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Coordinated Linear Algebra

Anna Davis, Paul Zachlin, Allan Donsig, Levi Heath, Mikkel Munkholm

Section 1.2 Unit Vectors, the Dot Product, and Orthogonal Projection

Subsection 1.2.1 Standard Unit Vectors in \(\R^2\) and \(\R^3\)

A unit vector is a vector of length 1. A unit vector in the positive direction of a coordinate axis is called a standard unit vector. There are two standard unit vectors in \(\R^2\text{.}\) The vector \(\mathbf{i}=\begin{bmatrix} 1\\ 0 \end{bmatrix}\) is parallel the \(x\)-axis, and the vector \(\mathbf{j}=\begin{bmatrix} 0\\ 1 \end{bmatrix}\) is parallel the \(y\)-axis.
Standard unit vectors in the real plane.
Figure 1.2.1. The standard unit vectors in \(\mathbb{R}^2\text{.}\)
Vector names \(\mathbf{i}\) and \(\mathbf{j}\) are reserved for standard unit vectors in the direction of \(x\) and \(y\) axes, respectively. We chose to express \(\mathbf{i}\) and \(\mathbf{j}\) as column vectors, instead of row vectors, because the context in which we will encounter them in the future will require them to be column vectors. You may see them presented as row vectors in a different course.
There are three standard unit vectors in \(\R^3\text{:}\)
\begin{equation*} \mathbf{i} = \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}, \quad \mathbf{j}=\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}, \quad\mathbf{k}=\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \end{equation*}
Standard unit vectors in 3-dimensional real space.

Subsection 1.2.2 A Vector as a Linear Combination of Standard Unit Vectors

Every vector in \(\R^2\) and \(\R^3\) can be written as a sum of scalar multiples of \(\mathbf{i}\text{,}\) \(\mathbf{j}\) and \(\mathbf{k}\text{.}\) For example, if \(\mathbf{v}=\begin{bmatrix} 3\\ -2\\ 7 \end{bmatrix}\text{,}\) then
\begin{equation*} \mathbf{v}=\begin{bmatrix} 3\\ -2\\ 7 \end{bmatrix} = \begin{bmatrix} 3\\ 0\\ 0 \end{bmatrix} + \begin{bmatrix} 0\\ -2\\ 0 \end{bmatrix} + \begin{bmatrix} 0\\ 0\\ 7 \end{bmatrix} = 3\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} + (-2)\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix} + 7\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} = 3\mathbf{i}-2\mathbf{j}+7\mathbf{k} \end{equation*}
The expression \(3\mathbf{i}-2\mathbf{j}+7\mathbf{k}\) is called a linear combination of \(\mathbf{i}\text{,}\) \(\mathbf{j}\) and \(\mathbf{k}\text{.}\)

Subsection 1.2.3 Standard Unit Vectors in \(\R^n\)

When working with vectors in \(\R^n\text{,}\) we often use a different notation to denote the standard unit vectors.

Definition 1.2.2.

Let \(\mathbf{e}_i\) denote a vector that has \(1\) as the \(i^{th}\) component and zeros elsewhere. In other words,
\begin{equation*} \mathbf{e}_i=\begin{bmatrix} 0\\ 0\\ \vdots\\ 1\\ \vdots\\ 0 \end{bmatrix} \end{equation*}
where \(1\) is in the \(i^{th}\) position. We say that \(\mathbf{e}_i\) is a standard unit vector of \(\R^n\).

Subsection 1.2.4 Unit Vector in the Direction of a Given Vector

Recall that a unit vector is a vector of length 1. Given a non-zero vector \(\mathbf{v}\text{,}\) we can find a unit vector in the same direction by multiplying \(\mathbf{v}\) by an appropriate scalar. For example, if \(\mathbf{v}=\begin{bmatrix}a\\b\end{bmatrix}\) and \(\norm{\mathbf{v}}=3\text{,}\) then a unit vector \(\mathbf{u}\) in the same direction is given by \(\mathbf{u}=\begin{bmatrix}a/3\\b/3\end{bmatrix}=\begin{bmatrix}a/\norm{\mathbf{v}}\\b/\norm{\mathbf{v}}\end{bmatrix}\text{.}\)
In general, dividing a non-zero vector by its own magnitude produces a unit vector in the same direction. We summarize this observation in a theorem.

Proof.

Because \(\mathbf{u}\) is a positive scalar multiple of \(\mathbf{v}\text{,}\) \(\mathbf{u}\) points in the direction of \(\mathbf{v}\text{.}\) We now show that \(\norm{\mathbf{u}}=1\text{.}\)
\begin{align*} \norm{\mathbf{u}}\amp=\amp\sqrt{ \Big(\frac{v_1}{\norm{\mathbf{v}}}\Big)^2+\Big(\frac{v_2}{\norm{\mathbf{v}}}\Big)^2+\ldots +\Big(\frac{v_n}{\norm{\mathbf{v}}}\Big)^2} \\ \amp=\amp\frac{1}{\norm{\mathbf{v}}}\sqrt{v_1^2+v_2^2+\ldots +v_n^2} \\ \amp=\amp\frac{\norm{\mathbf{v}}}{\norm{\mathbf{v}}}=1 \end{align*}
The abstract proof can it everything seem a little odd. In practice, the actual computation is relatively straightforward. Here is an example:

Example 1.2.4.

Find a unit vector in the direction of \(\mathbf{v}=\begin{bmatrix}2\\-3\\1\\0\\1\end{bmatrix}\text{.}\)
We first compute \(\norm{\mathbf{v}}\text{.}\)
\begin{equation*} \norm{\mathbf{v}}=\sqrt{4+9+1+1}=\sqrt{15} \end{equation*}
By Theorem 1.2.3,
\begin{equation*} \mathbf{u}=\begin{bmatrix}2/\sqrt{15}\\-3/\sqrt{15}\\1/\sqrt{15}\\0\\1/\sqrt{15}\end{bmatrix}=\frac{1}{\sqrt{15}}\begin{bmatrix}2\\-3\\1\\0\\1\end{bmatrix} \end{equation*}

Subsection 1.2.5 The Dot Product

Definition 1.2.5.

Let \(\mathbf{u}\) and \(\mathbf{v}\) be vectors in \(\R^n\text{.}\) The dot product of \(\mathbf{u}\) and \(\mathbf{v}\text{,}\) denoted by \(\mathbf{u}\cdot \mathbf{v}\text{,}\) is given by
\begin{equation*} \mathbf{u}\cdot\mathbf{v}=\begin{bmatrix}u_1\\u_2\\\vdots\\u_n\end{bmatrix}\cdot\begin{bmatrix}v_1\\v_2\\\vdots\\v_n\end{bmatrix}=u_1v_1+u_2v_2+\ldots+u_nv_n. \end{equation*}
We will see the dot product reppear over and over again. As such, an example of a computation is reasonable to start with:

Example 1.2.6.

Find \(\mathbf{u}\cdot \mathbf{v}\) if \(\mathbf{u}=\begin{bmatrix}-2\\0\\1\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}3\\2\\-4\end{bmatrix}\text{.}\)
\begin{equation*} \mathbf{u}\cdot\mathbf{v}=\begin{bmatrix}-2\\0\\1\end{bmatrix}\cdot\begin{bmatrix}3\\2\\-4\end{bmatrix}=(-2)(3)+(0)(2)+(1)(-4)=-6-4=-10 \end{equation*}
Note that the dot product of two vectors is a scalar. For this reason, the dot product is sometimes called a scalar product.
A quick examination of Example 1.2.6 will convince you that the dot product is commutative. In other words, \(\mathbf{u}\cdot\mathbf{v}=\mathbf{v}\cdot\mathbf{u}\text{.}\) This and other properties of the dot product are stated below.

Proof.

We will prove Item 2. The remaining properties are left as exercises.
\begin{align*} \left(\mathbf{u}+\mathbf{v}\right)\cdot \mathbf{w}\amp=\left(\begin{bmatrix} u_1\\ u_2\\ \vdots\\ u_n \end{bmatrix}+\begin{bmatrix} v_1\\ v_2\\ \vdots\\ v_n \end{bmatrix}\right)\cdot \begin{bmatrix}w_1\\w_2\\\vdots\\w_n\end{bmatrix}=\begin{bmatrix} u_1+v_1\\ u_2+v_2\\ \vdots\\ u_n+v_n \end{bmatrix}\cdot \begin{bmatrix}w_1\\w_2\\\vdots\\w_n\end{bmatrix} \\ \amp=(u_1+v_1)w_1 + (u_2+v_2)w_2 + \ldots + (u_n+v_n)w_n \\ \amp=u_1w_1 + v_1w_1 + u_2w_2+v_2w_2 + \ldots + u_nw_n+v_nw_n \\ \amp=(u_1w_1 + u_2w_2\ldots + u_nw_n) + (v_1w_1 + v_2w_2 + \ldots + v_nw_n) \\ \amp=\begin{bmatrix} u_1\\ u_2\\ \vdots\\ u_n \end{bmatrix} \cdot\begin{bmatrix}w_1\\w_2\\\vdots\\w_n\end{bmatrix} +\begin{bmatrix} v_1\\ v_2\\ \vdots\\ v_n \end{bmatrix} \cdot \begin{bmatrix}w_1\\w_2\\\vdots\\w_n\end{bmatrix} =\mathbf{u}\cdot\mathbf{w}+\mathbf{v}\cdot\mathbf{w} \end{align*}
We will illustrate Item 6 with an example.

Example 1.2.8.

Let \(\mathbf{u}=\begin{bmatrix}-2\\3\end{bmatrix}\text{.}\) Use \(\mathbf{u}\) to illustrate Item 6 of Theorem 1.2.7.
\begin{equation*} \norm{\mathbf{u}}^2=(-2)^2+3^2=(-2)(-2)+(3)(3)=\begin{bmatrix}-2\\3\end{bmatrix}\cdot\begin{bmatrix}-2\\3\end{bmatrix}=\mathbf{u}\cdot\mathbf{u} \end{equation*}
If we take the square root of both sides of the equation in Item 6, we get an alternative way to think of the length of a vector (Subsection 1.1.7).
Let’s also look at an example of Item 7.

Example 1.2.10.

Note that the standard unit vectors \(\mathbf{i},\mathbf{j}\) in \(\mathbb{R}^2\) are orthogonal--see Figure 1.2.1. Indeed,
\begin{equation*} \mathbf{i}\cdot\mathbf{j}=1(0)+0(1)=0. \end{equation*}
In general, by Item 7, for standard unit vectors \(\mathbf{e}_i,\mathbf{e}_j\) in \(\mathbb{R}^n\) with \(i\neq j\text{,}\) the vectors \(\mathbf{e}_i\) and \(\mathbf{e}_j\) are orthogonal since \(\mathbf{e}_i\cdot \mathbf{e}_j= 0\text{.}\)

Subsection 1.2.6 Orthogonal Projections

Given a line \(l\) and a vector \(\mathbf{v}\) emanating from a point on \(l\text{,}\) it is sometimes convenient to express \(\mathbf{v}\) as the sum of a vector \(\mathbf{v}_{\parallel}\text{,}\) parallel to \(l\text{,}\) and a vector \(\mathbf{v}_{\perp}\text{,}\) perpendicular to \(l\text{.}\) If you have taken a physics course, you may have seen a force vector decomposed into the sum of two components: one parallel and one perpendicular to the direction of motion.
Suppose \(\mathbf{d}\) is a direction vector for \(l\text{.}\) Then \(\mathbf{v}_{\parallel}=k\mathbf{d}\) for some scalar \(k\text{.}\) Our goal is to find \(k\text{.}\)
\begin{align*} \mathbf{v}\cdot\mathbf{d}\amp=(\mathbf{v}_{\parallel}+\mathbf{v}_{\perp})\cdot\mathbf{d} \\ \amp=(k\mathbf{d}+\mathbf{v}_{\perp})\cdot\mathbf{d} \\ \amp=k\mathbf{d}\cdot\mathbf{d}+\mathbf{v}_{\perp}\cdot\mathbf{d} \\ \amp=k\norm{\mathbf{d}}^2+0 \\ \amp=k\norm{\mathbf{d}}^2 \end{align*}
We conclude that
\begin{equation*} k=\frac{\mathbf{v}\cdot\mathbf{d}}{\norm{\mathbf{d}}^2} \end{equation*}
and
\begin{equation*} \mathbf{v}_{\parallel}=k\mathbf{d}=\left(\frac{\mathbf{v}\cdot\mathbf{d}}{\norm{\mathbf{d}}^2}\right)\mathbf{d} \end{equation*}
The vector \(\mathbf{v}_{\parallel}=\left(\frac{\mathbf{v}\cdot\mathbf{d}}{\norm{\mathbf{d}}^2}\right)\mathbf{d}\) is called the projection of \(\mathbf{v}\) onto \(\mathbf{d}\). In our discussion, \(\mathbf{d}\) is a direction vector for line \(l\text{.}\) So, we can also say that \(\mathbf{v}_{\parallel}\) is the projection of \(\mathbf{v}\) onto \(l\).
To find \(\mathbf{v}_{\perp}\text{,}\) observe that \(\mathbf{v}_{\perp}=\mathbf{v}-\mathbf{v}_{\parallel}\text{.}\)

Definition 1.2.11.

Let \(\mathbf{v}\) be a vector, and let \(\mathbf{d}\) be a non-zero vector. The projection of \(\mathbf{v}\) onto \(\mathbf{d}\) is given by
\begin{equation*} \text{proj}_{\mathbf{d}}\mathbf{v}=\left(\frac{\mathbf{v}\cdot\mathbf{d}}{\norm{\mathbf{d}}^2}\right)\mathbf{d} \end{equation*}
Although we will revisit the projection in greater detail much later, we illustrate it in an explicit case with a picture.

Example 1.2.12.

Find the projection of \(\mathbf{v}\text{,}\) shown below, onto the line given by \(y=\frac{1}{2}x-1\text{.}\)
We begin by finding vectors \(\mathbf{v}\) and \(\mathbf{d}\text{.}\) The tail of \(\mathbf{v}\) is located at \((-2, -2)\text{,}\) and the head of \(\mathbf{v}\) is at \((2, 4)\text{.}\) Using the ``head-tail" formula we get
\begin{equation*} \mathbf{v}=\begin{bmatrix}2-(-2)\\4-(-2)\end{bmatrix}=\begin{bmatrix}4\\6\end{bmatrix} \end{equation*}
The direction vector for the line \(y=\frac{1}{2}x-1\) is
\begin{equation*} \mathbf{d}=\begin{bmatrix}2\\1\end{bmatrix} \end{equation*}
We find that \(\mathbf{v}\cdot\mathbf{d}=14\) and \(\norm{\mathbf{d}}^2=5\text{.}\) Thus
\begin{equation*} \text{proj}_{\mathbf{d}}\mathbf{v}=\left(\frac{\mathbf{v}\cdot\mathbf{d}}{\norm{\mathbf{d}}^2}\right)\mathbf{d}=\frac{14}{5}\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}28/5\\14/5\end{bmatrix} \end{equation*}

Subsection 1.2.7 Distance from a Point to a Line

The shortest distance from a point to a line is the length of the perpendicular line segment dropped from the point to the line. Vector projection formula will help us find the length of such a perpendicular.

Example 1.2.13.

Let \(A(2, -1, 1)\) be a point in \(\R^3\text{.}\) Suppose line \(l\) is given by parametric equations
\begin{equation*} x=t+3 \end{equation*}
\begin{equation*} y=-t+1 \end{equation*}
\begin{equation*} z=t-2 \end{equation*}
Find the distance from \(A\) to \(l\text{.}\)
We will first construct a vector \(\mathbf{v}\) by picking an arbitrary point \(B\) on \(l\) to be the tail of \(\mathbf{v}\) and using point \(A\) as the head of \(\mathbf{v}\text{.}\) An easy point to choose on line \(l\) is the point \((3, 1, -2)\) that corresponds to \(t=0\text{.}\) Now we have
\begin{equation*} \mathbf{v}=\overrightarrow{BA}=\begin{bmatrix}2-3\\-1-1\\1-(-2)\end{bmatrix}=\begin{bmatrix}-1\\-2\\3\end{bmatrix} \end{equation*}
The line has a direction vector
\begin{equation*} \mathbf{d}=\begin{bmatrix}1\\-1\\1\end{bmatrix} \end{equation*}
We will now find the projection of \(\overrightarrow{BA}\) onto \(l\)
\begin{equation*} \text{proj}_{\mathbf{d}} \overrightarrow{BA}=\left(\frac{\mathbf{v}\cdot\mathbf{d}}{\norm{\mathbf{d}}^2}\right)\mathbf{d}=\frac{4}{3}\begin{bmatrix}1\\-1\\1\end{bmatrix}=\begin{bmatrix}4/3\\-4/3\\4/3\end{bmatrix} \end{equation*}
Next, we find \(\mathbf{v}_{\perp}\text{.}\)
\begin{equation*} \mathbf{v}_{\perp}=\mathbf{v}-\mathbf{v}_{\parallel}=\begin{bmatrix}-1\\-2\\3\end{bmatrix}-\begin{bmatrix}4/3\\-4/3\\4/3\end{bmatrix}=\begin{bmatrix}-7/3\\-2/3\\5/3\end{bmatrix} \end{equation*}
Finally, to find the distance between point \(A\) and line \(l\text{,}\) we find the magnitude of \(\mathbf{v}_{\perp}\text{.}\)
\begin{equation*} \norm{\mathbf{v}_{\perp}}=\frac{1}{3}\sqrt{49+4+25}=\frac{\sqrt{78}}{3} \end{equation*}

Exercises 1.2.8 Exercises

Exercise Group.

Express each of the following vectors as a linear combination of appropriate standard unit vectors.
1.
\begin{equation*} \mathbf{u}=\begin{bmatrix} 0\\ 4\\ -3 \end{bmatrix} \end{equation*}
Answer.
\begin{equation*} \mathbf{u}= 4\mathbf{j} + -3\mathbf{k} \end{equation*}
2.
\begin{equation*} \mathbf{v}=\begin{bmatrix} -1\\ 1 \end{bmatrix} \end{equation*}
Answer.
\begin{equation*} \mathbf{u}= -1\mathbf{i} + \mathbf{j} \end{equation*}
3.
\begin{equation*} \mathbf{w}=\begin{bmatrix} 5\\ -3\\ 1\\ 7 \end{bmatrix} \end{equation*}
Answer.
\begin{equation*} \mathbf{u}= 5\mathbf{e}_1 + -3\mathbf{e}_2 + \mathbf{e}_3 + 7\mathbf{e}_4 \end{equation*}

Exercise Group.

Express each given vector in component form.
4.
\(\mathbf{u}=\mathbf{i}+3\mathbf{j}\) is a vector in \(\R^2\text{.}\)
Answer.
\begin{equation*} \mathbf{u} = \begin{bmatrix} 1\\ 3\end{bmatrix} \end{equation*}
5.
\(\mathbf{v}=-\mathbf{j}+5\mathbf{k}\) is a vector in \(\R^3\text{.}\)
Answer.
\begin{equation*} \mathbf{v}=\begin{bmatrix} 0 \\ -1\\ 5 \end{bmatrix} \end{equation*}
6.
\(\mathbf{w}=\mathbf{e}_1-2\mathbf{e}_3+4\mathbf{e}_4\) is a vector in \(\R^4\text{.}\)
Answer.
\begin{equation*} \mathbf{w}=\begin{bmatrix} 1\\ 0\\ -2\\ 4\end{bmatrix} \end{equation*}

7.

Is it possible to express \(\mathbf{u}=\begin{bmatrix} -6\\ 1\\ 4 \end{bmatrix}\) as a linear combination of \(\mathbf{i}\) and \(\mathbf{j}\) alone, where \(\mathbf{i}\) and \(\mathbf{j}\) are in \(\R^3\text{?}\) Explain your reasoning.

Exercise Group.

Find a unit vector in the direction of the given vector \(\mathbf{v}\text{.}\)
8.
\(\mathbf{v}=\begin{bmatrix}-3\\4\end{bmatrix}\)
Answer.
\(\mathbf{u}=\begin{bmatrix} -0.6\\ 0.8 \end{bmatrix}\)
9.
\(\mathbf{v}=\begin{bmatrix}2\\-3\\6\end{bmatrix}\)
Answer.
\(\mathbf{u}=\begin{bmatrix} 2/7\\ -3/7\\ 6/7 \end{bmatrix}\)
10.
\(\mathbf{v}=\begin{bmatrix}1\\3\\-2\end{bmatrix}\)
Answer.
\(\mathbf{u}=\begin{bmatrix} 1/\sqrt{14}\\ 3/\sqrt{14}\\ -2/\sqrt{14}\end{bmatrix}\)

11.

Let \(\mathbf{v}=\begin{bmatrix}-1\\1\\\sqrt{7}\end{bmatrix}\text{.}\) Apply the concepts from this section to find a vector \(\mathbf{w}\) that points in the same direction as \(\mathbf{v}\) and whose length is 5.
Answer.
\(\mathbf{w}=\begin{bmatrix} -5/3\\ 5/3\\ 5\sqrt{7}/3 \end{bmatrix}\)

12.

Find the dot product of \(\mathbf{u}\) and \(\mathbf{v}\) if
\begin{equation*} \mathbf{u}=\begin{bmatrix}-1\\-2\\5\\4\end{bmatrix},\quad \mathbf{v}=\begin{bmatrix}2\\-2\\-3\\1\end{bmatrix} \end{equation*}
Answer.
\begin{equation*} \mathbf{u} \cdot \mathbf{v} = -9 \end{equation*}

13.

Find the dot product of \(\mathbf{u}\) and \(\mathbf{v}\) if
\begin{equation*} \mathbf{u}=\begin{bmatrix}1\\1/2\end{bmatrix},\quad \mathbf{v}=\begin{bmatrix}-2\\4\end{bmatrix} \end{equation*}
Answer.
\begin{equation*} \mathbf{u} \cdot \mathbf{v} = 0 \end{equation*}

14.

Use vector \(\mathbf{u}=\begin{bmatrix}2\\5\\-7\end{bmatrix}\) to illustrate Item 6 of Theorem 1.2.7.

16.

From the given list of vector pairs, identify ALL pairs of vectors that lie on perpendicular lines.
  • \(\mathbf{u}=\begin{bmatrix}1\\\frac{1}{2}\end{bmatrix}\text{,}\) \(\mathbf{v}=\begin{bmatrix}-2\\4\end{bmatrix}\)
  • \(\mathbf{u}=\begin{bmatrix}-1\\\frac{1}{2}\end{bmatrix}\text{,}\) \(\mathbf{v}=\begin{bmatrix}-2\\4\end{bmatrix}\)
  • \(\mathbf{u}=\begin{bmatrix}1\\\frac{1}{2}\end{bmatrix}\text{,}\) \(\mathbf{v}=\begin{bmatrix}1\\-2\end{bmatrix}\)
  • \(\mathbf{u}=\begin{bmatrix}-1\\-\frac{1}{2}\end{bmatrix}\text{,}\) \(\mathbf{v}=\begin{bmatrix}-2\\4\end{bmatrix}\)
Also compute \(\mathbf{u}\cdot\mathbf{v}\) for each pair. What do you observe?
Hint.
You may want to draw a picture and think about what you know about slopes of perpendicular lines.

Exercise Group.

For each problem below
  1. Find the value of \(x\) that will make vectors \(\mathbf{u}\) and \(\mathbf{v}\) perpendicular.
    Hint: Think of the \(x\)-component as the ``run" and the \(y\)-component as the ``rise", then use what you know about slopes of perpendicular lines.
  2. Find \(\mathbf{u}\cdot\mathbf{v}\text{.}\)
17.
\begin{equation*} \mathbf{u} = \begin{bmatrix}1\\2\end{bmatrix},\quad \mathbf{v}=\begin{bmatrix}2\\x\end{bmatrix} \end{equation*}
Answer.
\begin{equation*} x = -1 \end{equation*}
\begin{equation*} \mathbf{u}\cdot\mathbf{v}=0 \end{equation*}
18.
\begin{equation*} \mathbf{u} = \begin{bmatrix}5\\2\end{bmatrix},\quad \mathbf{v}=\begin{bmatrix}x\\-4\end{bmatrix} \end{equation*}
Answer.
\begin{equation*} x = 8/5 \end{equation*}
\begin{equation*} \mathbf{u}\cdot\mathbf{v}=0 \end{equation*}
19.
\begin{equation*} \mathbf{u} = \begin{bmatrix} 4\\-3\end{bmatrix},\quad \mathbf{v} =\begin{bmatrix}6\\x\end{bmatrix} \end{equation*}
Answer.
\begin{equation*} x = 8 \end{equation*}
\begin{equation*} \mathbf{u}\cdot\mathbf{v}=0 \end{equation*}

20.

  1. Vector \(\mathbf{u}\) that lies on the line \(y=mx\text{,}\) has the form \(\mathbf{u}=k\begin{bmatrix}1\\m\end{bmatrix}\text{.}\) Assuming that \(m\neq 0\text{,}\) find the general form for a vector \(\mathbf{v}\) that lies on a line perpendicular to \(y=mx\text{.}\)
  2. Find \(\mathbf{u}\cdot \mathbf{v}\text{.}\)
  3. Formulate a conjecture about the dot product of perpendicular vectors.
Hint.
What do you know about the slopes of perpendicular lines?

Exercise Group.

Find \(\text{proj}_{\mathbf{d}}\mathbf{v}\text{.}\)
21.
Let \(\mathbf{d}=\begin{bmatrix}-1\\3\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}1\\4\end{bmatrix}\text{.}\) Compute \(\text{proj}_{\mathbf{d}}\mathbf{v}\text{.}\)
Answer.
\begin{equation*} \text{proj}_{\mathbf{d}}\mathbf{v}=\begin{bmatrix} -1.1\\ 3.3\end{bmatrix} \end{equation*}
22.
Suppose \(\mathbf{d}=\begin{bmatrix}0\\2\\1\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}-1\\-4\\2\end{bmatrix}\text{.}\) Compute the projection of \(\mathbf{v}\) onto \(\mathbf{d}\text{.}\)
Answer.
\begin{equation*} \text{proj}_{\mathbf{d}}\mathbf{v}=\begin{bmatrix} 0\\ -2.4\\ -1.2\end{bmatrix} \end{equation*}

23.

Find the projection of vector \(\mathbf{v}\) onto line \(l\text{.}\) (If entering answers in decimal form, round to the nearest one hundredth.)
Answer.
\begin{equation*} \begin{bmatrix}-95/26\\ 19/26\end{bmatrix} \end{equation*}

24.

Find the distance between point \(A\) and line \(l\text{.}\)
Answer.
\(\sqrt{3.2}\)

25.

Show that \(\text{proj}_{\mathbf{d}}\mathbf{v}\) does not depend on the length of \(\mathbf{d}\) by proving that \(\text{proj}_{\mathbf{d}}\mathbf{v}=\text{proj}_{k\mathbf{d}}\mathbf{v}\) for \(k\neq 0\text{.}\) What does this result mean geometrically? Illustrate your response with a diagram.

26.

Find the radius of a circle centered at \((4, 2)\) if the line \(y=\frac{3}{2}x+3\) is tangent to the circle. Enter your response as a fraction.
Answer.
\begin{equation*} r=\sqrt{196/13} \end{equation*}
The graph below shows the line \(y=\frac{3}{2}x+3\) together with a circle of radius \(1\text{.}\) Change the value of \(r\) to the radius you have found to visualize the correct answer.
Figure 1.2.14. A larger version of this activity is available here
 1 
https://www.geogebra.org/calculator/bngnjxee
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