Section 12.3 Concepts
Subsection 12.3.1 Vectors
A directed line segment (or arrow) could be thought of dynamically as describing a change in position, from the initial point to the terminal point. A two-dimensional vector in the plane or a three-dimensional vector in space captures just the change part of “change in position,” leaving the position part (that is, the initial and terminal points) unspecified. For example, in the plane, the instructions “move two units right and three units down” describe a way to change positions, but don’t actually specify from where or to where the change in position is occurring. So a vector corresponds to an infinite number of directed line segments, where each of these directed line segments has a different initial point but all of them require the same “change” to change positions from initial point to terminal point. Continuing our example, every change in position between some initial and terminal points in the plane that requires moving two units right and three units down can be represented by the same vector.
A diagram consisting of four right triangles in the \(xy\)-plane, each with a directed line segment for the hypotenuse. The triangles are all identical, both in that they are all congruent but also that they are similarly oriented in the plane. Each triangle can identically be described as follows. A directed line segment extends downwards and rightwards from an initial point \(P_i\) to a terminal point \(Q_i\) (with \(i = 1, 2, 3, 4\) indexing the triangles), representing a vector labelled \(\uvec{u}\text{.}\) (This vector label is identical, without indexing subscript, for each triangle.) A dashed line segment extends horizontally \(2\) units rightwards of \(P_i\text{,}\) and from there another dashed line segment extends vertically \(3\) units downwards where it precisely meets \(Q_i\) to complete the triangle. These two line segments necessarily meet at a right angle.
We describe a two-dimensional vector in the plane with a pair of numerical components: the change in \(x\) and the change in \(y\text{.}\) If \(P(x_1,y_1)\) and \(Q(x_2,y_2)\) are points in the plane, then the vector associated to the directed line segment \(\abray{PQ}\) has components \(\uvec{v} = (\change{x},\change{y}) = (x_2 - x_1, y_2 - y_1)\text{.}\) (A three-dimensional vector in space requires a third component as well: the change in \(z\text{.}\))
Notice what happens when we use the origin \(O(0,0)\) as the initial point and an arbitrary point \(R(x,y)\) as the terminal point in a directed line segment: the vector associated to \(\abray{OR}\) is then \((x-0,y-0) = (x,y)\text{.}\) So when the initial point is the origin, the components of the vector are exactly the coordinates of the terminal point. In Discovery 12.1 we saw that this works in reverse as well. That is, if we have a vector \(\uvec{v} = (v_1,v_2)\text{,}\) then we could consider the numbers \(v_1,v_2\) as coordinates of a point \(R(x,y)\) with \(x=v_1\) and \(y=v_2\text{,}\) and then the vector associated to \(\abray{OR}\) is just \(\uvec{v}\) again. In this way, every vector corresponds to one unique directed line segment with initial point at the origin, and so that is sort of the “natural position” of the vector as a directed line segment. But we will find that it is often convenient to consider other directed line segments that correspond to a particular vector.
We live in a three-dimensional world (or, at least, it appears that way to us), and our little human brains cannot visualize points or arrows in four- or higher-dimensional spaces. However, we can still describe such imaginary objects using our experience from two- and three-dimensional points and vectors. For example, if we had two points \(P\) and \(Q\) in an imaginary four-dimensional space, they would each require four coordinates, so we would describe them as \(P(w_1,x_1,y_1,z_1)\) and \(Q(w_2,x_2,y_2,z_2)\text{.}\) Then the vector corresponding to the directed line segment \(\abray{PQ}\) would have four components and we would compute it as \(\uvec{v} = (\change{w}, \change{x}, \change{y}, \change{z}) = (w_2 - w_1, x_2 - x_1, y_2 - y_1, z_2 - z_1)\text{.}\)
Subsection 12.3.2 Vector addition
A vector describes a change in position. If we chain two changes in position together, by making the initial point of the second vector the same as the terminal point of the first vector, then we could consider the overall change in position.
A diagram of three vectors in a triangular configuration. Three points are plotted: \(P\) in the lower-left of the diagram, \(Q\) at the top of the diagram, just left of centre, and \(R\) at the far right of the diagram, slightly above centre. Between these points are drawn three directed line segments: \(\abray{PQ}\) representing a vector labelled \(\uvec{u}\text{,}\) \(\abray{QR}\) representing a vector labelled \(\uvec{v}\text{,}\) and \(\abray{PR}\) drawn with a dashed-line shaft and labelled with a question mark.
If these are points and vectors in the plane, then clearly the change in position from \(P\) to \(R\) will be described by the total net change in \(x\) and the total net change in \(y\text{,}\) as we discovered in Discovery 12.2. So, in Figure 12.3.2, we obtain the components for the vector corresponding to \(\abray{PR}\) by adding corresponding components of \(\uvec{u}\) and \(\uvec{v}\text{.}\) That is, if \(\uvec{u} = (u_1,u_2)\) and \(\uvec{v} = (v_1,v_2)\text{,}\) then the components of the dashed arrow labelled with a question mark are \((u_1+v_1,u_2+v_2)\text{.}\) For obvious reasons, we call this the sum of \(\uvec{u}\) and \(\uvec{v}\text{.}\)
A diagram of three vectors arranged in a triangular configuration to illustrate vector addition. The diagram of Figure 12.3.2 is reproduced, except that \(\abray{PR}\) is drawn with a solid-line shaft and represents the sum vector \(\uvec{u} + \uvec{v}\text{.}\)
Also, below the vector triangle appears a horizontal bar extending from the horizontal position of \(P\) to the horizontal position of \(Q\) and labelled \(u_1\text{,}\) and adjacent to it appears another bar extending from the horizontal position of \(Q\) to the horizontal position of \(R\) and labelled \(v_1\text{.}\) Below these is a third horizontal bar extending from the horizontal position of \(P\) all the way to the horizontal position of \(R\) and labelled \(u_1 + v_1\text{.}\)
Similarly, to the right of the vector triangle appears a vertical bar extending from the vertical position of \(Q\) to the vertical position of \(R\) and labelled \(\abs{v_2}\text{,}\) and adjacent to it appears another bar extending from the vertical position of \(R\) to the vertical position of \(P\) and labelled \(u_2 + v_2\text{.}\) Further to the right is a third vertical bar extending from the vertical position of \(Q\) all the way down to the vertical position of \(P\) and labelled \(u_2\text{.}\)
In Discovery 12.2 we also considered the result of interchanging the order of a pair of vectors that have been chained together.
A diagram of six vectors arranged into a pair of mirror-image vector-addition triangles. Four points \(P, Q, R, Q'\) are plotted so that they form the vertices of a parallelogram \(\abcdquad{PQRQ'}\text{.}\) Six directed line segments are drawn between these points:
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Parallel segments \(\abray{PQ}\) and \(\abray{Q'R}\) forming opposite sides of parallelogram \(\abcdquad{PQRQ'}\text{,}\) both representing the same vector \(\uvec{u}\text{.}\)
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Parallel segments \(\abray{PQ'}\) and \(\abray{QR}\) forming opposite sides of parallelogram \(\abcdquad{PQRQ'}\text{,}\) both representing the same vector \(\uvec{v}\text{.}\)
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The directed line segment \(\abray{PR}\text{,}\) a diagonal in parallelogram \(\abcdquad{PQRQ'}\text{,}\) is drawn twice: one copy representing the sum vector \(\uvec{u} + \uvec{v}\) as part of the vector-addition triangle \(\abctriangle{PQR}\text{,}\) and the other copy representing the sum vector \(\uvec{v} + \uvec{u}\) as part of the vector-addition triangle \(\abctriangle{PQ'R}\text{.}\)
In Figure 12.3.4, the vector for \(\abray{PQ'}\) is the same as that for \(\abray{QR}\text{,}\) because they represent the same change in position, just with a different initial point. Accordingly, we have labelled both vectors with \(\uvec{v}\text{.}\) And the same applies to \(\uvec{u}\) with respect to \(\abray{PQ}\) and \(\abray{Q'R}\text{.}\)
The diagram in Figure 12.3.4 illustrates that if we start at \(P\) and chain together the change-of-position instructions contained in vectors \(\uvec{u}\) and \(\uvec{v}\text{,}\) the order that we do so does not matter — the overall change in position will be from \(P\) to \(R\text{.}\) Thus, the order of vector addition doesn’t matter. Algebraically, we could have predicted that this would be the case because it doesn’t matter what order you add components: the identities \(u_1 + v_1 = v_1 + u_1\) and \(u_2 + v_2 = v_2 + v_2\) are both valid. But it’s useful conceptually to have these geometric pictures of vector addition because, whether you believe this about yourself or not, humans are spatial thinkers. And the geometric version of the vector identity \(\uvec{v} + \uvec{u} = \uvec{u} + \uvec{v}\) makes a pretty picture of a parallelogram, so we call it the parallelogram rule.
For three-dimensional vectors, we can imagine diagrams like the ones in Figure 12.3.3 and Figure 12.3.4 floating in space, and the parallelogram rule would hold there as well. In higher dimensions, we cannot draw pictures, but we could imagine that they are similar. At any rate, the algebra of vector addition is the same in any dimension: for \(\uvec{u} = (u_1,u_2,\dotsc,u_n)\) and \(\uvec{v} = (v_1,v_2,\dotsc,v_n)\) in \(\R^n\text{,}\) we have
\begin{equation*}
\uvec{u} + \uvec{v} = (u_1+v_1,u_2+v_2,\dotsc,u_n+v_n) \text{.}
\end{equation*}
Subsection 12.3.3 The zero vector
There is one special change in position that is unlike any other — the one where the initial and terminal points are the same, so that there is actually no change in position. In two dimensions, this means there is no change in either \(x\) or \(y\text{,}\) so the components are \((0,0)\text{.}\) Similarly, in any number of dimensions we have the zero vector \(\zerovec = (0,0,\dotsc,0)\text{.}\)
As we explored in Discovery 12.3, if we chain together a vector \(\uvec{v}\text{,}\) representing some change in position, with the zero vector, which represents no change, then the net result is just the change of \(\uvec{v}\text{.}\) That is, \(\uvec{v} + \zerovec = \uvec{v}\text{,}\) and also \(\zerovec + \uvec{v} = \uvec{v}\text{.}\)
Subsection 12.3.4 Vector negatives and vector subtraction
Subsubsection 12.3.4.1 Negatives
If we move from \(P\) to \(Q\text{,}\) and from there move from \(Q\) back to \(P\text{,}\) the net result is no change in position, which is represented by the zero vector. This means if we add the vector corresponding to \(\abray{PQ}\) to the one corresponding to \(\abray{QP}\text{,}\) the result is \(\zerovec\text{.}\) So if we label the vector for \(\abray{PQ}\) as \(\uvec{v}\text{,}\) it seems reasonable to label the vector for \(\abray{QP}\) as \(-\uvec{v}\text{,}\) the negative of \(\uvec{v}\text{,}\) so that we have \(\uvec{v} + (-\uvec{v}) = \zerovec\text{.}\) (See Figure 12.3.5.(a).)
If we are to have \(\uvec{v} + (-\uvec{v}) = \zerovec\text{,}\) and the components of \(\zerovec\) are all \(0\text{,}\) then since we add vectors by adding corresponding components, the components of \(-\uvec{v}\) must be the negatives of the components of \(\uvec{v}\text{.}\) For example, if \(\uvec{v} = (v_1,v_2)\) in the plane, then \(-\uvec{v} = (-v_1,-v_2)\text{.}\) In any dimension, we have
\begin{align*}
\uvec{v} \amp= (v_1,v_2,\dotsc,v_n) \amp
\amp\implies \amp
-\uvec{v} \amp= (-v_1,-v_2,\dotsc,-v_n)\text{.}
\end{align*}
This relationship between the components of \(\uvec{v}\) and \(-\uvec{v}\) will later lead to an identity between the negative and a certain scalar multiple of a vector in Subsection 12.3.5.
Aside: A look ahead.
A diagram illustrating a negative vector as the vector that sums with the original to the zero vector. Points \(P\) and \(Q\) are plotted, and directed line segment \(\abray{PQ}\) is drawn, representing a vector labelled \(\uvec{v}\text{.}\) The directed line segment \(\abray{QP}\) is drawn in parallel, representing the negative vector \(- \uvec{v}\text{.}\) If the point \(P\) can be thought of as representing the zero vector, completing the vector addition “triangle” \(\abctriangle{PQP}\text{.}\)
A diagram illustrating a negative vector as an oppositely directed line segment. Collinear points \(R', O, R\) are plotted so that \(O\) is the midpoint between \(R\) and \(R'\text{.}\) Directed line segments \(\abray{OR}\) and \(\abray{OR'}\) are drawn, with \(\abray{OR}\) representing a vector \(\uvec{v}\) and \(\abray{OR'}\) representing the negative vector \(- \uvec{v}\text{.}\)
Remembering that the “natural” position for a vector is with its tail at the origin, it’s useful to visualize negatives as in Figure 12.3.5.(b). That is, the negative of a vector will change positions by the same distance but in the opposite direction.
Subsubsection 12.3.4.2 Subtraction
To subtract vectors, we add to a vector the negative of another. In Figure 12.3.6, the vector arrows forming triangle \(\abctriangle{OPQ}\) represent a vector-addition triangle, so that the vector labelled \(\uvec{u} - \uvec{v}\) is obtained by adding \(\uvec{u}\) and \(-\uvec{v}\text{.}\)
A diagram illustrating vector subtraction as vector addition of a vector and the negative of another. Collinear points \(R', O, R\) are plotted so that \(O\) is the midpoint between \(R\) and \(R'\text{.}\) Directed line segments \(\abray{OR}\) and \(\abray{OR'}\) are drawn, with \(\abray{OR}\) representing a vector \(\uvec{v}\) and \(\abray{OR'}\) representing the negative vector \(- \uvec{v}\text{.}\) External to the line through those three points are plotted two more points \(P\) and \(Q\) so that \(\abcdquad{O P Q R'}\) is a parallelogram. The directed line segment \(\abray{PQ}\) is drawn, again representing the negative vector \(- \uvec{v}\text{,}\) as it is the side of parallelogram \(\abcdquad{O P Q R'}\) opposite \(\abray{OR'}\text{.}\) The directed line segment \(\abray{OP}\) is also drawn, forming another side of parallelogram \(\abcdquad{O P Q R'}\) and representing a vector labelled \(\uvec{u}\text{.}\) Finally, the directed line segment \(\abray{OQ}\) is drawn, forming a diagonal of parallelogram \(\abcdquad{O P Q R'}\) and completing the vector-addition triangle \(\abctriangle{OPQ}\text{.}\) This final directed line segment represents the difference vector \(\uvec{u} - \uvec{v}\text{.}\)
As we explored in Discovery 12.5, we get an interesting pattern if we draw in another copy of the vector labelled \(\uvec{u} - \uvec{v}\) with its initial point at \(R\text{.}\) In Figure 12.3.7, triangle \(\abctriangle{RP'P}\) creates the vector addition pattern \(\uvec{u} + (-\uvec{v}) = \uvec{u} - \uvec{v}\text{.}\) But notice that \(\abctriangle{ORP}\) creates a vector addition pattern starting at \(O\) and ending up at \(P\text{,}\) by \(\uvec{v} + (\uvec{u} - \uvec{v}) = \uvec{u}\text{.}\) So we can think of a difference of two vectors as a vector that runs between the heads of the two vectors in the difference when they share the same initial point. Algebraically, we can think of the \(\uvec{v}\) and \(-\uvec{v}\) cancelling in the expression \(\uvec{v} + (\uvec{u} - \uvec{v})\text{,}\) leaving just \(\uvec{u}\text{.}\)
A diagram using a parallelogram to simultaneously illustrate vector subtraction interpreted as vector addition with a negative vector and as the displacement between the two terminal points of the vectors in the subtraction operation. The diagram from Figure 12.3.6 is replicated, except that the directed line segments \(\abray{OR'}\) and \(\abray{PQ}\text{,}\) both representing the negative vector \(- \uvec{v}\text{,}\) as well the directed line segment \(\abray{OQ}\text{,}\) representing the difference vector \(\uvec{u} - \uvec{v}\text{,}\) all have shafts drawn as dashed lines.
In addition, a point \(P'\) is plotted collinear with \(P\) and \(Q\) so that \(P\) is the midpoint between \(Q\) and \(P'\text{,}\) making \(\abcdquad{ORP'P}\) a parallelogram congruent to \(\abcdquad{OPQR'}\text{.}\) Directed line segment \(\abray{P'P}\) is drawn with a dashed-line shaft, again representing the negative vector \(- \uvec{v}\text{,}\) as it has the same direction and length as \(\abray{PQ}\text{.}\) Directed line segment \(\abray{RP'}\) is also drawn with a dashed-line shaft, and again represents the vector \(\uvec{u}\text{,}\) as it forms the side of parallelogram \(\abcdquad{ORP'P}\) opposite \(\abray{OP}\text{.}\) Directed line segment \(\abray{P'P}\) is drawn with a dashed-line shaft, again representing the vector \(\uvec{u}\) as it forms the side of parallelogram \(\abcdquad{ORP'P}\) opposite \(\abray{OP}\text{.}\) Finally, the directed line segment \(\abray{RP}\) is drawn, forming a diagonal of parallelogram \(\abcdquad{ORP'P}\) and completing the vector-addition triangle \(\abctriangle{RP'P}\text{.}\) This final directed line segment again represents the difference vector \(\uvec{u} - \uvec{v}\text{,}\) but this time is placed so that its initial point \(R\) coincides with the terminal point of the \(\abray{OR}\) representation of \(\uvec{v}\) and its terminal point \(P\) coincides with the terminal point of the \(\abray{OP}\) representation of \(\uvec{u}\text{.}\)
Of course, there are two vectors that run between the heads of \(\uvec{u}\) and \(\uvec{v}\text{,}\) namely \(\uvec{u} - \uvec{v}\) and its negative.
A diagram placing a difference vector as a directed line segment between the terminal points of the two vectors in the subtraction operation, oppositely oriented compared to Figure 12.3.7. Four points \(O, P, P', R\) are plotted to form a parallelogram. The four sides are drawn as directed line segments:
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\(\abray{OP}\) to represent a vector labelled \(\uvec{u}\text{,}\) and the opposite side \(\abray{P'R}\) (drawn with a dashed-line shaft) to represent the negative vector \(- \uvec{v}\text{.}\)
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\(\abray{OR}\) and opposite side \(\abray{PP'}\) (the latter drawn with a dashed-line shaft), both representing a vector labelled \(\uvec{v}\text{.}\)
Finally, the directed line segment \(\abray{PR}\) is drawn to form a diagonal in the parallelogram, running from the terminal point of \(\uvec{u}\) at \(P\) to the terminal point of \(\uvec{v}\) at \(R\text{,}\) and representing the difference vector \(\uvec{v} - \uvec{u}\text{.}\)
Now \(\abctriangle{PP'R}\) creates a vector addition pattern starting at \(P\) and ending up at \(R\text{,}\) so that \(\uvec{v} + (-\uvec{u}) = \uvec{v} - \uvec{u}\text{.}\) But also \(\abctriangle{OPR}\) creates a vector addition pattern starting at \(O\) and ending up at \(R\text{,}\) so that \(\uvec{u} + (\uvec{v} - \uvec{u}) = \uvec{v}\text{.}\) And finally, the fact that \(\uvec{u} - \uvec{v}\) and \(\uvec{v} - \uvec{u}\) both run between \(P\) and \(R\text{,}\) but in opposite directions, verifies geometrically that \(-(\uvec{u} - \uvec{v}) = \uvec{v} - \uvec{u}\text{,}\) as we would expect algebraically.
Subsection 12.3.5 Scalar multiplication
Geometrically, when we scalar multiply a vector we “stretch” or scale its length by the scale factor. (If this scale factor is negative, then we also flip the vector around in the opposite direction.) Here are some examples.
A point is plotted near the centre of the diagram, an from it emanate several parallel directed line segments:
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A directed line segment representing a vector labelled \(\uvec{v}\text{.}\)
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A directed line segment in the same direction as \(\uvec{v}\) but twice as long, representing the scalar multiple vector \(2 \uvec{v}\text{.}\)
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A directed line segment in the same direction as \(\uvec{v}\) but half as long, representing the scalar multiple vector \(\frac{1}{2} \uvec{v}\text{.}\)
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A directed line segment the same length as \(\uvec{v}\) but in the opposite direction, representing the scalar multiple vector \((-1) \uvec{v}\text{.}\)
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A directed line segment in the opposite direction to \(\uvec{v}\) and close to one-and-a-half times as long, representing the scalar multiple vector \((-\sqrt{2}) \uvec{v}\text{.}\)
Notice how each of the vectors in Figure 12.3.9 either points in the same direction as or in the opposite direction to \(\uvec{v}\text{.}\) In particular, they are all parallel to one another. This happens precisely when the vectors are scalar multiples of one another.
Thinking of the vectors in Figure 12.3.9 as vectors in the plane, if we scale \(\uvec{v}\) by a factor of \(2\text{,}\) then our knowledge of similar triangles tells us that the change in both \(x\) and \(y\) must be double.
A directed line segment representing a vector labelled \(\uvec{v}\) runs upwards and rightwards from a point at the left edge of the diagram, just below centre, to a point just above and left of the centre of the diagram. A second directed line segment representing the scalar multiple vector \(2 \uvec{v}\) runs parallel to and from the same initial point as \(\uvec{v}\text{,}\) but is twice as long, reaching to the top of the diagram.
A dashed horizontal line runs from the initial point of \(\uvec{v}\) to the point directly below the terminal point of the scalar multiple \(2 \uvec{v}\text{.}\) A dashed vertical line runs from each of the the terminal points of \(\uvec{v}\) and \(2 \uvec{v}\) down to this horizontal line, creating two right triangles, one inside the other, with the vectors \(\uvec{v}\) and \(2 \uvec{v}\) as the hypotenuses. The horizontal and vertical legs of the smaller right triangle (with \(\uvec{x}\) as hypotenuse) are labelled \(\change{x}\) and \(\change{y}\text{,}\) respectively, and the horizontal and vertical legs of the larger right triangle (with \(2 \uvec{x}\) as hypotenuse) are labelled \(2 \change{x}\) and \(2 \change{y}\text{,}\) respectively.
So if \(\uvec{v} = (v_1,v_2)\text{,}\) then \(2\uvec{v} = (2v_1,2v_2)\text{.}\) This relationship between original vector \(\uvec{v}\) and scaled vector \(k\uvec{v}\) holds in general, in any dimension, and even for negative \(k\text{:}\)
\begin{align*}
\uvec{v} \amp= (v_1,v_2,\dotsc,v_n) \amp
\amp\implies \amp
k\uvec{v} \amp= (kv_1,kv_2,\dotsc,kv_n)\text{.}
\end{align*}
In the case that \(k = -1\text{,}\) we obtain the identity \((-1)\uvec{v} = -\uvec{v}\text{,}\) as promised earlier.
Remark 12.3.11.
It may seem redundant to write \((-1)\uvec{v} = -\uvec{v}\text{,}\) don’t both sides mean the same thing? In terms of the effect on components of \(\uvec{v}\text{,}\) yes they are the same. However, when we explore abstract vectors in Chapter 16, we won’t have components or the geometric notion of “opposite direction” as means of seeing this equality, and so there will initially be a subtle difference between the idea of a vector having an additive negative (so that \(\uvec{v} + (-\uvec{v}) = \zerovec\)) and the operation of scalar multiplying a vector by the particular scalar \(-1\text{.}\)
We can connect scalar multiplication to addition, as we explored in Discovery 12.6. If we add a vector to itself, then the sum vector will be twice as long as the original, so that \(\uvec{v} + \uvec{v} = 2 \uvec{v}\text{.}\)
A diagram of a vector added to itself. Collinear points \(P, Q, R\) are plotted so that \(Q\) is the midpoint of \(P\) and \(R\text{.}\) Directed line segments \(\abray{PQ}\) and \(\abray{QR}\) are plotted, both representing the same vector \(\uvec{v}\text{.}\) The directed line segment \(\abray{PR}\) is drawn in parallel, completing the vector addition “triangle” \(\abctriangle{PQR}\) (which is not actually a triangle as the three points are collinear). This third directed line segment represents both the sum vector \(\uvec{v} + \uvec{v}\) and the scalar multiple vector \(2 \uvec{v}\text{.}\)
Subsection 12.3.6 Vector algebra
We have already discovered a few rules of vector algebra, such as
\begin{align*}
\uvec{v} + \uvec{u} \amp= \uvec{u} + \uvec{v}, \amp
-(\uvec{v} - \uvec{u}) \amp= \uvec{u} - \uvec{v}, \amp
\uvec{v} + \uvec{v} \amp= 2\uvec{v}, \amp
(-1)\uvec{v} = -\uvec{v}.
\end{align*}
In Discovery 12.7, we explored a version of the distributive rule \(k (\uvec{u} + \uvec{v}) = k \uvec{u} + k \uvec{v}\) in the case \(k=2\text{.}\)
Points \(P, Q', R'\) are arranged in a triangular configuration. Points \(Q, S, R\) are plotted so that \(Q\) is the midpoint between \(P\) and \(Q'\text{,}\) \(S\) is the midpoint between \(Q'\) and \(R'\text{,}\) and \(R\) is the midpoint between \(P\) and \(R'\text{.}\) Directed line segments \(\abray{PQ}\text{,}\) \(\abray{QR}\text{,}\) and \(\abray{PR}\) are drawn to form a vector-addition triangle, with \(\abray{PQ}\) representing a vector labelled \(\uvec{u}\text{,}\) \(\abray{QR}\) representing a vector labelled \(\uvec{v}\text{,}\) and \(\abray{PR}\) representing the sum vector \(\uvec{u} + \uvec{v}\text{.}\) An identical vector-addition triangle is formed by directed line segments \(\abray{RS}\text{,}\) \(\abray{SR}'\text{,}\) and \(\abray{RR'}\text{,}\) with \(\abray{RS}\) again representing \(\uvec{u}\text{,}\) \(\abray{SR'}\) again representing \(\uvec{v}\text{,}\) and \(\abray{RR'}\) again representing the sum vector \(\uvec{u} + \uvec{v}\text{.}\) Directed line segments \(\abray{QQ'}\) and \(\abray{Q'S}\) are drawn to complete the parallelogram \(\abcdquad{QQ'SR}\text{,}\) with \(\abray{QQ'}\) again representing \(\uvec{u}\) and \(\abray{Q'S}\) again representing \(\uvec{v}\text{.}\)
Directed line segments \(\abray{PQ'}\text{,}\) \(\abray{Q'R'}\text{,}\) and \(\abray{PR'}\) are drawn to represent the scalar multiple vectors \(2 \uvec{u}\text{,}\) \(2 \uvec{v}\text{,}\) and \(2 (\uvec{u} + \uvec{v})\text{:}\)
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Points \(P, Q, Q'\) are collinear, with \(Q\) the midpoint between the other two points, and directed segments \(\abray{PQ}\) and \(\abray{QQ'}\) both represent the same vector \(\uvec{u}\text{.}\) Hence directed segment \(\abray{PQ'}\) represents the scalar multiple vector \(2 \uvec{u}\text{.}\)
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Points \(Q', S, R'\) are collinear, with \(S\) the midpoint between the other two points, and directed segments \(\abray{Q'S}\) and \(\abray{SR'}\) both represent the same vector \(\uvec{v}\text{.}\) Hence directed segment \(\abray{Q'R'}\) represents the scalar multiple vector \(2 \uvec{v}\text{.}\)
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Points \(P, R, R'\) are collinear, with \(R\) the midpoint between the other two points, and directed segments \(\abray{PR}\) and \(\abray{RR'}\) both represent the sum vector \(\uvec{u} + \uvec{v}\text{.}\) Hence directed segment \(\abray{PR'}\) represents the scalar multiple vector \(2 (\uvec{u} + \uvec{v})\text{.}\)
With \(\abray{PQ'}\) and \(\abray{Q'R'}\) representing vectors \(2 \uvec{u}\) and \(2 \uvec{v}\text{,}\) triangle \(\abctriangle{PQ'R'}\) now forms a vector-addition triangle, so that directed segment \(\abray{PR'}\) represents sum vector \(2 \uvec{u} + 2 \uvec{v}\) (as well as representing \(2 (\uvec{u} + \uvec{v})\)).
We will provide more rules of vector algebra as Proposition 12.5.1 in Subsection 12.5.1. In Discovery 12.12, we decided that the algebra of vectors is the same as the algebra of column matrices (which we have already been referring to as column vectors), so we should be able to anticipate a number of the vector algebra rules that will appear in that proposition.
Aside: A look ahead.
Warning 12.3.14.
There is no multiplication operation for vectors!
Algebraically, vectors in \(\R^n\) are the same as column matrices, and you cannot multiply two column matrices together because their sizes do not match up (except in \(\R^1\text{,}\) but let’s ignore that for now). This also means that you cannot square a vector, you cannot square-root a vector, you cannot invert a vector, and you cannot divide by a vector. Do not try to use any of these operations in vector algebra! In Chapters 13–14, we will encounter some operations tied to the geometry of vectors that we will call “vector products” and for which we will use multiplication-like notation, but they will be for very specific geometric purposes and do not really correspond to our idea of multiplication in the algebra of numbers.
Subsection 12.3.7 The standard basis vectors
In Discovery 12.8, we encountered two very special vectors in the plane, \(\uvec{e}_1 = (1,0)\) and \(\uvec{e}_2 = (0,1)\text{.}\) These two vectors could be considered the fundamental changes in position in the plane — \(\uvec{e}_1\) represents a change by one unit right, and \(\uvec{e}_2\) represents a change by one unit up.
The standard basis vectors \(\uvec{e}_1 = (1,0)\) and \(\uvec{e}_2 = (0,1)\) are drawn as directed line segments on a set of \(xy\)-axes, with initial points at the origin and terminal points at \((1,0)\) and \((0,1)\text{,}\) respectively, so that \(\uvec{e}_1\) lies along the \(x\)-axis and \(\uvec{e}_2\) lies along the \(y\)-axis.
Any change in position can be built out of these two fundamental changes in position. Using the example in Discovery 12.8, the vector \(\uvec{v} = (5,2)\) represents a change in position by \(5\) units right and \(2\) units up. We can achieve the “\(5\) units right” part with \(5 \uvec{e}_1 = (5,0)\) and the “\(2\) units up part” with \(2 \uvec{e}_2 = (0,2)\text{.}\) To get the total change in position represented by \(\uvec{v}\text{,}\) we can combine these two building blocks in the linear combination \(\uvec{v} = 5 \uvec{e}_1 + 2 \uvec{e}_2\text{.}\)
Diagram illustrating a vector decomposition as a linear combination of standard basis vectors \(\uvec{e}_1\) and \(\uvec{e}_2\) in \(\R^2\text{.}\) A right triangle appears, with a slant hypotenuse and the right angle between a horizontal leg and a vertical leg. The hypotenuse is a directed line segment representing the vector \(\uvec{v} = (5,2)\text{,}\) so that its terminal point is \(5\) units to the right and \(2\) units up from its initial point. The horizontal leg is a chain of five directed line segments, each representing the standard basis vector \(\uvec{e}_1\text{,}\) where the first in the chain shares its initial point with \(\uvec{v}\text{,}\) and each subsequent segment in the chain as its initial point at the terminal point of the previous. Similarly, the vertical leg is a chain of two directed line segments, each representing the standard basis vector \(\uvec{e}_2\text{,}\) where the first has its initial point at the terminal point of the last segment in the horizontal chain, and the second has its initial point at the terminal point of the first and shares its terminal point with \(\uvec{v}\text{.}\)
As you can imagine, every vector in the plane can be decomposed into a linear combination of \(\uvec{e}_1\) and \(\uvec{e}_2\) in this manner: for \(\uvec{v} = (v_1,v_2)\text{,}\) we have \(\uvec{v} = v_1\,\uvec{e}_1 + v_2\,\uvec{e}_2\text{.}\) For this reason, the two vectors \(\uvec{e}_1,\uvec{e}_2\) together are called the standard basis vectors in \(\R^2\text{,}\) as they form a basis from which every other vector can be constructed. To use an analogy with chemistry, these two vectors are the basic atoms of \(\R^2\text{,}\) and every other vector in \(\R^2\) is a molecule built out of a specific combination of these atoms. Since there are only two fundamental directions in \(\R^2\) (right/left and up/down), it is not surprising that we need only two basis vectors to represent all possible directions. This is the reason we call vectors in \(\R^2\) two-dimensional vectors.
Aside: Note.
In \(\R^3\text{,}\) there are three fundamental directions, two horizontal and one vertical. We might use navigational terminology for the two horizontal directions and describe them as north/south and east/west, and then we can refer to the vertical direction as up/down. So we need three standard basis vectors in \(\R^3\text{,}\)
\begin{align*}
\uvec{e}_1 \amp= (1,0,0), \amp
\uvec{e}_2 \amp= (0,1,0), \amp
\uvec{e}_3 \amp= (0,0,1),
\end{align*}
which we can visualize as in Figure 12.3.17.
The standard basis vectors \(\uvec{e}_1 = (1,0,0)\text{,}\) \(\uvec{e}_2 = (0,1,0)\text{,}\) and \(\uvec{e}_3 = (0,0,1)\) are drawn as directed line segments on a set of \(xyz\)-axes, with initial points at the origin and terminal points at \((1,0,0)\text{,}\) \((0,1,0)\text{,}\) and \((0,0,1)\text{,}\) respectively, so that \(\uvec{e}_1\) lies along the \(x\)-axis, \(\uvec{e}_2\) lies along the \(y\)-axis, and \(\uvec{e}_3\) lies along the \(z\)-axis.
As before, any vector in \(\R^3\) can be decomposed as a linear combination of these three fundamental vectors. For example, the vector \((1,-1,2)\) decomposes as
\begin{equation*}
(1,-1,2) = 1\uvec{e}_1 + (-1)\uvec{e}_2 + 2\uvec{e}_3 \text{.}
\end{equation*}
And we can repeat all this in \(\R^n\) for any value of \(n\text{,}\) where there are \(n\) standard basis vectors,
\begin{align*}
\uvec{e}_1 \amp= (1,0,0,\dotsc,0), \amp
\uvec{e}_2 \amp= (0,1,0,\dotsc,0), \amp
\uvec{e}_n \amp= (0,0,\dotsc,0,1).
\end{align*}
In fact, given a vector \(\uvec{v} = (v_1,v_2,\dotsc,v_n)\) in \(\R^n\) (whether \(n=2\) or \(n=3\) or higher), when we try to decompose
\begin{equation*}
\uvec{v} =
\fillinmath{XX}\uvec{e}_1
+ \fillinmath{XX}\uvec{e}_2
+ \dotsb
+ \fillinmath{XX}\uvec{e}_n\text{,}
\end{equation*}
we find that there is only one unique combination of scalar values that can fill in the blanks and create an equality between \(\uvec{v}\) on the left and the linear combination on the right:
\begin{equation*}
\uvec{v} =
v_1\uvec{e}_1
+ v_2\uvec{e}_2
\dotsb
+ v_n\uvec{e}_n\text{.}
\end{equation*}
