\rt$ with equation $ax+by+cz=d$, so that $\def\nn{\vec n}\nn=\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c abc$ is a normal vector to $\Pi$. Also let $A$ be any point in $\rt$.

The shortest path from $A$ to a point in $\Pi$ goes in the same direction as $\nn$. Let $B$ be any point in the plane $\Pi$.

From the diagram, we see that the shortest distance from $A$ to $\Pi$ is given by \[ \text{dist}(A,\Pi)=\|\def\pp{\vec p}\pp\|\] where \[ \pp=\text{proj}_{\nn}{\vec{AB}}.\] Using the formula for $\text{proj}_{\vec w}\vec v$ and the fact that $\|c\vec v\|=|c|\,\|\vec v\|$ where $c$ is a scalar and $\vec v$ is a vector, we obtain the formula \[ \text{dist}(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}.\]

Example

To find the distance from $A=(1,-4,3)$ to the plane $\Pi:2x-3y+6z=1$, choose any point $B$ in $\Pi$; for example, let $B=(2,1,0)$. Then $\nn=\c2{-3}6$ and $\vec{AB}=\c15{-3}$, so \[ \def\dist{\text{dist}}\dist(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}=\frac{|2(1)+(-3)5+6(-3)|}{\sqrt{2^2+(-3)^2+6^2}}=\frac{|-31|}{\sqrt{49}}=\frac{31}7.\]

Remark: the distance from the origin to a plane

If we write $\def\rt{\mathbb{R}^3}\def\dist{\text{dist}}\def\cp#1#2#3#4#5#6{\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\#1&#2&#3\\#4&#5&#6\end{vmatrix}}\def\nn{\vec n}\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}0=(0,0,0)$ for the origin in $\rt$ and apply the formula above to the plane $\Pi:ax+by+cz=d$ with $B=(d/a,0,0)$ (assuming that $a\ne 0$) then we obtain \[ \dist(0,\Pi)=\frac{|d|}{\|\nn\|}\] where $\nn$ is the normal vector $\nn=\c abc$.

So as $d$ varies (with the normal vector $\nn$ fixed), we obtain parallel planes at different distances to the origin $0$; the larger $d$ is, the further the plane is from $0$.

The distance between parallel planes

If $\Pi_1$ and $\Pi_2$ are parallel planes, then the shortest distance between them is given by \[ \dist(\Pi_1,\Pi_2)=\dist(A,\Pi_2)\] for any point $A$ is $\Pi_1$. The reason is that for parallel planes, changing $A$ to a different point in $\Pi_1$ does not change $\dist(A,\Pi_2)$.

Of course, if the planes $\Pi_1$ and $\Pi_2$ are not parallel, then they intersect (in many points: in a whole line). So for non-parallel planes we always have $\dist(\Pi_1,\Pi_2)=0$.

Example

The distance between the planes $3x+4y-2z=5$ and $3x+4y-3z=1$ is $0$, since the normal vectors $\c34{-2}$ and $\c34{-3}$ are not scalar multiples of one another, so they are in different directions, so the planes are not parallel.

Example

The planes $\Pi_1:3x+4y-2z=5$ and $\Pi_2:3x+4y-2z=1$ have the same normal vector $\c34{-2}$, so they are parallel. Their distance is given by $\dist(A,\Pi_2)$ where $A$ is any point in $\Pi_1$, and to find this we also need a point $B$ in $\Pi_2$.

We can choose $A=(1,0,-1)\in \Pi_1$ and $B=(1,0,1)\in \Pi_2$. (Of course, there are lots of different possible choices here, but they should all give the same answer!) Then $\vec {AB}=\c002$ and \[ \dist(\Pi_1,\Pi_2)=\dist(A,\Pi_2) = \frac{|\nn\cdot \vec{AB}|}{\|n\|}=\frac{|0+0+(-2)2|}{\sqrt{3^2+4^2+(-2)^2}} = \frac4{\sqrt{29}}.\]

Exercise: a formula for the distance between parallel planes

Show that the distance between the parallel planes $\Pi_1:ax+by+cz=d_1$ and $\Pi_2:ax+by+cz=d_2$ is \[\dist(\Pi_1,\Pi_2)=\frac{|d_2-d_1|}{\|\nn\|},\] where $\nn=\c abc$.

Example

To find the distance between $x+3y-5z=4$ and $2x+6y-10z=11$ we can rewrite the second equation as $x+3y-5z=11/2$ to see that this is a parallel plane to the first, with common normal vector $\nn=\c13{-5}$. By the formula in the exercise the distance between these planes is \[ \frac{|\tfrac{11}2-4|}{\|\nn\|} = \frac{|\tfrac 32|}{\sqrt{1^2+3^2+(-5)^2}} = \frac3{2\sqrt{35}}.\]

Lines in $\mathbb{R}^3$

A line $L$ in $\rt$ has an equation of the form \[ L: \c xyz=\c abc+t\c def, \quad t\in \mathbb R\] where $a,b,c,d,e,f$ are fixed numbers.

The variable $t$ is called a “free parameter”: it's “free” because it can take any value, and it's a “parameter” because this is just another name for a variable.

The equation above is called a parametric equation for the line $L$, because of the free parameter $t$.

What do $a,b,c,d,e,f$ mean?

• Set $t=0$: $A=(a,b,c)$ is a point in $L$
• Set $t=1$: $B=(a+d,b+e,c+f)$ is another
• So $\vec {AB}=\c def$ is a direction along $L$.

Example

Find the parametric equation of the line $L$ in $\rt$ which passes through $A=(2,1,-3)$ and $B=(4,-1,5)$.

• Note that $\vec{AB}=\c 2{-2}{8}$
• So this is a direction vector along the line $L$
• So the equation is $ L: \c xyz=\c 21{-3}+t\c 2{-2}8,\quad t\in \mathbb{R}$.
• (Same as $ L: \c xyz=\c 4{-1}5+t\c 2{-2}8,\quad t\in \mathbb{R}$).

The distance from a point to a line

Cross product method

Suppose $L$ is a line in $\def\rt{\mathbb{R}^3}\def\rn{\mathbb{R}^n}\rt$. Let $A$ be a point on $L$ and let $\def\vv{\vec v}\vv$ be a direction vector along $L$.

Given a point $B$, how can we find $d=\text{dist}(B,L)$, the (shortest) distance from the point $B$ to the line $L$?

Let $A$ be any point in $L$ and let $\theta$ be the angle between $AB$ and $\vv$. We have \[ d=\|\vec{AB}\|\,\sin \theta = \frac{\|\vec{AB}\|\,\|\vv\|\sin \theta}{\|\vv\|} = \frac{\|\vec{AB}\times \vv\|}{\|\vv\|}.\] So \[ \text{dist}(B,L) = \frac{\|\vec{AB}\times \vv\|}{\|\vv\|}\] where $A$ is any point in $L$.

Example

To find the distance from the point $B=(1,2,3)$ to the line \[L:\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c xyz=\c10{-1}+t\c41{-5},\quad t\in\mathbb{R}\] we can choose $A=(1,0,{-1})$ so that $\vec{AB}=\c024$ and taking $\vv=\c41{-5}$, we obtain \[ \vec{AB}\times \vv = \begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\0&2&4\\4&1&-5\end{vmatrix}=\c{-14}{16}{-8}=2\c{-7}{8}{-4}\] so \[ \def\dist{\text{dist}}\dist(B,L)=\frac{\|\vec{AB}\times\vv\|}{\|\vv\|}=\frac{2\sqrt{7^2+8^2+4^2}}{\sqrt{4^2+1^2+5^2}} = \frac{2\sqrt{129}}{\sqrt{42}} \approx 3.5051.\]

Dot product method

The method above relies on the cross product, so only works in $\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\def\rt{\mathbb{R}^3}\def\rn{\mathbb{R}^n}\def\vv{\vec v}\def\dist{\text{dist}}\rt$. The following alternative method works in $\rn$ for any $n$.

Observe that $\dist(B,L)$ is the length of the vector $\def\nn{\vec n}\nn=\vec{AB}-\def\pp{\vec p}\pp$ where $\pp=\def\proj{\text{proj}}\proj_{\vv}\vec{AB}$.

Example

Let's redo the previous example using this method.

We have $\vec{AB}=\c024$ and $\vv=\c41{-5}$, so \[\pp=\proj_{\vv}\vec{AB} = \left(\frac{\vec{AB}\cdot \vv}{\|\vv\|^2}\right)\vv = \frac{0(4)+2(1)+4(-5)}{4^2+1^2+5^2}\c41{-5} = -\frac{18}{42}\c41{-5}=-\frac{3}{7}\c41{-5},\] so \[ \nn=\vec{AB}-\pp=\c024-(-\frac37)\c41{-5}=\c024+\frac37\c41{-5}=\frac17\c{12}{17}{13}\] so \[ \dist(B,L)=\|\nn\|=\frac17\sqrt{12^2+17^2+13^2} = \frac17\sqrt{602} \approx 3.5051.\]

The distance between lines in $\mathbb{R}^3$

Skew lines

Suppose that $L_1$ and $L_2$ are skew lines in $\rt$: lines which are not parallel and do not cross.

Let $\vv_1$ be a direction vector along $L_1$, and let $\vv_2$ be a direction vector along $L_2$.

The shortest distance from $L_1$ and $L_2$ is measured along the direction orthogonal to both $\vv_1$ and $\vv_2$, namely the direction of $\nn=\vv_1\times\vv_2$.

Let $\Pi$ be the plane with normal vector $\nn$ which contains $L_1$.

For any point $B$ in $L_2$, we have \[\dist(L_1,L_2)=\dist(B,\Pi) = \frac{|\vec{AB}\cdot \nn|}{\|\nn\|}\] where $A$ is any point in $\Pi$; for example, we can take $A$ to be any point in $L_1$.

To summarise: for skew lines $L_1$ and $L_2$ with direction vectors $\vv_1$ and $\vv_2$, we have \[ \dist(L_1,L_2)=\frac{|\vec{AB}\cdot \nn|}{\|\nn\|}\] where $\nn=\vv_1\times\vv_2$ and $A$ and $B$ are points with one in $L_1$ and the other in $L_2$.

Remark

What about the distance between lines which are not skew? This means that either they are non-parallel and they intersect (so that the distance between them will be zero), or they are parallel lines.

• The same method and formula work if $L_1$ and $L_2$ are non-parallel lines which intersect, and you get $\dist(L_1,L_2)=0$ in this case. The reason is that in this case $L_1$ and $L_2$ will lie in one plane, $\Pi$, and $\vec{AB}$ will also be in $\Pi$, and $\vec n$ will be orthogonal to $\Pi$. So $\frac{\vec{AB}\cdot \vec n}{\|\vec n\|}=\frac{0}{\|\vec n\|}=0=\dist(L_1,L_2)$.
• If $L_1$ and $L_2$ are parallel lines (i.e., if the vectors $\vec v_1$ and $\vec v_2$ along the lines are in the same direction), then $\vv_1\times \vv_2=0$ which isn't helpful, so this method won't work here. In this case, observe that $\dist(L_1,L_2)=\dist(A,L_2)$ where $A$ is any point in $L_1$ (because the lines are parallel), so you can use one of the formulae above for the distance from a point to a line.

Example

Consider the skew lines \begin{align*} x&=1+t_1\\L_1:\quad y&=2t_1\qquad (t_1\in \mathbb{R})\\z&=1+3t_1 \end{align*} and \[ L_2:\c xyz=\c 321+t_2\c1{-1}1,\quad t_2\in\mathbb{R}.\]

Note that we can rewrite the equation of $L_1$ in “vector form”, which is easier to digest: \[ L_1:\c xyz=\c 101+t_1\c123,\quad t_1\in\mathbb{R}\] The direction vectors are $\c123$ and $\c1{-1}1$, so we take $\nn$ to be their cross product: \[ \nn=\c123\times\c1{-1}1=\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\1&2&3\\1&-1&1\end{vmatrix}=\c52{-3}\] and if $A=(1,0,1)$ and $B=(3,2,1)$ then $A$ and $B$ are points with one in $L_1$ and the other in $L_2$, and $\vec{AB}=\c 220$. Hence \[\dist(L_1,L_2)=\frac{|\vec{AB}\cdot \nn|}{\|\nn\|}=\frac{2(5)+2(2)+0(-3)}{\sqrt{5^2+2^2+3^2}} = \frac{14}{\sqrt{38}}.\]

Distance between lines in $\mathbb{R}^3$ in general

The formula $\dist(L_1,L_2)=\frac{|\vec{AB}\cdot \nn|}{\|\nn\|}$ where $\nn=\vec v_1\times \vec v_2$ works for

• skew lines (not parallel, not intersecting), as we saw above,
• and actually: any non-parallel lines $L_1$, $L_2$. We can see this by noticing that if the lines above intersect, then $L_1$ lies in $\Pi$, so the formula gives $\dist(L_1,L_2)=\dist(B,\Pi)=0$ which is the correct answer.

What about parallel lines?

• The formula can't work because we'd have $\vec v_1=\vec v_2$ so $\vec n=\vec v_1\times \vec v_2=\vec 0$
• Instead: observe that when $L_1$ and $L_2$ are parallel, we have $\dist(L_1,L_2)=\dist(A,L_2)$ for any point $A$ in $L_1$
• So we can use one of of the point-to-line distance formulae we saw earlier.