Discovery 13.1.
Based on the graph of \(y = \cos\theta\) on domain \(0 \le \theta \le \pi\) provided below, what can you say about \(\udotprod{u}{v}\) in the case that \(\theta\) is acute? … obtuse? … right?
Graph of \(y=\cos \theta\text{.}\)
Discovery guide 13.1 Discovery guide
Recall that for \(\uvec{u} = (u_1,u_2,\dotsc,u_n)\) and \(\uvec{v} = (v_1,v_2,\dotsc,v_n)\text{,}\) the dot product of \(\uvec{u}\) and \(\uvec{v}\) is defined by the formula given below on the left. It is an important formula because if \(\theta\) is the angle between two nonzero vectors \(\uvec{u}\) and \(\uvec{v}\text{,}\) then \(\theta\) satisfies both \(0 \le \theta \le \pi\) and the formula given below on the right.
\begin{align*}
\udotprod{u}{v} \amp= u_1 v_1 + u_2 v_2 + \dotsb + u_n v_n \amp
\cos\theta \amp= \frac{\udotprod{u}{v}}{\unorm{u}\unorm{v}}
\end{align*}
Extending the concept of perpendicular to higher dimensions, vectors \(\uvec{u}\) and \(\uvec{v}\) are called orthogonal if \(\udotprod{u}{v} = 0\text{.}\)
Discovery 13.2.
(a)
Can you guess a vector \(\uvec{v} = (v_1,v_2)\) that is orthogonal to \(\uvec{u} = (1,-3)\) in the plane? Make sure your guess satisfies the definition of orthogonal: you need \(\udotprod{u}{v} = 0\text{.}\)
(b)
What relationship to your initial guess \(\uvec{v}\) will other vectors in the plane that are orthogonal to \(\uvec{u}\) have?
Hint.
Draw a diagram of your vectors \(\uvec{u}\) and \(\uvec{v}\text{,}\) both with initial points at the origin. On your diagram, how can you modify your intial guess \(\uvec{v}\) geometrically while still maintaining orthogonality with \(\uvec{u}\text{?}\)
(c)
Turn the pattern of your guess from Task a into a general pattern for vectors in the plane: if \(\uvec{u} = (a,b)\text{,}\) then an example of a vector orthogonal to \(\uvec{u}\) is
\begin{equation*}
\uvec{v} = (\fillinmath{XXX},\fillinmath{XXX})\text{.}
\end{equation*}
Discovery 13.3.
(a)
Draw the vector \(\uvec{a} = (3,1)\) in the \(xy\)-plane with its tail at the origin. Now imagine you were to also draw in every possible scalar multiple of \(\uvec{a}\) (positive, negative, zero, fractional, etc.). What geometric shape would these scalar multiples of \(\uvec{a}\) trace out? Draw this shape on your diagram.
(b)
Plot the point \(Q(4,4)\) on your diagram. On the line defined by \(\uvec{a}\) that you drew in the first part of this activity, draw in the point that you think is closest to \(Q\text{.}\) Label this point \(P\text{.}\) Now draw \(\abray{PQ}\text{,}\) and label this vector as \(\uvec{n}\text{.}\)
What is the relationship between \(\uvec{n}\) and the line? What is the value of \(\udotprod{n}{a}\text{?}\)
(c)
Vector \(\abray{OP}\) is parallel to \(\uvec{a}\text{,}\) so \(\abray{OP}\) is a scalar multiple of \(\uvec{a}\text{.}\) Our goal is to determine the scalar \(k\) so that the head of \(k\uvec{a}\) lies at \(P\text{.}\) Complete the triangle in your diagram by drawing in the vector \(\uvec{u} = \abray{OQ}\text{.}\)
Then express \(\uvec{n}\) as a combination of \(\uvec{u}\) and \(k\uvec{a}\text{.}\)
(d)
Substitute your expression for \(\uvec{n}\) from Task c into your equation for \(\udotprod{n}{a}\) from Task b, and then solve for \(k\) as a formula in \(\uvec{u}\) and \(\uvec{a}\text{.}\)
Now complete the general formula:
\begin{equation*}
k\uvec{a}=\left(\fillinmath{XXXXXX}\right)\uvec{a}
\end{equation*}
(where in the brackets you should fill in a formula in the variable letters \(\uvec{u}\) and \(\uvec{a}\text{,}\) without using their actual numerical components, that describes how to compute \(k\uvec{a}\) from \(\uvec{u}\) and \(\uvec{a}\)).
The vector \(k\uvec{a}\) in Discovery 13.3 is called the orthogonal projection of \(\uvec{u}\) onto \(\uvec{a}\), and we write \(\uproj{u}{a}\) to mean this vector. It is also sometimes called the vector component of \(\uvec{u}\) parallel to \(\uvec{a}\). The vector \(\uvec{n} = \uvec{u}-\uproj{u}{a}\) is called the vector component of \(\uvec{u}\) orthogonal to \(\uvec{a}\).
The same problem can be solved in higher dimensions by the same formula for \(\uproj{u}{a}\text{.}\)
Discovery 13.4.
(a)
Suppose \(\uvec{u}\) is orthogonal to \(\uvec{a}\text{.}\) What is \(\uproj{u}{a}\text{?}\) What is the component of \(\uvec{u}\) orthogonal to \(\uvec{a}\text{?}\)
(b)
Answer the same two questions in the case that \(\uvec{u}\) is parallel to \(\uvec{a}\text{.}\)
Discovery 13.5.
If \(\ell\) is the line through the origin and parallel to a vector \(\uvec{a}\text{,}\) and \(\uvec{u}\) is some other vector, then our construction in Discovery 13.3 guarantees that \(\proj_{\uvec{a}} \uvec{u}\) represents the closest point on \(\ell\) to the terminal point of \(\uvec{u}\text{.}\)
The distance between a point and a line is defined as the shortest (i.e. perpendicular) distance between the two. Use the orthogonal projection to come up with a procedure to determine the distance between the line \(\ell\colon\,y=x/2\) and the point \(Q(2,4)\text{.}\)
Discovery 13.6.
The homogeneous linear equation \(2 x + 3 y = 0\) defines a line through the origin in \(\R^2\) (i.e. the \(xy\)-plane).
(a)
Recall that a point \((x,y)\) lies on the line if and only if its coordinates satisfy the given equation. Let’s consider such a point as the terminal point of the vector \(\uvec{x} = (x,y)\) with its initial point at the origin. Does the left-hand side of the equation for the line look like the formula for some quantity related to \(\uvec{x}\) and some other vector? Perhaps some quantity that we’ve been exploring in detail recently?
(b)
In light of the first part of this activity, what does the right-hand side of the equation for the line say about the relationship between a vector \(\uvec{x}=(x,y)\) that lies along the line and the other special vector you identified in the previous part?
Discovery 13.7.
The non-homogeneous linear equation \(2 x + 3 y = 8\) defines a line through the point \(P(1,2)\) in \(\R^2\text{.}\)
(a)
Draw the line and label the point \(P(1,2)\text{.}\) Choose another arbitrary point on the line and label it \(Q(x,y)\text{.}\) Draw the vector \(\uvec{v} = \abray{PQ}\) along the line. Express the components of \(\uvec{v}\) as formulas in \(x\) and \(y\text{.}\)
(b)
Draw the vector \(\uvec{n} = (2,3)\) (from the coefficients in the line equation, just as in Discovery 13.6) with its tail at \(P\text{.}\) What do you notice about the relationship between this normal vector and the vector \(\uvec{v}\) parallel to the line? Express this relationship in terms of the dot product, and then expand out this dot product.
The same sort of analysis can be carried out for a plane in space determined by algebraic equation \(a x + b y + c z = d\text{.}\) The coefficients form a normal vector \(\uvec{n} = (a,b,c)\) that is perpendicular to the plane (i.e. orthogonal to every vector that is parallel to the plane), and given some specific point \(\uvec{x}_0 = (x_0,y_0,z_0)\) that lies on the plane, the plane can be described by the point-normal form \(\dotprod{\uvec{n}}{(\uvec{x}-\uvec{x}_0)} = 0\text{.}\)
Discovery 13.8.
Consider the planes \(\Pi_1\text{,}\) \(\Pi_2\text{,}\) and \(\Pi_3\) described algebraically below.
\begin{align*}
\Pi_1 \amp \colon x - y + 2 z = 2 \amp
\Pi_2 \amp \colon 2 x - 2 y + 4 z = 7 \amp
\Pi_3 \amp \colon x - y + 3 z = 2
\end{align*}
Use the concept of normal vector to justify the claim that \(\Pi_1\) and \(\Pi_2\) are parallel, but that \(\Pi_3\) is not parallel to either of \(\Pi_1\) or \(\Pi_2\text{.}\)
Orthogonal projection onto a plane in space is a little more complicated, and is likely something you would learn about in a second course in linear algebra. But it’s possible to use a different strategy to determine the distance between a point and a plane by using the fact that a plane has one unique normal “direction.”
Discovery 13.9.
(a)
Using the diagram below as inspiration, come up with a procedure to determine the distance \(d\) between a point \(Q\) and a plane \(\Pi\text{.}\)
Hint.
Determine a vector that represents an equivalent distance, and then \(d\) will be the norm of this vector.
(b)
Come up with a procedure using vectors to determine the distance between parallel planes. Do not assume that either of the planes passes through the origin.
Hint.
Find a way to reduce this problem to the problem in the first part of this activity.
