Section 11.4 Examples
Subsection 11.4.1 Vectors in \(\R^n\)
Example 11.4.1. Computing components of a vector from displacement in \(\R^2\).
Following Discovery 11.1, 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.)
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 \(\uvec{v}\text{.}\)
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 \(\abray{OR}\text{.}\)
-
Points \(S(-2,1)\) and \(T(0,-2)\text{,}\) along with directed segment \(\abray{ST}\text{.}\)
Both of these new directed segments are also labelled as representing vector \(\uvec{v}\text{.}\)
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.
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.
Notice that these different representations of the vector \(\uvec{v}\) are parallel and have the same length.
Vectors can be similarly computed from pairs of points in any dimension by subtracting coordinates.
Example 11.4.2. 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*}
Example 11.4.3. 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*}
Subsection 11.4.2 Vector operations
Here we’ll work through some of the computations of Discovery guide 11.1, and provide the accompanying diagrams.
Example 11.4.4. Vector addition in \(\R^2\).
In Discovery 11.2, we were tasked with geometrically adding vectors \(\uvec{u} = (2,3)\) and \(\uvec{v} = (3,-1)\) in the plane, starting at initial point \(P(1,1)\text{.}\)
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 \(\uvec{u}+\uvec{v} = (5,2)\text{.}\)
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 \(\uvec{u} + \uvec{v}\text{.}\)
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.
Example 11.4.5. 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*}
Example 11.4.6. Negative vectors.
In Discovery 11.4, 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.
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 \(- \uvec{v}\text{.}\)
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 \(- \uvec{v}\text{.}\)
As in Subsection 11.3.4, 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.
Example 11.4.7. Scalar multiplication.
In Discovery 11.6, we explored scalar multiplication geometrically in the plane using \(\uvec{v} = (2,1)\text{,}\) initially by relating scalar multiplication to addition.
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 \(2 \uvec{v}\text{.}\)
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*}
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*}
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.
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.
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*}
