DLA Examples

Section 12.4 Examples

In this section.

Subsection 12.4.1 Vectors in $\R^n$

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Subsection 12.4.2 Vector operations

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Subsection 12.4.3 Vectors versus lines and planes

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Subsection 12.4.1 Vectors in $\R^n$

Example 12.4.1. Computing components of a vector from displacement in $\R^2$.

Following Discovery 12.1, consider the vector $\uvec{v}$ in $\R^2$ (that is, in the $xy$-plane) that represents displacement from position $P(1,2)$ to position $Q(3,-1)\text{.}$ Its components are computed as the change in $x$ and the change in $y$ in moving from $P$ to $Q\text{:}$

\begin{equation*} \uvec{v} = (\change{x}, \change{y}) = (3 - 1, -1 - 2) = (2, -3) \text{.} \end{equation*}

By looking at the points’ coordinates we see that moving from point $P$ to $Q$ requires moving $2$ units right to get from $x=1$ to $x=3$ and moving $3$ units down to get from $y=2$ to $y=-1\text{,}$ and our calculation of $\uvec{v}$ above agrees. (See the diagram below left.)

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Diagram of a directed line segment in the plane.

The points $P(1,2)$ and $Q(3,-1)$ are plotted in the $xy$-plane, with $P$ in the first quadrant and $Q$ in the fourth. The directed line segment $\abray{PQ}$ is drawn and is labelled as representing the vector $\uvec{v}\text{.}$

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Diagram of three directed line segments in the plane, all of which represent the same vector.

Points $P(1,2)$ and $Q(3,-1)$ are plotted in the $xy$-plane, and the corresponding directed line segment $\abray{PQ}$ is again drawn and labelled as representing vector $\uvec{v}\text{.}$ The following pairs of points and corresponding directed line segments are also drawn:

Points $O(0,0)$ and $R(2,-3)\text{,}$ along with directed segment $\abray{OR}\text{.}$

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Points $S(-2,1)$ and $T(0,-2)\text{,}$ along with directed segment $\abray{ST}\text{.}$

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Both of these new directed segments are also labelled as representing vector $\uvec{v}\text{.}$

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The same vector with some other initial point will also have a terminal point that is $2$ units to the right and $3$ units down from the initial point. The diagram above right illustrates several such examples.

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For example, if we take the initial point to be the origin, to create a directed line segment $\abray{OR}$ that again represents $\uvec{v}\text{,}$ the terminal point should have coordinates $(2,-3)$ so that

\begin{equation*} (\change{x}, \change{y}) = (2 - 0, -3 - 0) = (2, -3) = \uvec{v} \text{.} \end{equation*}

Similarly, if we again compute the components of the vector represented by directed line segment $\abray{ST}$ by computing displacement from position $S(-2,1)$ to position $T(0,-2)\text{,}$ we obtain

\begin{equation*} (\change{x}, \change{y}) = \bbrac{0 - (-2), -2 - 1} = (2, -3) = \uvec{v} \text{.} \end{equation*}

These calculations verify that directed line segments $\abray{PQ}\text{,}$ $\abray{OR}\text{,}$ and $\abray{ST}$ do indeed all represent the same vector.

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Notice that these different representations of the vector $\uvec{v}$ are parallel and have the same length.

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Vectors can be similarly computed from pairs of points in any dimension by subtracting coordinates.

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Example 12.4.2. Computing components of a vector from displacement in $\R^3$.

Compute the vector associated to the directed line segment $\abray{PQ}$ representing displacement in space from position $P(1,2,-3)$ to position $Q(3,-1,0)$ by

\begin{align*} \uvec{v} \amp = (\change{x}, \change{y}, \change{z}) \\ \amp = \bbrac{3-1, -1-2, 0-(-3)} \\ \amp = (2,-3, 3) \text{.} \end{align*}

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Example 12.4.3. Computing components of a vector from displacement in $\R^4$.

We compute the vector associated to the directed line segment $\abray{PQ}$ that represents displacement in space from “position” $P(1,2,-3,-4)$ to “position” $Q(1,-1,1,-1)$ by

\begin{align*} \uvec{v} = \abray{PQ} \amp = (\change{x_1}, \change{x_2}, \change{x_3}, \change{x_4}) \\ \amp = \bbrac{1-1, -1-2, 1-(-3), -1-(-4)} \\ \amp = (0,-3, 4, 3) \text{.} \end{align*}

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Subsection 12.4.2 Vector operations

Here we’ll work through some of the computations of Discovery guide 12.1, and provide the accompanying diagrams.

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Example 12.4.4. Vector addition in $\R^2$.

In Discovery 12.2, we were tasked with geometrically adding vectors $\uvec{u} = (2,3)$ and $\uvec{v} = (3,-1)$ in the plane, starting at initial point $P(1,1)\text{.}$

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To add vectors geometrically, we put them head-to-tail, as in the diagram at right. The vector $\uvec{u} = (2,3)$ instructs us to move $2$ units right and $3$ units up, so starting at $P(1,1)$ we end up at $Q(3,4)\text{.}$ Then the vector $\uvec{v} = (3,-1)$ instructs us to move $3$ units right and $1$ unit down, so starting at $Q$ we end up at $R(6,3)\text{.}$ The sum vector $\uvec{u} + \uvec{v}$ represents the overall change from $P$ to $R\text{,}$ which is $5$ units right and $2$ units up, so that $\uvec{u}+\uvec{v} = (5,2)\text{.}$

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A diagram of a vector-addition triangle in the first quadrant of the plane. Points $P(1,1)\text{,}$ $Q(3,4)\text{,}$ and $R(6,3)$ are plotted on a set of $xy$-axes. Directed line segments $\abray{PQ}$ and $\abray{QR}$ are drawn and labelled as representing vectors $\uvec{u}$ and $\uvec{v}\text{.}$ The directed line segment $\abray{PR}$ is drawn with a dashed-line shaft and is labelled as representing the sum vector $\uvec{u} + \uvec{v}\text{.}$

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We can also compute the sum vector numerically by

\begin{align*} \uvec{u} + \uvec{v} \amp = (2,3) + (3,-1) \\ \amp = \bbrac{2+3,3+(-1)} \\ \amp = (5,2) \text{.} \end{align*}

Adding the vectors numerically is obviously faster and easier than drawing a diagram, but it’s good to have a mental picture of the geometric version of addition — it will help conceptually later on.

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Example 12.4.5. Vector addition in higher dimensions.

Our geometric picture and numeric computation of addition are similar for three-dimensional vectors in space. In $\R^n$ with $n \gt 3\text{,}$ we can’t draw a picture but we could imagine vector addition would take same the familiar triangle shape, and the numeric computations are similar again. For example, in $\R^5$ we have

\begin{align*} (1,2,3,4,5) + (6,-2,4,0,1) \amp = \bbrac{1+6,2+(-2),3+4,4+0,5+1} \\ \amp = (7,0,7,4,6) \text{.} \end{align*}

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Example 12.4.6. Negative vectors.

In Discovery 12.4, we explored the concept of a negative vector as the vector that will return us to our initial point, after changing positions along vector $\uvec{v} = (2,1)$ in the plane, starting at the origin. Recall that if a vector has its initial point at the origin, then the terminal point has coordinates equal to the components of the vector. As $\uvec{v}$ represents moving $2$ units right and $1$ unit up, then to return to the initial position we must move $2$ units left and $1$ unit down, so that $- \uvec{v} = (-2,-1)\text{.}$ Of course, the components of $- \uvec{v}$ do not depend on what initial point we choose — we would need to make the same reverse change of position no matter where $\uvec{v}$ started.

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Diagram of a directed line segment in the plane representing a vector, with its initial point at the origin, along with the oppositely-directed line segment in parallel, with its terminal point at the origin and its initial point at the terminal point of the first directed line segment.

The point $R(2,1)$ is plotted on a set of $xy$-axes. The directed line segment $\abray{OR}$ is drawn, where $O$ represents the origin, to represent vector $\uvec{v}\text{.}$ The opposite directed line segment $\abray{RO}$ is also drawn, and is labelled as representing the negative vector $- \uvec{v}\text{.}$

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Diagram of a directed line segment and the corresponding oppositely-directed line segment in the plane, both with their initial points at the origin. The first represents a vector and the second represents the negative vector.

On a set of $xy$-axes, the point $R(2,1)$ is plotted in the first quadrant of the plane and the “mirror” point $R'(-2,-1)$ is plotted in the third quadrant. The directed line segment $\abray{OR}$ is drawn, where $O$ represents the origin, to represent vector $\uvec{v}\text{.}$ The oppositely-directed line segment $\abray{OR'}$ is also drawn, and is labelled as representing the negative vector $- \uvec{v}\text{.}$

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As in Subsection 12.3.4, it is helpful to have a mental picture of a negative vector where its initial point is the same as for the original vector. In this orientation, the vector and its negative are parallel but oppositely directed.

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Example 12.4.7. Scalar multiplication.

In Discovery 12.6, we explored scalar multiplication geometrically in the plane using $\uvec{v} = (2,1)\text{,}$ initially by relating scalar multiplication to addition.

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A diagram illustrating scalar multiplication as repeated addition, with two directed line segments of equal length chained together head-to-tail, and a third directed line segment that traverses the combined path of the first two.

Points $R(2,1)$ and $R'(4,2)$ are plotted on a set of $xy$-axes. These two points are collinear with the origin $O\text{,}$ and the directed line segments $\abray{OR}$ and $\abray{RR'}$ are drawn. Both segments are labelled as representing the vector $\uvec{v}\text{.}$ Directed line segment $\abray{OR'}$ is drawn in parallel and is labelled as representing the scalar multiple vector $2 \uvec{v}\text{.}$

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The diagram to the left illustrates that $\uvec{v} + \uvec{v} = 2\uvec{v}\text{,}$ which we can also confirm numerically:

\begin{align*} \uvec{v} + \uvec{v} \amp = (2 + 2, 1 + 1) \\ \amp = (2,4) \\ \amp = 2 (2,1) \\ \amp = 2 \uvec{v} \text{.} \end{align*}

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Geometrically, the scalar multiples $3 \uvec{v}\text{,}$ $-2 \uvec{v}\text{,}$ $\frac{1}{2}\uvec{v}\text{,}$ and $-\frac{5}{4} \uvec{v}$ are all parallel to $\uvec{v}$ but with lengths stretched or compressed by the scale factor. Additionally, a negative scalar multiple flips the vector around in the opposite direction. In the diagrams below, we have placed each vector with initial point at the origin, so that its components are equal to the coordinates of its terminal point. We may also calculate:

\begin{align*} 3 \uvec{v} \amp= 3 (2,1) = ( 6, 3), \amp -2 \uvec{v} \amp= -2 (2,1) = (-4,-2),\\ \frac{1}{2} \uvec{v} \amp= \frac{1}{2} (2,1) = \left( 1 , \frac{1}{2} \right), \amp -\frac{5}{4} \uvec{v} \amp= -\frac{5}{4} (2,1) = \left( -\frac{5}{2}, -\frac{5}{4} \right). \end{align*}

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A directed line segment in the plane representing a vector, along with two parallel directed line segments representing examples of integer scalar multiples of the first vector.

Points $R(2,1)$ and $S(6,3)$ are plotted in the first quadrant of the $xy$-plane, and point $S'(-4, -2)$ is plotted in the third quadrant of the $xy$-plane. The directed line segment $\abray{OR}$ is drawn and labelled as representing the vector $\uvec{v}$ (where $O$ is the origin). Parallel directed line segments $\abray{OS}$ and $\abray{OS'}$ are also drawn and labelled as representing the scalar multiple vectors $3 \uvec{v}$ and $(-2) \uvec{v}\text{,}$ respectively.

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$A directed line segment in the plane representing a vector, along with two parallel directed line segments representing examples of fractional scalar multiples of the first vector.$

Points $R(2,1)$ and $T(1,1/2)$ are plotted in the first quadrant of the $xy$-plane, and point $T'(-5/2, -5/4)$ is plotted in the third quadrant of the $xy$-plane. The directed line segment $\abray{OR}$ is drawn and labelled as representing the vector $\uvec{v}$ (where $O$ is the origin). Parallel directed line segments $\abray{OT}$ and $\abray{OT'}$ are also drawn and labelled as representing the scalar multiple vectors $\frac{1}{2} \uvec{v}$ and $\left(- \frac{5}{4}\right) \uvec{v}\text{,}$ respectively.

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In higher dimensions, scalar multiplication works in exactly the same way numerically — we just multiply each component of the vector by the scale factor. For example, for $\uvec{v} = (1,-2,3,-4,5)$ in $\R^5\text{,}$ we have

\begin{equation*} -17\uvec{v} = (-17, 34, -51, 68, -85) \text{.} \end{equation*}

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Example 12.4.8. Determining linear combinations.

In $\R^3 \text{,}$ can the vector $\uvec{w} = (5,8,-5) $ be expressed as a linear combination of the vectors

\begin{align*} \uvec{v}_1 \amp = ( 0, 1,-1) \text{,} \amp \uvec{v}_2 \amp = (-2,-4, 3) \text{,} \amp \uvec{v}_3 \amp = (-1,-7, 7) \text{?} \end{align*}

A linear combination of these vectors is of the form $k_1 \uvec{v}_1 + k_2 \uvec{v}_2 + k_3 \uvec{v}_3 $ for some collection of scalars $k_1, k_2, k_3 \text{.}$ So we are asking if it is possible to determine values of these scalars so that

\begin{equation} k_1 (0,1,-1) + k_2 (-2,-4,3) + k_3 (-1,-7,7) = (5,8,-5) \tag{✶} \end{equation}

becomes true.

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In the aspirational equality (✶), the right-hand side is a single vector. So let’s also compute the left-hand side to a single vector so that we can properly compare:

\begin{align*} \text{LHS} \amp = k_1 (0,1,-1) + k_2 (-2,-4,3) + k_3 (-1,-7,7) \\ \amp = (0, k_1, -k_1) + (-2 k_2, -4 k_2, 3 k_2) + (-k_3, -7 k_3, 7 k_3) \\ \amp = (-2 k_2 - k_3, k_1 - 4 k_2 - 7 k_3, -k_1 + 3 k_2 + 7 k_3) \text{.} \end{align*}

This three-dimensional vector and the three-dimensional vector $(5,8,-5) $ can only be equal if each of their individual components are equal:

\begin{equation*} \begin{sysofeqns}{rcrcrcr} \amp \amp -2 k_2 \amp - \amp k_3 \amp = \amp 5 \\ k_1 \amp - \amp 4 k_2 \amp - \amp 7 k_3 \amp = \amp 8 \\ - k_1 \amp + \amp 3 k_2 \amp + \amp 7 k_3 \amp = \amp -5 \end{sysofeqns}\text{.} \end{equation*}

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Our question about linear combinations has now turned into a question about a linear system: does the above system have solutions? We’ll turn it into an augmented matrix and reduce to answer this new question:

\begin{equation*} \begin{abmatrix}{rrr|r} 0 \amp -2 \amp -1 \amp 5 \\ 1 \amp -4 \amp -7 \amp 8 \\ -1 \amp 3 \amp 7 \amp -5 \end{abmatrix} \qquad\rowredarrow\qquad \begin{abmatrix}{rrr|r} 1 \amp 0 \amp 0 \amp 3 \\ 0 \amp 1 \amp 0 \amp -3 \\ 0 \amp 0 \amp 1 \amp 1 \end{abmatrix}\text{.} \end{equation*}

We now have our answer. Since the system has one unique solution, there is exactly one way to express $\uvec{w} $ as a linear combination of $\uvec{v}_1, \uvec{v}_2, \uvec{v}_3 \text{:}$

\begin{equation*} (5,8,-5) = 3 (0,1,-1) + (-3) (-2,-4,3) + 1 (-1,-7,7) \text{.} \end{equation*}

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Remark 12.4.9.

In the previous example, notice how the vectors $\uvec{v}_1, \uvec{v}_2, \uvec{v}_3, \uvec{w} $ became the columns of the augmented system matrix. Think about why this happened; this is a pattern that we will see repeated when we explore linear combinations in more abstract settings in Chapter 17 and beyond.

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Subsection 12.4.3 Vectors versus lines and planes

Example 12.4.10. Creating a vector parallel to a line in $\R^2$.

The graph of $3 x - 2 y = 2$ is a line in the Cartesian plane. The points $P(0,-1)$ and $Q(2,2)$ both lie on this line, as can be verified by substituting the coordinates of each point into the left-hand side of the line equation and checking that the result computes to $-1\text{.}$ The directed line segment $\abray{PQ}$ thus lies along the line, and the vector $\uvec{v}$ associated to that directed line segment, computed as

\begin{equation*} \uvec{v} = \bbrac{2 - 0, 2 - (-1)} = (2,3) \text{,} \end{equation*}

is parallel to the line.

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Diagram of a vector parallel to a line in the Cartesian plane, created as a directed line segment between two points on the line. — Figure 12.4.11. A directed line segment between two distinct points on a line creates a vector parallel to that line.
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Example 12.4.12. Creating a vector parallel to a plane in $\R^3$.

The graph of $2 x - y + 6 z = 8$ is a plane in three-dimensional space. The points $P(0,-2,1)$ and $Q(1,0,1)$ both lie on this line, as can be verified by substituting the coordinates of each point into the left-hand side of the plane equation and checking that the result computes to $8\text{.}$ The directed line segment $\abray{PQ}$ thus lies along the plane, and the vector $\uvec{v}$ associated to that directed line segment, computed as

\begin{equation*} \uvec{v} = \bbrac{1 - 0, 0 - (-2), 1 - 1} = (1,2,0) \text{,} \end{equation*}

is parallel to the plane.

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