66.16: Vector Arithmetic

A vector A may be written in cartesian (rectangular) form as

\[\mathbf{A}=A_{x} \mathbf{i}+A_{y} \mathbf{j}+A_{z} \mathbf{k}\]

where \(\mathbf{i}\) is a unit vector (a vector of magnitude 1) in the \(x\) direction, \(\mathbf{j}\) is a unit vector in the \(y\) direction, and \(\mathbf{k}\) is a unit vector in the \(z\) direction. \(A_{x}, A_{y}\), and \(A_{z}\) are called the \(x, y\), and \(z\) components (respectively) of vector \(\mathbf{A}\), and are the projections of the vector onto those axes.

The magnitude ("length") of vector \(\mathbf{A}\) is

\[\\mathbf{A}|=A=\sqrt{A_{x}^{2}+A_{y}^{2}+A_{z}^{2}}\]

For example, if \(\mathbf{A}=3 \mathbf{i}+5 \mathbf{j}+2 \mathbf{k}\), then \(|\mathbf{A}|=A=\sqrt{3^{2}+5^{2}+2^{2}}=\sqrt{38}\).

In two dimensions, a vector has no \(\mathbf{k}\) component: \(\mathbf{A}=A_{x} \mathbf{i}+A_{y} \mathbf{j}\).

Addition and Subtraction

To add two vectors, you add their components. Writing a second vector as \(\mathbf{B}=B_{x} \mathbf{i}+B_{y} \mathbf{j}+B_{z} \mathbf{k}\), we have

\[\mathbf{A}+\mathbf{B}=\left(A_{x}+B_{x}\right) \mathbf{i}+\left(A_{y}+B_{y}\right) \mathbf{j}+\left(A_{z}+B_{z}\right) \mathbf{k} .\]

For example, if \(\mathbf{A}=3 \mathbf{i}+5 \mathbf{j}+2 \mathbf{k}\) and \(\mathbf{B}=2 \mathbf{i}-\mathbf{j}+4 \mathbf{k}\), then \(\mathbf{A}+\mathbf{B}=5 \mathbf{i}+4 \mathbf{j}+6 \mathbf{k}\).

Subtraction of vectors is defined similarly:

\[\mathbf{A}-\mathbf{B}=\left(A_{x}-B_{x}\right) \mathbf{i}+\left(A_{y}-B_{y}\right) \mathbf{j}+\left(A_{z}-B_{z}\right) \mathbf{k} .\]

For example, if \(\mathbf{A}=3 \mathbf{i}+5 \mathbf{j}+2 \mathbf{k}\) and \(\mathbf{B}=2 \mathbf{i}-\mathbf{j}+4 \mathbf{k}\), then \(\mathbf{A}-\mathbf{B}=\mathbf{i}+6 \mathbf{j}-2 \mathbf{k}\).

Scalar Multiplication

To multiply a vector by a scalar, just multiply each component by the scalar. Thus if \(c\) is a scalar, then

\[c \mathbf{A}=c A_{x} \mathbf{i}+c A_{y} \mathbf{j}+c A_{z} \mathbf{k} .\]

For example, if \(\mathbf{A}=3 \mathbf{i}+5 \mathbf{j}+2 \mathbf{k}\), then \(7 \mathbf{A}=21 \mathbf{i}+35 \mathbf{j}+14 \mathbf{k}\).

Dot Product

It is possible to multiply a vector by another vector, but there is more than one kind of multiplication between vectors. One type of vector multiplication is called the dot product, in which a vector is multiplied by another vector to give a scalar result. The dot product (written with a dot operator, as in \(\mathbf{A} \cdot \mathbf{B}\) ) is

\[\mathbf{A} \cdot \mathbf{B}=A B \cos \theta=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z},\]

where \(\theta\) is the angle between vectors \(\mathbf{A}\) and \(\mathbf{B}\). For example, if \(\mathbf{A}=3 \mathbf{i}+5 \mathbf{j}+2 \mathbf{k}\) and \(\mathbf{B}=2 \mathbf{i}-\mathbf{j}+4 \mathbf{k}\), then \(\mathbf{A} \cdot \mathbf{B}=6-5+8=9\).

The dot product can be used to find the angle between two vectors. To do this, we solve Eq. (P.6) for \(\theta\) and find \(\cos \theta=\mathbf{A} \cdot \mathbf{B} /(A B)\). Applying this to the previous example, we get \(A=\sqrt{38}\) and \(B=\sqrt{21}\), so \(\cos \theta=9 /(\sqrt{38} \sqrt{21})\), and thus \(\theta=71.4^{\circ}\).

An immediate consequence of Eq. (P.6) is that two vectors are perpendicular if and only if their dot product is zero.

Cross Product

Figure \(\PageIndex{1}\): The vector cross product \(\mathbf{A} \times \mathbf{B}\) is perpendicular to the plane of \(\mathbf{A}\) and \(\mathbf{B}\), and in the right-hand sense. (Credit: “Connected Curriculum Project”, Duke University.)

Another kind of multiplication between vectors, called the cross product, involves multiplying one vector by another and giving another vector as a result. The cross product is written with a cross operator, as in \(\mathbf{A} \times \mathbf{B}\). It is defined by

\[
\begin{aligned}
\mathbf{A} \times \mathbf{B} & =(A B \sin \theta) \mathbf{u} \\
& =\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
A_{x} & A_{y} & A_{z} \\
B_{x} & B_{y} & B_{z}
\end{array}\right| \\
& =\left(A_{y} B_{z}-A_{z} B_{y}\right) \mathbf{i}-\left(A_{x} B_{z}-A_{z} B_{x}\right) \mathbf{j}+\left(A_{x} B_{y}-A_{y} B_{x}\right) \mathbf{k},
\end{aligned}
\]

where again \(\theta\) is the angle between the vectors, and \(\mathbf{u}\) is a unit vector pointing in a direction perpendicular to the plane containing \(\mathbf{A}\) and \(\mathbf{B}\), in a right-hand sense: if you curl the fingers of your right hand from \(\mathbf{A}\) into \(\mathbf{B}\), then the thumb of your right hand points in the direction of \(\mathbf{A} \times \mathbf{B}\) (Fig. P.1). As an example, if \(\mathbf{A}=3 \mathbf{i}+5 \mathbf{j}+2 \mathbf{k}\) and \(\mathbf{B}=2 \mathbf{i}-\mathbf{j}+4 \mathbf{k}\), then \(\mathbf{A} \times \mathbf{B}=(20-(-2)) \mathbf{i}-(12-4) \mathbf{j}+(-3-10) \mathbf{k}=22 \mathbf{i}-8 \mathbf{j}-13 \mathbf{k}\).

Rectangular and Polar Forms

A two-dimensional vector may be written in either rectangular form \(\mathbf{A}=A_{x} \mathbf{i}+A_{y} \mathbf{j}\) described earlier, or in polar form \(\mathbf{A}=A \angle \theta\), where \(A\) is the vector magnitude, and \(\theta\) is the direction measured counterclockwise from the \(+x\) axis. To convert from polar form to rectangular form, one finds

\[
\begin{aligned}
& A_{x}=A \cos \theta \\
& A_{y}=A \sin \theta
\end{aligned}
\]

Inverting these equations gives the expressions for converting from rectangular form to polar form:

\[
\begin{aligned}
A & =\sqrt{A_{x}^{2}+A_{y}^{2}} \\
\tan \theta & =\frac{A_{y}}{A_{x}}
\end{aligned}
\]