Chapter 2.2 Scaling Vectors and Subsets of the Plane
We studied movement using vectors. In order to study more complicated motion, we need to understand how to “stretch” or “shrink” vectors. In this section, we will learn how to scale vectors, understand circles, and how to scale general subsets of the plane.
Scaling Vectors
This is how we scale a vector.
Vector Scaled
For any real number \(r\), take the vector \(r\langle a, b\rangle\) to be given by \[r\langle a, b\rangle = \langle ra, rb\rangle.\] This is the vector \(V\) scaled by a factor of \(r\).
This agrees with our notion of addition of vectors, since \[nV = \underbrace{V + \cdots + V}_{n\text{ times}}.\]
This defines the scaling of a vector by a factor of \(r\) and gives rise to various scalings of the plane.
Note that for any real numbers \(a\) and \(b\) and for any vectors \(V\) and \(W\), \[(a+b)V = aV + bV\quad \text{and}\quad a(V+W) = aV + aW.\]
Example 1
Below is sketched an arrow that represents the vector \(V\). Draw an arrow that represents \(3V\) and \(\frac{1}{2}V\)
In the example below, let’s look at the interaction between vectors and a scalar.
Example 2
Compute \(2\langle 2,3\rangle - \langle 2, 0\rangle\). What does this vector look like? Show that it is equal to the vector \(2\left(\langle 2,3\rangle - \frac{1}{2}\langle 2, 0\rangle\right)\).
The vector \(2\langle 2,3\rangle - \langle 2, 0\rangle\) is equal to \(\langle 2,6\rangle\).
Simplify \(2\left(\langle 2,3\rangle - \frac{1}{2}\langle 2, 0\rangle\right)\), to get
\[\begin{align*} 2\left(\langle 2,3\rangle - \frac{1}{2}\langle 2, 0\rangle\right)&=2\left(\langle 2,3\rangle-\langle 1,0\rangle\right)\\ &=2\langle 1,3\rangle\\ &=\langle 2,6\rangle \end{align*}\]
As we can see, it is what we expected.
Circles and the Polar Form of a Vector
How do we define distance matematically? We are going to define it such as a way that makes intuitive sense with how we define distance in our real world.
To be able to do this, we will assume the basic notions of Euclidean Geometry and determine what distance in the plane must be in terms of the coordinates of points.
Suppose that the triangle below is a right triangle with side lengths \(a\), \(b\), and \(c\).
If \(c\) is the length of the hypotenuse, then \[a^2 + b^2 = c^2.\]
The Pythagorean theorem allows us to define a length of a vector in this way.
Magnitude
For any vector \(\langle a, b\rangle\), denote by \(||\langle a, b\rangle||\) the quantity \[||\langle a, b\rangle|| = \sqrt{a^2 + b^2}.\] Call \(||V||\) the magnitude of \(V\).
Let’s practice computing the length of a vector.
Example 3
What is the length of the vector \(\langle 5, 12\rangle\)?
The length is \(\|\langle 5, 12\rangle\|=\sqrt{5^2+12^2}=\sqrt{169}=13.\)
What about computing the distance between two points? It is simply the length of the vector that takes one point to the other.
Distance
For any two points \((a,b)\), \((c,d)\) in the plane, the distance \(D\) between \((a,b)\) and \((c,d)\) is, therefore, equal to \(||(c,d) - (a,b)||\)
Now let’s compute the distance between two points.
Example 4
What is the distance between \((1,3)\) and \((6, 15)\)?
The distance is \(\|(1,3)-(6, 15)\|=\|\langle-5,-12\rangle\|=\sqrt{(-5)^2+(-12)^2}=\sqrt{169}=13.\)
Here are some observations:
If \(V\) is a vector and \(c\) is a real number, then \[||cV|| = |c|||V||.\]
Scaling a vector \(V\) by a positive factor \(c\) scales the length of \(V\) by a factor of \(c\).
Multiplying \(V\) by \(-1\) reverses the direction of \(V\). That is, it produces the additive inverse.
Understand this with this example.
Example 5
What is the length of \(\langle 2, 5\rangle\)? What is the length of \(\langle 6, 15\rangle\)? What is the length of \(\langle -6, -15\rangle\)?
The length of \(\langle 2, 5\rangle\) is \(\|\langle 2, 5\rangle\|=\sqrt{2^2+5^2}=\sqrt{29}.\)
The length of \(\langle 6, 15\rangle\) is \(\|\langle 6, 15\rangle\|=\|3\langle 2, 5\rangle\|=|3|\cdot \|\langle 2,5\rangle \|=3\sqrt{29}\).
The length of \(\langle -6, -15\rangle\) is \(\|\langle -6, -15\rangle\|=\|-3\langle 2, 5\rangle\|=|-3|\cdot \|\langle 2,5\rangle \|=3\sqrt{29}\).
As we can see in the above example, the vectors are all related in the sense that they are simply scaled versions of the same vector.
We can encode this information by separating out the direction and length of a vector in the following way.
Unit Vector
For any non-zero vector \(V\), define \(\hat{V}\) (read: ``vee hat’’) by \[\hat{V} = \frac{1}{||V||}V.\]
The vector \(\hat{V}\) is a vector of unit length, a unit vector, that points in the same direction as \(V\).
The polar form of \(V\) is \(||V||\hat{V}\).
This form separates out this information: - \(\hat{V}\), which gives the direction in which \(V\) points and - \(||V||\), which is the distance that \(V\) moves points.
Example 6
What is the polar form of \(\langle 2, 7\rangle\)?
To get the polar form of \(V=\langle 2, 7\rangle\), calculate \(\|V\|\); \(\|V\|=\sqrt{53}\).
Next calculate \(\hat{V}=\frac{1}{\|V\|}V\), which in this case will be \(\hat{V}=\left\langle \frac{2}{\sqrt{53}},\frac{7}{\sqrt{53}}\right\rangle.\)
Hence the polar form of \(V\) is \(\sqrt{53}\left\langle \frac{2}{\sqrt{53}},\frac{7}{\sqrt{53}}\right\rangle\).
An important set we should know about is the circle.
Circle
A circle is the set of all points in the plane equidistant from a fixed point, the center.
The distance from the center to a point on a circle is the radius of the circle.
The distance formula implies the circle of radius \(r\) centered at \((h,k)\) is the set of points that satisfy \[(x-h)^2 + (y-k)^2 = r^2.\]
The distance formula written out like \(\|(x,y)-(h,k)\|=r\) or \(\|(x,y)-(h,k)\|^2=r^2\) gives us another way to look at a circle.
Use this to write an equation for a specific circle.
Example 7
Find the equation for the circle of radius \(2\) centered at \((3,-1)\).
A point \((x,y)\) will be on the circle of radius \(2\) centered at \((3,-1)\) if and only if its distance from \((3,-1)\) is \(2\).
Mathematically, this means the following:
\[ \begin{align*} \|(x,y)-(3,-1)\|&=2\\ \|\langle x-3,y+1\rangle\|&=2\\ \sqrt{(x-3)^2+(y+1)^2}&=2. \end{align*} \] Squaring both sides gives us the equation: \[(x-3)^2+(y+1)^2=4.\]
Given the center and a point on the circle, we can form the equation of the circle.
Example 8
Find an equation for the circle centered at \((2,5)\) that intersects the point \((5, 9)\).
The circle centered at \((2,5)\) intersects the point \((5,9)\). Calculate the radius from this information
\[ \begin{align*} r&=\|(5,9)-(2,5)\|\\ &=\|\langle 3,4\rangle\|\\ &=\sqrt{3^2+4^2}\\ &=\sqrt{25}\\ &=5 \end{align*} \] A point \((x,y)\) will be on the circle of radius \(5\) centered at \((2,5)\) if and only if its distance from \((2,5)\) is \(5\).
Mathematically, this means the following:
\[ \begin{align*} \|(x,y)-(2,5)\|&=5\\ \|\langle x-2,y-5\rangle\|&=5\\ \sqrt{(x-2)^2+(y-5)^2}&=5. \end{align*} \] Hence the equation is \[(x-2)^2+(y-5)^2=25.\]
What kind of points are on a unit circle? Remember, it is all points whose distance from \((0,0)\) is exactly \(1\), so not every point in the plane will be on the circle.
Example 9
Can a point on the unit circle have an \(x\)-coordinate equal to 3?
No because its distance from \((0,0)\) will not equal \(1\).
Given a point not on the unit circle, there is a point on the unit circle both points point in the same direction relative to the origin, which is the center of the point.
Projection Onto Unit Circle
Given a point \(p\) not equal to the origin \(O\) (take \(O\) to be the point \((0,0)\)), let \(V\) be the vector \(p - O\).
The point \(\hat{V} + O\) is the projection of \(p\) onto the unit circle.
The point \(p\) lies on the line intersecting \(O\) and \(\hat{V}+O\).
Any point in the plane, except the origin, \(O\), can be projected onto the unit circle.
Here is an example.
Example 10
Find the projection of the point \((3, 4)\) onto the unit circle.
The projection of \(V=(3,4)\) onto \(C\) is \(\hat{V}+(0,0)\).
Calculate \(\hat{V}=\frac{1}{\|V\|}V\) to get \(\hat{V}=\left\langle \frac{3}{5},\frac{4}{5} \right\rangle .\)
Hence the projection is \[p=\hat{V}+(0,0)=\left ( \frac{3}{5},\frac{4}{5} \right).\]
The equation for the unit circle and knowledge about projections onto the unit circle allow us to locate points on the unit circle with a variety of specified properties. For example, there will be at most two points on the unit circle with one specified coordinate.
Example 11
Can a point on the unit circle have a \(y\) coordinate equal to \(\frac{3}{5}\)? If so, what will its \(x\) coordinate be if the point lies in the first quadrant? Is there a second point with this given \(y\) coordinate?
The point \(\left(x,\frac{3}{5}\right)\) is on the unit circle if and only if \(\left(\frac{3}{5}\right)^2+x^2=1\). So \[ \begin{align*} x^2&=1-\frac{9}{25}\\ &=\frac{25-9}{25}\\ &=\frac{16}{25} \end{align*} \] implies that \[ \begin{align*} x&=\pm\sqrt{\frac{16}{25}}\\ &=\pm\frac{\sqrt{16}}{\sqrt{25}}\\ &=\pm \frac{4}{5}. \end{align*} \] This means there are two points on the unit circle with a \(y\) coordinate equal to \(\frac{3}{5}\): \(\left(\frac{4}{5},\frac{3}{5}\right)\) and \(\left(-\frac{4}{5},\frac{3}{5}\right)\). The point \(\left(\frac{4}{5},\frac{3}{5}\right)\) lies in the first quadrant.
There are exactly two points on the unit circle that lie on the same line through the origin.
Example 12
The line \(L\) has slope 3 and intersects the origin. Where does \(L\) intersect the unit circle?
Because \(L\) has a slope of \(3\) and intersects the origin, this means its equation is given by \(y=3x\). From this equation, find other points on \(L\). In particular \((1,3)\) is a point on \(L\).
Take \[V= \langle 1,3 \rangle .\]
The line \(L\) will intersect the unit circle at \(p=\hat{V}+(0,0)\) and \(q=-\hat{V}+(0,0)\).
The points are \(p=\hat{V}+(0,0)=\left(\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right)\) and \(q=-\hat{V}+(0,0)=\left(-\frac{1}{\sqrt{10}},-\frac{3}{\sqrt{10}}\right)\),
Scaling Subsets of the Plane
We now discuss how to scale points in the plane.
Symmetrically Scaled
Suppose that \(A\) is a subset of the plane and \(c\) is a positive real number.
Define by \(cA\) the set \[cA = \{c(a-(0,0)) + (0,0) \colon a\in A\}\] and define the function \(\sigma_c\) by \[\sigma_c(a) = c(a-(0,0)) + (0,0).\]
If \(p\) and \(q\) are in \(A\) and the distance between them is \(d\), then \(\sigma_c(p)\) and \(\sigma_c(q)\) are in \(cA\) and the distance between them is \(cd\).
The set \(cA\) looks just like \(A\), but is “symmetrically rescaled”.
Practice scaling points with this example.
Example 13
What is the distance between \((1,3)\) and \((2, 5)\)? Calculate \(\sigma_2(1,3)\) and \(\sigma_2(2,5)\) as well as the distance between these points.
The distance between \((1,3)\) and \((2,5)\) is \(\|(2,5)-(1,3)\|=\|\langle 1,2\rangle\|=\sqrt{5}.\)
Thus \[\sigma_2(1,3)=2\left((1,3)-(0,0)\right)+(0,0)=2\langle 1,3\rangle +(0,0)=(2,6),\] and \[\sigma_2(2,5)=2\left((2,5)-(0,0)\right)+(0,0)=2\langle 2,5\rangle +(0,0)=(4,10).\] The distance between \((2,6)\) and \((4,10)\) is \(\|(4,10)-(2,6)\|=\|\langle 2,4\rangle\|=2\sqrt{5}.\)
Now we look at scaling that is not uniformed.
Asymmetric Scaling
Suppose that \(A\) is a subset of the plane and \(c\) is a positive real number.
Asymmetrically scale \(A\) by scaling the axes by different scaling factors.
Define functions \(X_c\) and \(Y_c\) on vectors by \[X_c(\langle a,b\rangle) = \langle ca, b\rangle \quad {\rm and} \quad Y_c(\langle a,b\rangle) = \langle a, cb\rangle.\]
Extend these transformations to sets by defining by \(X_cA\) (the \(x\)-axis scaling of \(A\)) and \(Y_cA\) (the \(y\)-axis scaling of \(A\)) by \[X_cA = \{X_c(a-(0,0)) + (0,0) \colon a\in A\} \quad {\rm and}\quad Y_cA = \{Y_c(a-(0,0)) + (0,0) \colon a\in A\}.\]
Practice scaling with this example.
Example 14
Calculate \(X_4(1,5)\) and \(Y_3(2,3)\) and sketch how this transformations move the two points below.
Calculate to get \[X_4(1,5)=X_4((1,5)-(0,0))+(0,0)=X_4\langle 1,5\rangle+(0,0)=\langle 4,5\rangle +(0,0)=(4,5),\]
and
\[Y_3(2,3)=Y_3((2,3)-(0,0))+(0,0)=Y_3\langle 2,3\rangle+(0,0)=\langle 2,9\rangle +(0,0)=(2,9).\]
Now, let’s look at how scaling affects functions.
Write the function \(f\) as \[f = \{(x,f(x))\colon x\in \mathcal D(f)\}.\]
Symmetric scaling: \[\sigma_cf = \{(cx, cf(x))\colon x\in \mathcal D(f)\}, \quad {\rm so}\quad \sigma_c f(x) = cf\Big(\frac{x}{c}\Big).\]
\(y\)-Axis scaling: \[Y_cf = \{(x, cf(x))\colon x\in \mathcal D(f)\}, \quad {\rm so}\quad Y_cf(x) = cf(x).\]
\(x\)-Axis scaling: \[X_cf = \{(cx, f(x))\colon x\in \mathcal D(f)\}, \quad {\rm so}\quad X_cf(x) = f\Big(\frac{x}{c}\Big).\]
Now we will determine a formula for a function that has had more than one transformation applied to it.
To make it easier to write out, we define two functions that give rise to the shifts and scalings that we have previously discussed as applied to functions.
For each real \(h\), define the shift function \(T_h\) by \[T_h(x) = x+ h.\]
For each positive real \(a\), define the scale function \(S_a\) by \[S_a(x) = ax.\] Note that \(a\) can also be negative and we will discuss the meaning of this shortly.
The composite \(T_k\circ S_b \circ f \circ S_a\circ T_h\) is given by \[(T_k\circ S_b \circ f \circ S_a\circ T_h)(x) = bf(a(x+h)) + k.\]
Functions to the right of \(f\) in the composite shift or scale the \(x\)–axis and functions to the left of \(f\) in the composite shift or scale the \(y\)–axis, but they do so differently!
Here is what we mean:
\(+h\) indicates we move to the right by \(h\)
\(a\) indicates we scale \(x\)-axis by \(\frac{1}{a}\)
\(b\) indicates we scale \(y\)-axis by \(b\)
\(+k\) indicates we move up by \(k\).
Practice by doing this next example.
Example 15
Write \(g\) as a composite function using the function \({\rm pow}_1\), where \(g\) is given by \[g(x) = 2x+1.\]
Write \(g\) as \(T_1\circ S_2\circ f\):
\[ \begin{align*} \left(T_1\circ S_2\circ f\right)(x)&=T_1\circ \left(S_2( f(x))\right)\\ &=T_1\circ (2x)\\ &=(2x)+1\\ &=2x+1. \end{align*} \] The function \(g\) is \(f\) but scaled in the \(y\)-axis by \(2\) and shifted up by \(1\).
In this example, try to decompose this more complicated function.
Example 16
Write \(g\) as a composite function using the function \({\rm pow}_2\), where \(g\) is given by \[g(x) = 4(x-5)^2 + 3.\]
Write \(g\) as \(T_3\circ S_4\circ f\circ T_{-5}\):
\[ \begin{align*} \left(T_3\circ S_4\circ f\circ T_{-5}\right)(x)&=T_3\circ S_4\circ \left(f(T_{-5}(x))\right)\\ &=T_3\circ S_4\circ(x-5)^2\\ &=T_3\circ (4(x-5)^2)\\ &=4(x-5)^2+3. \end{align*} \] The function \(g\) is \(f\) but shifted right by 5, scaled in the \(y\)-axis by \(4\) and shifted up by \(3\).
In this example, we will revist lines but through a transformational lense.
Example 17
Find an equation for the line passing through the points \((1,2)\) and \((5,7)\) using only a transformational approach. Note that, up to scaling and shifting, there is only one line given by a function.
Use \({\rm pow_1}=x\) and transform it to get the equation.
First, \(L\) passing through \((1,2)\) and \((5,7)\) means that its slope is \(m=\frac{7-2}{5-1}=\frac{5}{4}\).
Because \(L\) passes through \((1,2)\) and has slope \(m=\frac{5}{4}\), obtain the equation of \(L\) from \({\rm pow_1}=x\) by scaling the \(y\)-axis by \(\frac{5}{4}\) and moving \({\rm pow_1}=x\) by \(\langle 1,2 \rangle .\)
Hence the equation for the line is \(f(x)=\frac{5}{4}(x-1)+2\).