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Vector arithmetic (addition, subtraction, and multiplication by a scalar) is performed componentwise. Table 1.2.1 illustrates these three operations for vectors in the Cartesian plane, with the obvious generalization to vectors in higher dimensions.
Operation

Algorithm

Scalar Multiplication

$c\left[\begin{array}{c}{a}_{1}\\ {a}_{2}\end{array}\right]\=\left[\begin{array}{c}c{a}_{1}\\ c{a}_{2}\end{array}\right]$ ${}$

Addition

$\left[\begin{array}{c}{a}_{1}\\ {a}_{2}\end{array}\right]\+\left[\begin{array}{c}{b}_{1}\\ {b}_{2}\end{array}\right]\=\left[\begin{array}{c}{a}_{1}\+{b}_{1}\\ {a}_{2}\+{b}_{2}\end{array}\right]$

Subtraction

$\left[\begin{array}{c}{a}_{1}\\ {a}_{2}\end{array}\right]\left[\begin{array}{c}{b}_{1}\\ {b}_{2}\end{array}\right]equals;\left[\begin{array}{c}{a}_{1}{b}_{1}\\ {a}_{2}{b}_{2}\end{array}\right]$

Table 1.2.1 Componentwise vector arithmetic



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The following calculation shows that scalar multiplication is nothing more than a scaling of the length of a vector.
$\u2225c\left[\begin{array}{c}{a}_{1}\\ {a}_{2}\end{array}\right]\u2225$ = $\u2225\left[\begin{array}{c}c{a}_{1}\\ c{a}_{2}\end{array}\right]\u2225$ = $\sqrt{{\left(c{a}_{1}\right)}^{2}plus;{\left(c{a}_{2}\right)}^{2}}equals;\sqrt{{c}^{2}\left({a}_{1}^{2}plus;{a}_{2}^{2}\right)}equals;\leftc\right\sqrt{{a}_{1}^{2}plus;{a}_{2}^{2}}$ = $\leftc\right$ $\parallel \left[\begin{array}{c}{a}_{1}\\ {a}_{2}\end{array}\right]\parallel$
In other words, $\u2225c\mathbf{A}\u2225$ = $\leftc\right$$\u2225\mathbf{A}\u2225$. If $c<0$ then multiplication by $c$ is both a scaling and a reversal of direction.
Consequently, the vector $\mathbf{A}\/\u2225\mathbf{A}\u2225$ has magnitude (length) 1; dividing a vector by its magnitude to produce a unit vector with the same direction is said to normalize the vector.
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Figures 1.2.12 provide graphical interpretations of vector addition. In Figure 1.2.1, vector A (thick red) is added to vector B (thick green) to produce the resultant $\mathbf{C}\=\mathbf{A}\+\mathbf{B}$ (black). The resultant is the diagonal of the parallelogram formed by adjoining to A and B, the vectors $\mathbf{A}\prime$ and $\mathbf{B}\prime$, respectively the translate of A to the tip of B, and and the translate of B to the tip of A. The translates are drawn with thinner arrows.
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Figure 1.2.1 Parallelogram of vector addition





Figure 1.2.2 Triangle for vector addition






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Figure 1.2.2 is simply one of the two triangles formed in the parallelogram of addition (Figure 1.2.1) by the resultant (black diagonal). Note that Figure 1.2.2 translates B to the tip of A, and the original location of vector B is never displayed. However, the translation of B is considered to be B itself. In this interpretation of vector addition, B is translated to the tip of A and the third side of the "triangle" so formed is the sum of A and B.
Figure 1.2.2 implies $\mathbf{C}\=\mathbf{A}\+\mathbf{B}$, from which follows $\mathbf{B}\=\mathbf{C}\mathbf{A}$. In other words, the vector from the tip of A to the tip of C is $\mathit{C}\mathit{A}$. It is exceedingly useful to remember this "reversal", namely, that the vector from P to Q is $\mathbf{Q}\mathbf{P}$.
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Unit vectors along the three coordinate axes are typically given the names i, j, and k. , but other texts, especially physics texts, tend to use the notation $\stackrel{\^}{\mathbf{i}}$, $\stackrel{\^}{\mathbf{j}}$, and $\stackrel{\^}{\mathbf{k}}$. Representations of these unit basis vectors are given in Table 1.2.2. Of course, in the plane there are only two such vectors, i and j, and then they have only the first two components of the vectors in Table 1.2.2.
$\mathbf{i}\=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$

$\mathbf{j}\=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$

$\mathbf{k}\=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$

Table 1.2.2 Unit basis vectors along coordinate axes



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Since vector addition is componentwise, the vector whose components are $a\,b$, and $c$, can be represented as the sum in Table 1.2.3.
$\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$ = $a\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]\+b\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]\+c\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$ = $a\mathbf{i}plus;b\mathbf{j}plus;c\mathbf{k}$ ${}$

Table 1.2.3 Representation of a vector in terms of the unit basis vectors



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Only Maple's Physics package implements $\left\{\stackrel{\^}{\mathbf{i}}\,\stackrel{\^}{\mathbf{j}}\,\stackrel{\^}{\mathbf{k}}\right\}$ as the unit basis vectors. Hence, throughout the remainder of this Study Guide, the representation shown at the right in Table 1.2.3 will be for notational purposes only. No meaningful computations can be done in Maple (outside of the Physics package) with such symbols.