Chapter 2.4 Orthogonality and Reflection
In this section, we will discuss how vectors can be used to describe the concept of perpendicular lines and also reflecting across a line.
Orthogonality of Vectors and Lines
First, we define what it means for two vectors to be perpendicular.
Perpendicular Vectors
Vectors \(\langle a,b\rangle\) and \(\langle c,d\rangle\) are perpendicular if the triangle below is right and the edge with endpoints \((0,0)\) and \((a+c, b+d)\) is the hypotenuse
It turns out there is a formula for finding a vector perpendicular to another.
Perpendicular Vector Coordinates
Take \(V\) to be the vector \(\langle a, b\rangle\) and define \(V_\perp\) (read: “vee perp”) to be the vector \[V_\perp = \langle -b, a\rangle.\]
The vector \(V_\perp\) is perpendicular to \(V\), has the same length as \(V\), and points to the left of \(V\).
Here is an example.
Example 1
Take \(V=\langle 3,-2\rangle\). The vector \(V_\perp=\langle 2,3\rangle\) is perpendicular to \(V\).
Why should we study perpendicular vectors? It is because it will allow us to do many things that would have otherwise required solving equations.
For example, if we have some information about a right triangle and know two of the three vertices, we can determine the other vertex by interpreting the relative orientation.
Example 2
Take \((1,3)\), \((5, 5)\), and \(p\) to be vertices of a right triangle. The edges \(\overline{(1,3)(5,5)}\) and \(\overline{(5,5)p}\) are legs of the triangle. Suppose that \(\overline{(5,5)p}\) is half the length of \(\overline{(1,3)(5,5)}\) and that \(p\) lies to the right of \((5,5)\). Find the coordinates of \(p\).
Based on the picture above, \(p=-\frac{1}{2}V_\perp+(5,5).\)
Notice that \(V=(5,5)-(1,3)=\langle 4,2\rangle\), \(V_\perp=\langle -2,4\rangle,\) and \(-\frac{1}{2}V_\perp=\langle 1,-2\rangle.\)
Hence \(p=\langle 1,-2\rangle +(5,5)=(6,3)\).
The other usefulness of perpendicular vectors is that we can reinterpret the idea of perpendicular lines using vectors.
Example 3
Two lines are perpendicular if any two nonzero vectors that move points along the two lines are perpendicular.
Take \(m\) to be a non-zero real number and take \(L\) to be the line given by \[y = mx+b.\]
If \(L_\perp\) is perpendicular to \(L\), then the slope of \(L_\perp\) is \(-\frac{1}{m}\).
Why is this true?
Notice that \[V = \langle 1, m \rangle \] moves points along \(L\). If \(V_{\perp}=\langle -m ,1\rangle\) moves points along \(L_{\perp}\) so slope is \(\frac{1}{-m}=-\frac{1}{m}.\)
Let’s practice with an example.
Example 4
Take \(L\) to be the line that intersects \((2,5)\) and \((7, 1)\). Find an equation of the line that intersects \((3,1)\) and that is perpendicular to \(L\).
The slope of \(L\) is \(m=-\frac{4}{5}\). To get a perpendicular line, the slope must be \(m=\frac{5}{4}\).
Hence, the equation for \(L_{\perp}\) is \(y=\frac{5}{4}(x-3)+1\).
Transformation can get us the equation: \(y=\frac{5}{4}x\) is shifted by \(\langle 3,1\rangle\).
Distance from Points to Line
Another thing perpendicular vectors can do is help us solve the following question:
Take \(p\) to be the point in the plane and take \(L\) to be a line that does not contain \(p\). Find the point \(q\) on \(L\) that is closest to \(p\).
Here is how we can figure this out:
If \(q\) is the point on \(L\) that is closest to \(p\), then the line containing both \(p\) and \(q\) is perpendicular to \(L\).
If there was any other point \(r\) that is closer, then a right triangle can be formed using \(p\), \(q\) and \(r\). The distance between \(p\) and \(r\) would correspond to the hypothenus, which would be greater than the length of the leg that corresponds to \(p\) and \(q\). So, there is no other possibility.
Use this next example to see how we find the closest point with an actual line and given point.
Example 5
Take \(L\) to be the line given by \[y = 3x +2.\] Find the point on \(L\) that is closest to the point \((1, 7)\).
Construct a line \(L_{\perp}\) that passes through \(p\):
\[ \begin{align*} L_{\perp}&=-\frac{1}{3}(x-1)+7. \end{align*} \]
The point in which \(L\) and \(L_{\perp}\) intersect at will be on \(L\) that is closest to the point \((1,7)\).
\[ \begin{align*} L&=L_{\perp}\\ 3x+2&=-\frac{1}{3}(x-1)+7\\ 3x+2&=-\frac{1}{3}x+\frac{1}{3}+7\\ 3x+2&=-\frac{1}{3}x+\frac{22}{3}\\ 3x+\frac{1}{3}x&=\frac{22}{3}-2\\ \frac{10}{3}x&=\frac{16}{3}\\ x&=\frac{16}{10}\\ x&=\frac{8}{5}. \end{align*} \]
To get the \(y\)-coordinate, evaluate \(L\) at \(x=\frac{8}{5}\) to get \(y=\frac{34}{5}.\)
Hence \(q=\left(\frac{8}{5},\frac{34}{5}\right).\)
Reflecting Sets across Lines
The next thing we can do is define reflection.
Take \(L\) to be a line and \(p\) to be a point in the plane.
Take \(L_{\perp}\) to be the line that intersects \(p\) and is perpendicular to \(L\).
The line \(L_{\perp}\) intersects \(L\) at a point \(M\).
There is a unique point \(q\) such that \(M\) is the midpoint of the line segment \(\overline{pq}\).
This point \(q\) is defined to be the reflection of \(p\) across \(L\).
In the next example, we implement this idea.
Example 6
Reflect the point \((1,5)\) across the line \(L\) given by \[y = 2x -4.\]
The equation for the line \(L_{\perp}\) that crosses \((1,5)\) and is perpendicular to \(L\) is \(y=-\frac{1}{2}(x-1)+5\).
Obtain the reflection of \((1,5)\) across the line \(L\) by using the fact that the lines intersect at a point \(M\),
\[\begin{align*} 2x-4&=-\frac{1}{2}(x-1)+5\\ 2x-4&=-\frac{1}{2}x+\frac{1}{2}+5\\ 2x-4&=-\frac{1}{2}x+\frac{11}{2}\\ 2x+\frac{1}{2}x&=\frac{11}{2}+4\\ \frac{5}{2}x&=\frac{19}{2}\\ x&=\frac{19}{5} \end{align*}\]
Find the \(y\)-coordinate of \(M\) by evaluating \(L\) at \(x=\frac{19}{5}\). Calculate it to get the full coordinates of \(M\) to be \(M=\left(\frac{19}{5},\frac{18}{5}\right).\)
Take \(p\) in the direction of \(V=M-p\) to obtain \(q\): \[V=\left(\frac{19}{5},\frac{18}{5}\right)-(1,5)=\left\langle \frac{14}{5},-\frac{7}{5} \right\rangle.\] Therefore,
\[ \begin{align*} q&=2V+p\\ &=2\left\langle \frac{14}{5},-\frac{7}{5} \right\rangle+(1,5)\\ &=\left(\frac{33}{5},\frac{11}{5}\right). \end{align*} \]
The final thing we do revist an idea you may have learned before: reflecting across the line \(y=x\). You may have learned before that a point \((x,y)\) reflected across \(y=x\) will be the point \((y,x).\) We will now derive the formula.
Example 7
Take \(p\) to be the point in the plane with coordinates \((a,b)\).
Take \(L\) to be the line given by \[y =x.\]
Show that the reflection of \(p\) across \(L\) has coordinates \((b,a)\).
The line \(L_{\perp}\) that is perpendicular to \(L\) is crosses at \((a,b)\) is \(y=-(x-a)+b.\)
Obtain the reflection of \((a,b)\) across the line \(L\) by using the fact that the lines will intersect at a point \(M\).
\[\begin{align*} x&=-(x-a)+b\\ x&=-x+a+b\\ 2x&=a+b\\ x&=\frac{a+b}{2} \end{align*}\]
Find the \(y\)-coordinate of \(M\) by evaluating \(L\) at \(x=\frac{a+b}{2}\). Calculate to get the full coordinates of \(M\) to be \(M=\left(\frac{a+b}{2},\frac{a+b}{2}\right).\)
Take \(p\) in the direction of \(V=M-p\) to obtain \(q\): \[V=\left(\frac{a+b}{2},\frac{a+b}{2}\right)-(a,b)=\left\langle \frac{b-a}{2},\frac{a-b}{2} \right\rangle.\] Therefore,
\[ \begin{align*} q&=2V+p\\ &=2\left\langle \frac{b-a}{2},\frac{a-b}{2} \right\rangle+(a,b)\\ &=\left(b,a\right). \end{align*} \]
We can use reflections to create new functions.
Given a function \(f\) and point \((x,f(x))\) in \(f\), we can reflect it across \(y=x\) to get the point \((f(x),x)\). If \(f\) is one-to-one, then the collection of points \((f(x),x)\) is a new function: the inverse function!