Elementary Vector Analysis

$\newcommand{\vecb}[1]{{\bf #1}} \newcommand{\ihat}{\hat{\vecb{i}}} \newcommand{\jhat}{\hat{\vecb{j}}} \newcommand{\khat}{\hat{\vecb{k}}}$ In order to measure many physical quantities, such as force or velocity, we need to determine both a magnitude and a direction. Such quantities are conveniently represented as vectors.

The direction of a vector $\vecb{v}$ in 3-space is specified by its components in the $x$, $y$, and $z$ directions, respectively: $$(x,y,z) \quad {\small\textrm{or}} \quad x\ihat + y\jhat + z\khat,$$
where $\ihat$, $\jhat$, and $\khat$ are the coordinate vectors along the $x$, $y$, and $z$-axes.
 $\ihat=(1,0,0)$ $\jhat=(0,1,0)$ $\khat=(0,0,1)$

The magnitude of a vector $\vecb{v}=(x,y,z)$, also called its length or norm, is given by $$\left\| \vecb{v} \right\| = \sqrt{x^{2}+y^{2}+z^{2}}.$$

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Notes

• Vectors can be defined in any number of dimensions, though we focus here only on 3-space.

• When drawing a vector in 3-space, where you position the vector is unimportant; the vector's essential properties are just its magnitude and its direction. Two vectors are equal if and only if corresponding components are equal.

• A vector of norm 1 is called a unit vector. The coordinate vectors are examples of unit vectors.

• The zero vector, $\vecb{0} = (0,0,0)$, is the only vector with magnitude 0.

Basic Operations on Vectors

 To add or subtract vectors $\vecb{u} = (u_{1},u_{2},u_{3})$ and $\vecb{v} = (v_{1},v_{2},v_{3})$, add or subract the corresponding coordinates: \begin{eqnarray*} \vecb{u}+\vecb{v} &= & (u_{1}+v_{1},u_{2}+v_{2},u_{3}+v_{3}) \\ \vecb{u}-\vecb{v} &= & (u_{1}-v_{1},u_{2}-v_{2},u_{3}-v_{3}). \end{eqnarray*} $\quad$

To multiply vector $\vecb{u}$ by a scalar $k$, multiply each coordinate of $\vecb{u}$ by $k$: $$k\vecb{u}=(ku_{1},ku_{2},ku_{3}).$$

Example

The vector $\vecb{v}= (2,1,-2) = 2\ihat + \jhat -2\khat$ has magnitude $$\left\| \vecb{v} \right\| = \sqrt{2^2 +1^2 -(-2)^2} = 3.$$

$\quad$

Thus, the vector $\frac{1}{3}\vecb{v} = \left(\frac{2}{3},\frac{1}{3},\frac{-2}{3}\right)$ is a unit vector in the same direction as $\vecb{v}$.

In general, for $\vecb{v} \not= \vecb{0}$, we can scale (or normalize) $\vecb{v}$ to the unit vector $\frac{\vecb{v}}{\left\| \vecb{v} \right\|}$ pointing in the same direction as $\vecb{v}$.

Dot Product

Let $\vecb{u} = (u_{1},u_{2},u_{3})$ and $\vecb{v} = (v_{1},v_{2},v_{3})$. The dot product $\vecb{u} \cdot \vecb{v}$ (also called the scalar product or Euclidean inner product) of $\vecb{u}$ and $\vecb{v}$ is defined in two distinct (though equivalent) ways:

 \begin{eqnarray*} \vecb{u} \cdot \vecb{v} & = & u_1v_1+u_2v_2+u_3v_3 \\ & = & \left\{ \begin{array}{cl} \left\| \vecb{u} \right\| \left\| \vecb{v} \right\| \cos \theta & {\small\textrm{if }} \vecb{u} \not= \vecb{0}, \vecb{v} \not= \vecb{0}\\ 0 & {\small\textrm{if }} \vecb{u} = \vecb{0} {\small\textrm{ or }} \vecb{v} = \vecb{0}\\ \end{array} \right.\\ & & \qquad{\small\textrm{where }} 0 \le \theta \le \pi {\small\textrm{ is the angle between }} \vecb{u} {\small\textrm{ and }} \vecb{v} . \end{eqnarray*} $\quad$

Properties of the Dot Product

• $\vecb{u} \cdot \vecb{v} = \vecb{v} \cdot \vecb{u}$

• $\vecb{u} \cdot (\vecb{v} + \vecb{w}) = (\vecb{u} \cdot \vecb{v}) + (\vecb{u} \cdot \vecb{w})$

• $\vecb{u} \cdot \vecb{u} = \left\| \vecb{u} \right\|^{2}$

See if you can verify each of these!

Example

If $\vecb{u}=(1,-2,2)$ and $\vecb{v}=(-4,0,2)$, then $\begin{array}{rcl} \vecb{u} \cdot \vecb{v} &=& (1)(-4)+(-2)(0)+(2)(2)\\ &=&-1+0+4\\ &=&0. \end{array}$

Using the second definition of the dot product with $\left\| \vecb{u} \right\|=3$ and $\left\| \vecb{v} \right\|=2\sqrt{5}$, $$\vecb{u} \cdot \vecb{v} = 0 = 6\sqrt{5}\cos\theta$$ so $\cos\theta=0$, yielding $\theta = \frac{\pi}{2}$.

Though we might not have guessed it, $\vecb{u}$ and $\vecb{v}$ are perpendicular to each other!

In general,

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$\qquad$ Two non-zero vectors $\vecb{u}$ and $\vecb{v}$ are perpendicular (or orthogonal) if and only if $\vecb{u} \cdot \vecb{v} = 0$.

Proof

Projection of a Vector

It is often useful to resolve a vector $\vecb{v}$ into the sum of vector components parallel and perpendicular to a vector $\vecb{u}$.

Consider first the parallel component, which is called the projection of $\vecb{v}$ onto $\vecb{u}$. This projection should be in the direction of $\vecb{u}$ and should have magnitude $\left\| \vecb{v} \right\|\cos\theta$, where $0 \le \theta \le \pi$ is the angle between $\vecb{u}$ and $\vecb{v}$. Let's normalize $\vecb{u}$ to $\frac{\vecb{u}}{\left\| \vecb{u} \right\|}$ and then scale this by the magnitude $\left\| \vecb{v} \right\|\cos\theta$:

 $\qquad$ projection of $\vecb{v}$ onto $\vecb{u}$ $\begin{array}{rl} = & \left(\left\| \vecb{v} \right\|\cos\theta\right)\frac{\vecb{u}}{\left\| \vecb{u} \right\|} \\ = & \frac{\left\| \vecb{v} \right\|\left\| \vecb{u}\right\|\cos\theta}{\left\| \vecb{u} \right\|^{2}}\vecb{u}\\ = & \frac{\vecb{v} \cdot \vecb{u}}{\left\| \vecb{u} \right\|^{2}}\vecb{u}. \end{array}$

$\quad$

The perpendicular vector component of $\vecb{v}$ is then just the difference between $\vecb{v}$ and the projection of $\vecb{v}$ onto $\vecb{u}$.

In summary,
 $\qquad$ projection of $\vecb{v}$ onto $\vecb{u}$ $\quad = \quad$ $\frac{\vecb{v} \cdot \vecb{u}}{\left\| \vecb{u} \right\|^{2}}\vecb{u}$ $\qquad$ vector component of $\vecb{v}$ perpendicular to $\vecb{u}$ $\quad = \quad$ $\vecb{v}-\frac{\vecb{v} \cdot \vecb{u}}{\left\| \vecb{u} \right\|^{2}}\vecb{u}.$

Cross Product

Let $\vecb{u} = (u_{1},u_{2},u_{3})$ and $\vecb{v} = (v_{1},v_{2},v_{3})$. The cross product $\vecb{u} \times \vecb{v}$ yields a vector perpendicular to both $\vecb{u}$ and $\vecb{v}$ with direction determined by the right-hand rule. Specifically, $$\vecb{u} \times \vecb{v} = (u_{2}v_3-u_3v_2)\ihat - (u_1v_3-u_3v_1)\jhat + (u_1v_2-u_2v_1)\khat.$$ It can also be shown that $$\left\| \vecb{u} \times \vecb{v} \right\| = \left\| \vecb{u} \right\|\left\| \vecb{v} \right\|\sin\theta \quad {\small\textrm{for }} \vecb{u} \not= \vecb{0}, \quad \vecb{v} \not= \vecb{0}$$
 where $0 \le \theta \le \pi$ is the angle between $\vecb{u}$ and $\vecb{v}$.

Proof

Thus, the magnitude $\left\| \vecb{u} \times \vecb{v} \right\|$ gives the area of the parallelogram formed by $\vecb{u}$ and $\vecb{v}$.

As implied by the geometric interpretation,

 Non zero vectors $\vecb{u}$ and $\vecb{v}$ are parallel if and only if $\vecb{u} \times \vecb{v}=\vecb{0}$.

Proof

Properties of the Cross Product

• $\vecb{u} \times \vecb{v} = - \left( \vecb{v} \times \vecb{u} \right)$

• $\vecb{u} \times \left( \vecb{v} + \vecb{w} \right) = \left(\vecb{u} \times \vecb{v} \right) + \left( \vecb{u} \times \vecb{w} \right)$

• $\vecb{u} \times \vecb{u} = \vecb{0}$

Again, see if you can verify each of these.

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Connections between the Dot Product and Cross Product

Key Concepts

Let $\vecb{u} = (u_{1},u_{2},u_{3})$ and $\vecb{v} = (v_{1},v_{2},v_{3})$.
• Basic Operations, Norm of a vector

\begin{eqnarray*} \vecb{u}+\vecb{v} &= & (u_{1}+v_{1},u_{2}+v_{2},u_{3}+v_{3}) \\ \vecb{u}-\vecb{v} &= & (u_{1}-v_{1},u_{2}-v_{2},u_{3}-v_{3}) \\ k\vecb{u} & = & (ku_{1},ku_{2},ku_{3}) \\ \left\| \vecb{v} \right\| & = & \sqrt{x^{2}+y^{2}+z^{2}} \end{eqnarray*}

• Dot Product

\begin{eqnarray*} \vecb{u} \cdot \vecb{v} & = & u_1v_1+u_2v_2+u_3v_3 \\ & = & \left\{ \begin{array}{cl} \left\| \vecb{u} \right\| \left\| \vecb{v} \right\| \cos \theta & {\small\textrm{if }} \vecb{u} \not= \vecb{0}, \vecb{v} \not= \vecb{0}\\ 0 & {\small\textrm{if }} \vecb{u} = \vecb{0} {\small\textrm{ or }} \vecb{v} = \vecb{0}\\ \end{array} \right.\\ & & \qquad{\small\textrm{where }} 0 \le \theta \le \pi {\small\textrm{ is the angle between }} \vecb{u} {\small\textrm{ and }} \vecb{v} \end{eqnarray*}

$\qquad$ For $\vecb{u} \not= \vecb{0}, \quad \vecb{v} \not= \vecb{0}$,

$\qquad\qquad \vecb{u} \cdot \vecb{v} = 0$ if and only if $\vecb{u}$ is orthogonal to $\vecb{v}$.

• Projection of a Vector

 $\qquad$ projection of $\vecb{v}$ onto $\vecb{u}$ $\quad = \quad$ $\frac{\vecb{v} \cdot \vecb{u}}{\left\| \vecb{u} \right\|^{2}}\vecb{u}$ $\qquad$ vector component of $\vecb{v}$ perpendicular to $\vecb{u}$ $\quad = \quad$ $\vecb{v}-\frac{\vecb{v} \cdot \vecb{u}}{\left\| \vecb{u} \right\|^{2}}\vecb{u}.$

• Cross Product

\begin{eqnarray*} \vecb{u} \times \vecb{v} & = & (u_{2}v_3-u_3v_2)\ihat - (u_1v_3-u_3v_1)\jhat + (u_1v_2-u_2v_1)\khat\\ \left\| \vecb{u} \times \vecb{v} \right\| & = & \left\| \vecb{u} \right\|\left\| \vecb{v} \right\|\sin\theta \quad {\small\textrm{for }} \vecb{u} \not= \vecb{0}, \quad \vecb{v} \not= \vecb{0} \end{eqnarray*}  where $0 \le \theta \le \pi$ is the angle between $\vecb{u}$ and $\vecb{v}$.