Section 11.4 Examples
Subsection 11.4.1 Vectors in \(\R^n\)
Following Discovery 11.1, consider the vector \(\uvec{v}\) in \(\R^2\) (that is, in the plane) that represents changing position from \(P(1,2)\) to \(Q(3,-1)\text{.}\)
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} = (\Delta x,\Delta 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. In particular, if we take the initial point to be the origin, then the terminal point will have coordinates \((2,-3)\text{,}\) same as the components of \(\uvec{v}\text{.}\)
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 by subtracting coordinates in any dimension. For example, we compute the vector that represents changing position in space from \(P(1,2,-3)\) to \(Q(3,-1,0)\) by
\begin{align*}
\uvec{v} \amp = (\Delta x, \Delta y, \Delta z) \\
\amp = \bbrac{3-1, -1-2, 0-(-3)} \\
\amp = (2,-3, 3) \text{.}
\end{align*}
Another example in four-dimensional space, with
\begin{align*}
P \amp (1,2,-3,-4) \text{,}
\amp
Q \amp (1,-1,1,-1) \text{,}
\end{align*}
yields
\begin{align*}
\uvec{v} = \abray{PQ} \amp = (\Delta x_1, \Delta x_2, \Delta x_3, \Delta 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 11.4.1. 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. 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 11.4.2. 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 11.4.3. 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.
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 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 11.4.4. 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.
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.
Since the initial point is the origin, each vector 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*}
