OBJECTIVES

When you finish this section, you should be able to:

1. Find the dot product of two vectors (p. 715)
2. Find the angle between two vectors (p. 716)
3. Determine whether two vectors are orthogonal (p. 718)
4. Find a vector in space from its magnitude and direction (p. 718)
5. Find the projection of a vector (p. 719)
6. Compute work (p. 720)

DEFINITION Dot Product

If $v=v_{1} {\bf i}+v_{2} \mathbf{j}$ and $\mathbf{w}=w_{1} \mathbf{i}+w_{2}\mathbf{j}$ are two vectors in the plane, the dot product $\mathbf{v\,{\cdot}\, w}$ is defined as $$\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{v\,{\cdot}\, w}=v_{1}w_{1}+v_{2}w_{2} }$$

Similarly, if $\mathbf{v}=v_{1} \mathbf{i}+v_{2} \mathbf{j}+v_{3}\mathbf{k}$ and $\mathbf{w}=w_{1} \mathbf{i}+w_{2} \mathbf{j}+w_{3}\mathbf{k}$ are two vectors in space, the dot product $\mathbf{v\,{\cdot}\, w}$ is defined as $$\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{v\,{\cdot}\, w}=v_{1}w_{1}+v_{2}w_{2}+v_{3}w_{3} }$$

NOTE

The dot product of two vectors is not a vector. It is a scalar. The dot product is sometimes called the scalar product.

(a) If $\mathbf{v}=2\mathbf{i}-3\mathbf{j}$ and $\mathbf{w}= \mathbf{i}+\mathbf{j}$, then $$\begin{equation*} \begin{array}{rcl@{\quad}rcl} \mathbf{v}\,{\cdot}\, \mathbf{w} &=& (2)(1)+(-3)(1)=2-3=-1 & \mathbf{w\,{\cdot}\, v} &=& (1) (2) +(1) (-3) =2-3=-1 \\ \mathbf{v\,{\cdot}\, v} &=& 2^{2}+ (-3) ^{2}=4+9=13 & \mathbf{w\,{\cdot}\, w} &=& 1^{2}+1^{2}=2 \end{array} \end{equation*}$$

(b) If $\mathbf{v}=2\mathbf{i}-\mathbf{j}+\mathbf{k}$ and $\mathbf{w}=4\mathbf{i}+2\mathbf{j}-\mathbf{k}$, then $$\begin{array}{rcl@{\quad}rcl} \mathbf{v}\,{\cdot}\, \mathbf{w}&=&(2)(4)+(-1)(2)+(1)(-1) & \mathbf{w\,{\cdot}\, v}&=&(4) (2) +(2) (-1) +(-1) (1) \\ &=&8-2-1=5 & &=&8-2-1=5\\ \mathbf{v}\,{\cdot}\, \mathbf{v}&=&4+1+1=6 & \mathbf{w}\,{\cdot}\, \mathbf{w}&=&16+4+1=21 \end{array}$$

NOW WORK

Problem 7(a).

THEOREM Properties of the Dot Product

If $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ are vectors and $a$ is any scalar, then:

• Magnitude property: $$\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{v}\,{\cdot}\, \mathbf{v} =\left\Vert \mathbf{v}\right\Vert ^{2} }$$
• Commutative property: $$\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{u\,{\cdot}\, v=v\,{\cdot}\, u} }$$
• Distributive property: $$\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{u}\,{\cdot}\, (\mathbf{v}+\mathbf{w})=\mathbf{u}\,{\cdot}\, \mathbf{v}+\mathbf{u}\,{\cdot}\, \mathbf{w} }$$
• Linearity property: $$\bbox[5px, border:1px solid black, #F9F7ED]{ a ( \mathbf{u\,{\cdot}\, v}) =( a\mathbf{u}) \,{\cdot}\, \mathbf{v} }$$
• Zero-vector property: $$\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{0}\,{\cdot}\, \mathbf{v}=0 }$$

Proof

To show $\mathbf{v}\,{\cdot}\, \mathbf{v}=\left\Vert \mathbf{v} \right\Vert ^{2}$, we let $\mathbf{v}=v_{1}\mathbf{i}+v_{2}\mathbf{j}+v_{3} \mathbf{k}$. Then $$\begin{equation*} \mathbf{v}\,{\cdot}\, \mathbf{v} =v_{1}v_{1}+v_{2}v_{2}+v_{3}v_{3}=v_{1}^{2}+v_{2}^{2}+v_{3}^{2}=\left\Vert \mathbf{v}\right\Vert ^{2} \end{equation*}$$

Commutative Property: Let $\mathbf{u}=u_{1}\mathbf{i}+u_{2}\mathbf{j} +u_{3}\mathbf{k}$ and $\mathbf{v}=v_{1}\mathbf{i}+v_{2}\mathbf{j} +v_{3}\mathbf{k}$. Then $$\mathbf{u}\,{\cdot}\, \mathbf{v} =u_{1}v_{1}+u_{2}v_{2}+u_{3}v_{3}=v_{1}u_{1}+v_{2}u_{2}+v_{3}u_{3}=\mathbf{v} \,{\cdot}\, \mathbf{u}$$

A consequence of the property $\mathbf{v}\,{\cdot}\, \mathbf{v}=\left\Vert \mathbf{ v}\right\Vert ^{2}$ is that for any vector $\mathbf{v}$, $$\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{v}\,{\cdot}\, \mathbf{v}\geq 0\quad\hbox{and}\ \quad \mathbf{v }\,{\cdot}\, \mathbf{v}=0\quad\hbox{if and only if}\quad \mathbf{v}=\mathbf{0} }$$

NEED TO REVIEW?

The Law of Cosines states that $c^{2}=a^{2}+b^{2}-2ab \cos \theta$, where $a,b$, and $c$ , are the lengths of the sides of a triangle and $\theta$ is the angle opposite the side of length $c$. The Law of the Cosines is discussed in Appendix A.4, p. A-37.

NOTE

If $\mathbf{v}\,{\cdot}\, \mathbf{w} >0,$ the angle between $\mathbf{v}$ and $\mathbf{w}$ is acute. If $\mathbf{v}\,{\cdot}\, \mathbf{w} \lt 0,$ the angle between $\mathbf{v}$ and $\mathbf{w}$ is obtuse.

THEOREM Angle Between Vectors

The angle $\theta$, where $0\leq \theta \leq \pi$, between two nonzero vectors $\mathbf{v}$ and $\mathbf{w}$ is given by the formula $$\bbox[5px, border:1px solid black, #F9F7ED]{ \cos \theta =\dfrac{\mathbf{v }\,{\cdot}\, \mathbf{w}}{\Vert \mathbf{v}\Vert \Vert \mathbf{w}\Vert } }$$

Find the angle between the vectors $\mathbf{v}=2\mathbf{i}-\mathbf{j}+ \mathbf{k}$ and $\mathbf{w}=- \mathbf{i}+\mathbf{j}$.

Solution We find $\mathbf{v}\,{\cdot}\, \mathbf{w}$, $\Vert \mathbf{v} \Vert ,$ and $\Vert \mathbf{w}\Vert.$ $$\begin{eqnarray*} \mathbf{v}\,{\cdot}\, \mathbf{w} &=&(2) (-1) +(-1) (1) +(1) (0) =-2-1+0=-3 \\[4pt] \Vert \mathbf{v}\Vert &=&\sqrt{2^{2}+(-1) ^{2}+1^{2}}=\sqrt{6} \\[4pt] \Vert \mathbf{w}\Vert &=&\sqrt{(-1) ^{2}+1^{2}}=\sqrt{2} \end{eqnarray*}$$

Figure 36

Then if $\theta$ is the angle between $\mathbf{v}$ and $\mathbf{w}$, $$\begin{equation*} \cos \theta =\dfrac{\mathbf{v}\,{\cdot}\, \mathbf{w}}{\Vert \mathbf{v}\Vert \Vert \mathbf{w}\Vert }=\dfrac{-3}{( \sqrt{6}) (\sqrt{2}) }= \dfrac{-3}{\sqrt{12}}=-\dfrac{\sqrt{3}}{2} \end{equation*}$$

Since $0\leq \theta \leq \pi$, the angle $\theta$ between $\mathbf{v}$ and $\mathbf{w}$ is $\dfrac{5\pi}{6}$ radians. See Figure 36.

NOW WORK

Problem 7(b).

An airplane has an air speed of $400$ km/h and is headed east. Find the true direction of the airplane relative to the ground if there is a northwesterly wind of $80~\text{km}/~\text{h}$.

Solution This is the same situation from Example 8 of Section 10.3. There, we found the velocity of the airplane relative to the ground is $\mathbf{v}_{\mathrm{g}}=( 400+40\sqrt{2}) \mathbf{i}-40\sqrt{2 }\mathbf{j}$.

The angle $\theta$ between $\mathbf{v}_{\mathrm{g}}$ and the vector $\mathbf{i}$ (the positive $x$-axis) is given by $$\begin{eqnarray*} \cos \theta &=&\dfrac{\mathbf{v}_{\mathrm{g}}\,{\cdot}\, \mathbf{i}}{\Vert \mathbf{v}_{\mathrm{g}}\Vert \Vert \mathbf{i}\Vert }&=&\dfrac{ 400+40\sqrt{2}}{460.06}\approx 0.9924 \\[5pt] \theta &\approx &\cos ^{-1}(0.9924) \approx 7.07^\circ \end{eqnarray*}$$

Figure 37

The true direction of the plane is approximately $7.07^\circ$ south of east. See Figure 37.

NOW WORK

Problem 35.

THEOREM Orthogonal Vectors

Two nonzero vectors $\mathbf{v}$ and $\mathbf{w}$ are orthogonal if and only if $$\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{v}\,{\cdot}\, \mathbf{w}=0 }$$

NOTE

Orthogonal, perpendicular, and normal are all terms that mean “meet at a right angle.” It is customary to refer to two vectors as orthogonal, two lines as perpendicular, and a line and a plane or a vector and a plane as normal.

Proof

We use the formula for the angle between two nonzero vectors, namely $\cos \theta =\dfrac{\mathbf{v}\,{\cdot}\, \mathbf{w}}{\left\Vert \mathbf{v}\right\Vert \left\Vert \mathbf{w}\right\Vert }.$

If the vectors $\mathbf{v}$ and $\mathbf{w}$ are orthogonal, then the angle $\theta$ between $\mathbf{v}$ and $\mathbf{w}$ is $\dfrac{\pi}{2}$. Since $\cos\dfrac{\pi}{2}=0$, it follows that $\mathbf{v}\,{\cdot}\, \mathbf{w}=0.$

Conversely, if $\mathbf{v}\,{\cdot}\, \mathbf{w}=0$, then $\mathbf{v}=\mathbf{0}$ or $\mathbf{w}=\mathbf{0}$, or $\cos \theta =0$. Since $\mathbf{v}$ and $\mathbf{w}$ are nonzero vectors, $\cos \theta =0,$ so $\theta =\dfrac{\pi }{2 }$ and $\mathbf{v}$ and $\mathbf{w}$ are orthogonal.

The vectors $\mathbf{v}=2\mathbf{i}-\mathbf{j}+5\mathbf{k}$ and $\ \mathbf{w}=3\mathbf{i}+\mathbf{j}-\mathbf{k}$ are orthogonal, since $$\mathbf{v}\,{\cdot}\, \mathbf{w}=6-1-5=0$$

Since $\mathbf{i}\,{\cdot}\, \mathbf{j}=1\,{\cdot}\, 0+ 0\,{\cdot}\, 1=0$, the standard basis vectors $\mathbf{i}$ and $\mathbf{j}$ in the plane are orthogonal. The standard basis vectors $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ in space are mutually orthogonal, since $\mathbf{i}\,{\cdot}\, \mathbf{j}=0,$ $\mathbf{j}\,{\cdot}\, \mathbf{k}=0,$ and $\mathbf{k}\,{\cdot}\, \mathbf{i}=0.$

Find a scalar $a$ so that the vectors $\mathbf{v}=2a\mathbf{i}+\mathbf{j}- \mathbf{k}$ and $\mathbf{w}=\mathbf{i}-a \mathbf{j}+\mathbf{k}$ are orthogonal.

Solution The vectors $\mathbf{v}$ and $\mathbf{w}$ are orthogonal if $\mathbf{v}\,{\cdot}\, \mathbf{w}=0.$ So, $$\begin{eqnarray*} \mathbf{v}\,{\cdot}\, \mathbf{w}=2a-a-1&=& 0 \\[3pt] a&=& 1 \end{eqnarray*}$$

The vectors $\mathbf{v}=2\mathbf{i}+\mathbf{j}-\mathbf{k}$ and $\mathbf{w}= \mathbf{i}-\mathbf{j}+\mathbf{k}$ are orthogonal.

NOW WORK

Problem 17.

THEOREM

If $\mathbf{v}=v_{1}\mathbf{i}+v_{2}\mathbf{j}+v_{3}\mathbf{k}$ is a nonzero vector in space, then $$\bbox[5px, border:1px solid black, #F9F7ED]{ \begin{array}{l} \cos \alpha =\dfrac{v_{1}}{\sqrt{v_{1}^{2}+v_{2}^{2}+v_{3}^{2}}}=\dfrac{ v_{1}}{\Vert \mathbf{v}\Vert }\qquad \cos \beta =\dfrac{v_{2}}{ \sqrt{v_{1}^{2}+v_{2}^{2}+v_{3}^{2}}}=\dfrac{v_{2}}{\Vert \mathbf{v}\Vert }\\ \cos \gamma =\dfrac{v_{3}}{\sqrt{v_{1}^{2}+v_{2}^{2}+v_{3}^{2} }}=\dfrac{v_{3}}{\Vert \mathbf{v}\Vert } \end{array} }$$ $$\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{v=}\left\Vert \mathbf{v}\right\Vert [ \cos \alpha \mathbf{i} +\cos \beta \mathbf{j}+\cos \gamma \mathbf{k}] }$$

Solution (a) The magnitude of $\mathbf{v}$ is $$\begin{equation*} \Vert \mathbf{v}\Vert =\sqrt{(-3)^{2}+2^{2}+(-6)^{2}}=\sqrt{49}=7 \end{equation*}$$

The direction cosines of the vector $\mathbf{v}$ are $$\cos \alpha =\dfrac{v_{1}}{\left\Vert \mathbf{v}\right\Vert }=\dfrac{-3}{7} \qquad \cos \beta =\dfrac{v_{2}}{\left\Vert \mathbf{v}\right\Vert }= \dfrac{2}{7} \qquad \cos \gamma =\dfrac{v_{3}}{\left\Vert \mathbf{v}\right\Vert }=\dfrac{-6}{7}$$

(b) Now $\mathbf{v=}\left\Vert \mathbf{v}\right\Vert \left[ \cos \alpha {\bf i} + \cos \beta \mathbf{j}+\cos \gamma \mathbf{k}\right] =7\!\left( -\dfrac{3}{7}\mathbf{i}+\dfrac{2}{7}\mathbf{j}-\dfrac{6}{7}\mathbf{k }\right)\! .$

NOW WORK

Problem 19.

NOTE

The vector $\mathbf{v}_{2}$ orthogonal to $\mathbf{w}$ is sometimes denoted by orth$_{\bf w} {\bf v}$.

THEOREM Vector Projection

If $\mathbf{v}$ and $\mathbf{w}$ are two nonzero vectors, the vector projection of $\mathbf{v}$ onto $\mathbf{w}$ is $$\bbox[5px, border:1px solid black, #F9F7ED]{ \hbox{proj}_{\mathbf{w}}\mathbf{v}=\dfrac{\mathbf{v}\,{\cdot}\, \mathbf{w}}{\Vert \mathbf{w}\Vert ^{2}}\mathbf{w} }$$

The decomposition of $\mathbf{v}$ into $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ , where $\mathbf{v}_{1}$ is parallel to $\mathbf{w}$ and $\mathbf{v}_{2}$ is orthogonal to $\mathbf{w}$, is $$\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{v}_{1}=\dfrac{\mathbf{v}\,{\cdot}\, \mathbf{w}}{\Vert \mathbf{w}\Vert ^{2}}\mathbf{w}\qquad \mathbf{v}_{2}= \mathbf{v-v}_{1} }$$

Find the vector projection of $\mathbf{v}=2\mathbf{i}-\mathbf{j}+\mathbf{k}$ onto $\mathbf{w}=\mathbf{i}+\mathbf{j}+\mathbf{k}$. Decompose $\mathbf{v}$ into two vectors $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$, where $\mathbf{v} _{1}$ is parallel to $\mathbf{w}$ and $\mathbf{v}_{2}$ is orthogonal to $\mathbf{w}$.

Solution We use the formula for the projection of $\mathbf{v}$ onto $\mathbf{w}.$ $$\begin{eqnarray*} \mathbf{v}_{1}& =&\hbox{proj}_{\mathbf{w}}\mathbf{v}=\frac{\mathbf{v}\,{\cdot}\, \mathbf{w}}{\Vert \mathbf{w}\Vert ^{2}}\mathbf{w} \underset{\underset{\underset{{\color{#0066A7}{\hbox{\(\Vert \mathbf{w}\Vert =\sqrt{1^{2}+1^{2} +1^{2}} =\sqrt{3}\)}}}} {\color{#0066A7}{\hbox{\(\mathbf{v}\,{\cdot}\, \mathbf{w=}2-1+1=2\)}}}} {\color{#0066A7}{\uparrow }}}{=}\frac{2}{(\sqrt{3})^{2}}\mathbf{w}=\frac{2}{3}( \mathbf{i}+\mathbf{j}+\mathbf{k})=\frac{2}{3}\,\mathbf{i}+\frac{2}{3}\, \mathbf{j}+\frac{2}{3}\,\mathbf{k} \\ \mathbf{v}_{2}& =&\mathbf{v}-\mathbf{v}_{1}=(2\mathbf{i}-\mathbf{j}+\mathbf{k} )-\left( \frac{2}{3}\,\mathbf{i}+\frac{2}{3}\,\mathbf{j}+\frac{2}{3}\, \mathbf{k}\right) =\frac{4}{3}\,\mathbf{i}-\frac{5}{3}\,\mathbf{j}+\frac{1}{3 }\,\mathbf{k} \end{eqnarray*}$$

NOW WORK

Problem 27.

Find the work done by a force of 2 newtons (N) acting in the direction $\mathbf{i}+\mathbf{j}+\mathbf{k}$ in moving an object 1 m from (0, 0, 0) to (1, 0, 0).

Solution We need to express the force $\mathbf{F}$ in terms of its magnitude and direction. The unit vector $\mathbf{u}$ in the direction $\mathbf{v}=\mathbf{i}+\mathbf{j}+\mathbf{k}$ is $$\mathbf{u}=\dfrac{\mathbf{v}}{\left\Vert \mathbf{v}\right\Vert }=\dfrac{ \mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{3}}=\dfrac{1}{\sqrt{3}}\mathbf{i}+ \dfrac{1}{\sqrt{3}}\mathbf{j}+\dfrac{1}{\sqrt{3}}\mathbf{k}$$

Since the force vector $\mathbf{F}$ has magnitude $2,$ we have $$\mathbf{F}=2\left( \frac{1}{\sqrt{3}}\mathbf{i}+\dfrac{1}{\sqrt{3}}\mathbf{j} +\dfrac{1}{\sqrt{3}}\mathbf{k}\right) =\dfrac{2}{\sqrt{3}}( \mathbf{i}+ \mathbf{j}+\mathbf{k})$$

The line of motion of the object from $( 0,0,0)$ to $\left( 1,0,0\right)$ is $\skew5\overrightarrow{\it AB}=\mathbf{i}$. The work $W$ is therefore $$W=\mathbf{F}\,{\cdot}\, \skew5\overrightarrow{\it AB}=\frac{2}{\sqrt{3}}(\mathbf{i}+\mathbf{j }+\mathbf{k})\,{\cdot}\ \mathbf{i}=\frac{2}{\sqrt{3}}\hbox{ joules}$$

RECALL

$1$ joule $=1$ newton$\cdot$meter; $1~\text {J} =1~\text {N}\cdot\ \text{m}$.

NOW WORK

Problem 43.

Figure 43 shows a man pushing on a lawn mower handle with a force of 30 lb. How much work is done in moving the lawn mower a distance of 75 ft if the handle makes an angle of $60^{\circ }$ with the ground?

Figure 43 $\skew5\overrightarrow{\it AB}=75\,\mathbf{i}$

Solution We set up the coordinate system so that the lawn mower is moved from $(0,0)$ to $(75,0)$. Then the motion occurs along $\skew5\overrightarrow{\it AB}=75\,\mathbf{i}$. The force vector $\mathbf{F}$, as shown in Figure 44, makes an angle of $300^\circ$ to the positive $x$-axis. Since $\Vert \mathbf{F} \Vert = 30$, $\mathbf{F}$ is given by $$\begin{equation*} \mathbf{F}=30[ (\cos 300^{\circ })\mathbf{i}+(\sin 300^{\circ })\mathbf{ j}] =30\left[ \dfrac{1}{2}\mathbf{i}-\dfrac{\sqrt{3}}{2}\mathbf{j} \right] =15\mathbf{i}-15\sqrt{3}\mathbf{j} \end{equation*}$$

Figure 44 $\Vert \mathbf{F} \Vert = 30$

Then the work $W$ done is $$W=\mathbf{F}\,{\cdot}\, \skew5\overrightarrow{\it AB}=(15\mathbf{i}-15\sqrt{3}\mathbf{j} )\,{\cdot}\, 75\mathbf{i}=1125 \hbox{ft-lb}$$