Section 12.4 Examples
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 \(Q(3,-1)\text{.}\)
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 \(\uvec{v}\text{.}\)
We can compute the components of \(\uvec{v}\) by computing 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*}
We can see by looking at their coordinates 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.
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 below illustrates several such examples.
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 \(\uvec{v}\text{.}\)
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 12.4.2. Computing components of a vector from displacement in \(\R^3\).
We compute the vector associated to the directed line segment \(\abray{PQ}\) that represents 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 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*}
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.
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 \(P(1,1)\text{.}\)
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 \(\uvec{u} + \uvec{v}\text{.}\)
To add vectors geometrically, we put them head-to-tail. 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 \(\uvec{u}+\uvec{v} = (5,2)\text{.}\) We can also add the vectors algebraically 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 algebraically 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 12.4.5. Vector addition in higher dimensions.
Our geometric picture and algebraic computation of addition are similar for three-dimensional vectors in space. In \(\R^n\) with \(n>3\text{,}\) we can’t draw a picture but we could imagine vector addition would take same the familiar triangle shape, and the algebraic computations are similar again. For example,
\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)
\end{align*}
in \(\R^5\text{.}\)
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.
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 \(- \uvec{v}\text{.}\)
If \(\uvec{v}\) represents moving \(1\) unit right and \(2\) units up, then to return to our original position we must move \(1\) unit left and \(2\) units 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.
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.
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 \(- \uvec{v}\text{.}\)
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.
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 above diagram illustrates that \(\uvec{v}+\uvec{v} = 2\uvec{v}\text{,}\) which we can also confirm algebraically:
\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.
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.
Since the initial point is the origin, each vector in the diagram above has components equal to the coordinates of its terminal point. In particular, we have
\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*}
In higher dimensions, scalar multiplication works in exactly the same way algebraically — 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*}
