Chapter 2.3 Movement Along Lines
In this section, we will learn how to use vectors to describe motion on a line. In fact, any line can be described using vectors.
Absolute and Relative Movement
Take \(L\) to be the line that passes through \((a,b)\) and \((c,d)\).
Suppose that \((a, b)\) and \((c,d)\) are distinct and take \(V\) to be the vector that is given by \[V = (c,d) - (a,b).\]
For any real number \(k\), the vector \(kV\) moves points in \(L\) along \(L\).
See it with this example.
Example 1
Take \(L\) to be the line that contains the points \((6,1)\) and \((8,4)\). Find a vector that moves points in \(L\) along \(L\).
The vector \(V=(8,4)-(6,1)=\langle 2,3\rangle\) moves points in \(L\) along \(L\).
Alternatively, this also works \(W=(6,1)-(8,4)=\langle-2,-3\rangle\).
An equation for the line \(L\) can be used to find the vector that moves points in \(L\) along \(L\).
Example 2
Take \(L\) to be the line that is given by the equation \[y = 4x-1.\] Find a vector that moves points in \(L\) along \(L\).
The slope of \(L\) is \(m=4\). So the vector \(V=\langle 1,4\rangle\) points in the direction of the line \(L\).
In general, if a line has a slope of \(m\), then \(\langle 1,m\rangle\) moves points on \(L\) along \(L\).
Moving a point along a line by a relative distance means moving the point by an amount that is a multiple of some given amount. Such movement involves scaling a given vector. When moving a point along a line by a given distance, it is helpful to use the polar form of the vector
Example 3
Suppose that \((1,2)\) and \((4,6)\) lie on the line \(L\). Find all points on \(L\) that are a distance of 3 from \((1,2)\).
The vector \(V=(4,6)-(1,2)=\langle 3,4\rangle\) is in the same direction as \(L\).
To find all points that are a distance of \(3\) from \((1,2)\), create a unit vector that is in the same direction as \(V\). In other words, calculate \(\hat{V}\), which will give us \[\hat{V}=\left\langle \frac{3}{5} ,\frac{4}{5} \right\rangle .\]
The points that are a distance of \(3\) will be \[p=3\hat{V}+(1,2)\] and \[q=-3\hat{V}+(1,2).\]
Work out the calculation to get \(p=\left(\frac{9}{5}+1,\frac{12}{5}+2\right)\) and \(q=\left(-\frac{9}{5}+1,-\frac{12}{5}+2\right).\)
For the next problem we will find a point on a line so that its distance from one of the end points satisfy certain distance requirements.
Example 4
Suppose that \((1,2)\) and \((4,6)\) lie on the line \(L\). Determine a point \(p\) in \(L\) so that the distance from \((1,2)\) to \(p\) is \(\frac{1}{4}\) times the distance from \((1,2)\) to \((4,6).\)
The vector \(V=(4,6)-(1,2)=\langle 3,4\rangle\) is in the same direction as \(L\). It moves points along \(L.\)
For example, \(V+(1,2)\) takes the point \((1,2)\) to \((4,6).\)
To get the right point \(p\), scale \(V\) by \(\frac{1}{4}:\)
\[p=\frac{1}{4}V+(1,2)=\left(\frac{3}{4}+1,3\right).\]
To confirm this is the right \(p\), calculate the distances from \((1,2)\) to \(p\) and \((1,2)\) to \((4,6):\)
\[\|(1,2)-(4,6)\|=5\quad\text{and}\quad \|(1,2)-p\|=\frac{5}{4},\]
so \[\|(1,2)-(4,6)\|=\frac{1}{4}\|(1,2)-p\|\] as desired.
In the next problem, we will find a point on a line so that its distance from one of the endpoints and its distance from the other endpoint satisfy certain distance requirements.
Example 5
The line segment \(L\) has endpoints \((1,2)\) and \((4,6)\). Determine a point \(p\) in \(L\) so that the distance from \((1,2)\) to \(p\) is \(3\) times the distance from \(p\) to \((4,6).\)
The vector \(V=(4,6)-(1,2)=\langle 3,4\rangle\) is in the same direction as \(L\). It moves points along \(L.\)
For example, \(V+(1,2)\) takes the point \((1,2)\) to \((4,6).\)
To get the right point \(p\), we need to scale \(V\) by the right scaling \(c\).
Define \[p=c V+(1,2).\]
The point \(p\) must satisfy the following requirements:
\[ \begin{cases} \|p-(1,2)\|=3\|p-(4,6)\|\\ \|p-(1,2)\|+\|p-(4,6)\|=\|V\|. \end{cases} \]
The second requirement is that the distances from \(p\) to each endpoint should add up to the distance from one endpoint of the line segment to the other line segment.
Use the information to notice that
\[ \begin{align*} \|p-(1,2)\|+\|p-(4,6)\|&=\|V\|\\ \|p-(1,2)\|+\frac{1}{3}\|p-(1,2)\|&=\|V\|\\ \frac{4}{3}\|p-(1,2)\|&=\|V\|\\ \|p-(1,2)\|&=\frac{3}{4}\|V\|. \end{align*} \]
The right scaling is \(c=\frac{3}{4}\), so \[p=\frac{3}{4}V+(1,2)=\left(\frac{9}{4}+1,5\right).\]
To confirm this is the right \(p\), calculate the distances from \((1,2)\) to \(p\) and \(p\) to \((4,6):\)
\[\|(1,2)-p\|=\frac{15}{4}\quad\text{and}\quad \|(4,6)-p\|=\frac{5}{4},\]
so \[\|(1,2)-p\|=3\|(4,6)-p\|\] as desired.
Any line can be represented using vectors in the following way.
Take \((a,b)\) and \((c,d)\) to be two distinct points on a line \(L\), and denote by \(V\) the vector \[V = (c,d) - (a,b).\]
For any real number \(t\), \(tV\) moves any point on \(L\) to another point on \(L\).
Furthermore, if \((x,y)\) is a point on \(L\), then there is a \(t\) so that \[tV + (a,b) = (x,y).\]
Example 6
Suppose that \((2,7)\) and \((4,11)\) lie on the line \(L\). Take \[V = (4,11) - (2,7).\] Find \(t\) so that \[tV + (2,7) = (8,19).\]
Note that \[ \begin{align*} tV + (2,7) &=t \langle 2, 4\rangle +(2,7)\\ &=\langle 2t, 4t\rangle +(2,7)\\ &=(2t+2,4t+7) \end{align*} \] Hence \[tV + (2,7) = (8,19)\] if and only if \[(2t+2,4t+7)=(8,19).\] So the following system of equation must be satisfied \[ \begin{cases} 2t+2&=8\\ 4t+7&=19 \end{cases} \] Solve to get \(t=3\) is a solution.
Parameterized Lines
Take \(L\) to be a line in the plane.
If \(V\) is a nonzero vector that moves points along \(L\), then \(V\) gives a way of identifying points on the real number line with points on \(L\).
Define by \(\ell\) the function that is given by \[\ell(t) = tV + p.\] This function is a parameterization of \(L\).
For any point \(p\) in \(L\), \[L = \{tV + p\colon t\in \mathbb R\}.\]
The given expression for \(L\) is more than just a set of points, it describes the set of points as a path parameterized by a variable representing time.
Movement along lines is, in this way, described by scaling.
Simply scale a vector that moves points along a line to reach any other point that lies along the line in the direction of the vector.
To reach points in the opposite direction, multiply the vector by \(-1\) and then scale this vector.
Example 7
Find a vector equation of the line that passes through \((2,3)\) and \((5,8)\). Why do we emphasis a?
Calculate \(V\) to get \[V=(5,8)-(2,3)=\langle 3, 5\rangle .\] Hence
\[L(t)=tV+(2,3)=t\langle 3,5\rangle +(2,3)=(3t+2,5t+3).\] Another possibility is \[-tV+(2,3)=(-3t+2,-5t+3)\] and also \[tV+(5,8)=t\langle 3,5\rangle +(5,8)=(3t+5,5t+8).\]
A vector equation contains information about the slope of the line. See how we obtain the information in this example.
Example 8
What is the slope of the line that is described by the set \[\{t\langle 3, 5\rangle + (1,7)\colon t\in \mathbb R\}?\] Describe this line in slope/\(y\)-intercept form.
The vector \(V=\langle 3,5\rangle\) tells us go right 3 and up 5, so \(m=\frac{5}{3}\) is the slope of the line.
The line intersects at \((1,7)\).
So the equation of the line in slope intercept form is \(y=\frac{5}{3}(x-1)+7\).
We will find the equation that describes the movement of a particle with constant velocity and other features.
Example 9
A particle that moves at a constant velocity is at \((4,1)\) at time \(5\), moves at a speed of \(2\), and intersects the point \((7, 3)\). Find an equation for the position of the particle at time \(t\).
The particle passes through \((4,1)\) and \((7,3)\), so it moves in the direction of \[V=(7,3)-(4,1)=\langle 3, 2\rangle .\]
To get a specific speed, create a unit vector in the direction of \(V\); \[\|V\|=\sqrt{13},\] so \[\hat{V}=\left\langle \frac{3}{\sqrt{13}},\frac{2}{\sqrt{13}}\right\rangle.\]
To get a particle with a speed of 1 and at \((4,1)\) at \(t=0\), then \(t\hat{V}+(4,1)\) works.
To get a particle with a speed of 2 and at \((4,1)\) at \(t=0\), then \(2t\hat{V}+(4,1)\) works.
To get a particle with a speed of 2 and at \((4,1)\) at \(t=5\), then \(2(t-5)\hat{V}+(4,1)\) works.
Hence our final answer is \[L(t)=2(t-5)\left\langle \frac{3}{\sqrt{13}},\frac{2}{\sqrt{13}}\right\rangle+(4,1).\]
Here is how to find midpoints of a line segment using vectors.
Example 10
Determine the midpoint of the line segment with endpoints \((a,b)\) and \((c,d)\).
The vector \(V=\langle c-a,d-b\rangle\) takes \((a,b)\) to \((c,d)\).
To get the midpoint,go “half” of \(V\) from \((a,b)\). Thus
\[ \begin{align*} \text{midpoint}&=\frac{1}{2}V+(a,b)\\ &=\frac{1}{2}\langle c-a,d-b\rangle+(a,b)\\ &=\left\langle \frac{c-a}{2},\frac{d-b}{2}\right\rangle+(a,b)\\ &=\left(\frac{c-a}{2}+a,\frac{d-b}{2}+b\right)\\ &=\left(\frac{c-a}{2}+\frac{2a}{2},\frac{d-b}{2}+\frac{2b}{2}\right)\\ &=\left(\frac{c+a}{2},\frac{d+b}{2}\right) \end{align*} \]
Particles that move in linear motion may or may not collide. We can figure this out by understanding what the vector equation tells us. First, however, we need to create an equation that describes the set.
Example 11
Two boats move at constant velocities. One boat starts at \((0,0)\) at time 0 and moves 5 miles per hour north and 3 miles per hour west. Another boat moves at 1 mile per hour north and 5 miles per hour east. After 3 hours, the boats collide. Where was the second boat at time 0? Note, \((1,0)\) is one mile to the East of a port and \((0,1)\) is one mile to the North of a port.
The first boat can be described by the equation \[A(t)=t\langle -3, 5\rangle +(0,0)=(-3t,5t).\]
Although where the boat is initially is unknown, it’s possible to write an equation for the boat: \[B(t)=t\langle 5,1\rangle +(a,b)=( 5t+a,t+b),\] where \(a\) and \(b\) will be determined later.
Since the boats collide after three hours, this means \(A(3)=B(3)\), where \(A(3)=(-9,15)\) and \(B(3)=(15+a,3+b)\).
So \(A(3)=B(3)\) implies that
\[ \begin{cases} 15+a&=-9\\ 3+b&=15 \end{cases} \] must hold. Solve the the system of equations to get \(a=-24\) and \(b=12.\)
Hence the initial position of the second boat is \(B(0)=( 5\cdot 0-24,0+12)=(-24,12).\)
Given two formulas that describe the position of particles, can we determine if they collide?
Example 12
At time \(t\), Particle A is at position \(A(t)\) and Particle B is at position \(B(t)\), where \[A(t) = t\langle 2, 5\rangle + (1,3)\quad \text{and}\quad B(t) = t\langle -1, 2\rangle + (4, 5).\] Do the particles collide?
The particles collide if there is a real number \(t\) so that \(A(t)=B(t).\)
Hence
\[ \begin{align*} t\langle 2, 5\rangle + (1,3)&=t\langle -1, 2\rangle + (4, 5)\\ (2t+1,5t+3)&=(-t+4,2t+5)\\ \end{align*} \] implies that
\[ \begin{cases} 2t+1&=-t+4\\ 5t+3&=2t+5 \end{cases}. \] However, the first equation has the solution \(t=1\) while the second equation has the solution \(t=\frac{2}{3}\). Hence the boats do not collide.
Next, we will formulate the equation of motion of a particle that moves in a piecewise linear fashion.
Example 13
A particle moves at a constant velocity on \([1, 4]\) and \((4, 9]\). It is at \((2, 5)\) at time \(1\), at \((5, 1)\) at time \(4\), and at \((6, 8)\) at time \(9\). Find an equation for the position, \(\ell(t)\), of the particle at time \(t\).
On the interval \([1,4]\), create a path for a particle with the following requirements:
goes from \((2,5)\) to \((5,1)\) with constant velocity
at \(t=1\), the particle is at \((2,5)\)
at \(t=4\), the particle is at \((5,1).\)
The direction part can be computed like this: \[V=(5,1)-(2,5)=\langle 3,-4\rangle .\]
For each unit of time, go in the direction of \(V\).
So the equation \(tV+(2,5)\) or \(t\langle 3,-4\rangle+(2,5)\) describes a particle with constant velocity that starts initially at \((2,5)\).
In order for the particle to start at \((2,5)\) at \(t=1\), shift the time:
\[(t-1)\langle 3,-4\rangle+(2,5).\]
This is the equation of a particle with constant velocity that is at position \((2,5)\) at \(t=1\). However, at \(t=4\) the particle is NOT in position \((5,1)\). To get it there at \(t=4\), scale the time by \(4-1=3\):
The particle can be described by the equation \[\frac{(t-1)}{4-1}\langle 3,-4\rangle+(2,5)\] or \[\frac{(t-1)}{3}\langle 3,-4\rangle+(2,5).\]
On the interval \((4,9]\), create a path for a particle with the following requirements:
goes from \((5,1)\) to \((6,8)\) with constant velocity
at \(t=4\), the particle is at \((5,1)\)
at \(t=9\), the particle is at \((6,8).\)
Follow similar calculations above with the appropriate changes to get an equation for the particle to be
\[\frac{t-4}{9-4}\langle 1,7\rangle +(5,1)\] or \[\frac{t-4}{5}\langle 1,7\rangle +(5,1).\] Putting everything together, the path is
\[ \ell(t)=\begin{cases} \frac{(t-1)}{3}\langle 3,-4\rangle +(2,5) &\text{ if } 1\leq t\leq 4\\ \frac{t-4}{5}\langle 1,7\rangle +(5,1) &\text{ if } 4<t<9 \end{cases} \]