OBJECTIVES

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

1. Find the cross product of two vectors (p. 725)
2. Prove algebraic properties of the cross product (p. 726)
3. Apply geometric properties of the cross product (p. 727)

spanDEFINITIONspan Cross Product

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 cross product \(\mathbf{v}\times \mathbf{w}\) is defined as the vector \[\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{v}\times \mathbf{w}=(v_{2}w_{3}-v_{3}w_{2})\mathbf{i}-(v_{1}w_{3}-v_{3}w_{1})\,\mathbf{j}+(v_{1}w_{2}-v_{2}w_{1}) \mathbf{k} } \]

Finding the Cross Product of Two Vectors

If \(\mathbf{v}=2\mathbf{i}-\mathbf{j}+\mathbf{k}\) and \(\mathbf{w}=4\mathbf{i} +2\mathbf{j}-\mathbf{k}\), then using the definition of the cross product, we get \[ \begin{eqnarray*} \mathbf{v}\times \mathbf{w}&=&[(-1)(-1)-(1)(2)]\mathbf{i}-[(2)(-1)-(1)(4)] \mathbf{j}+[(2)(2)-(-1)(4)]\mathbf{k}\\[4pt] &=&-\mathbf{i}+6\mathbf{j}+8\mathbf{k} \end{eqnarray*} \]

NOW WORK

Problem 13(a).

NOTE

Notice the minus sign in the middle term.

Evaluating Determinants

1. \(\left\vert \begin{array}{r@{\quad}r} 2 & 3 \\ -1 & 2 \end{array} \right\vert =(2)(2)-(-1)(3)=4+3=7\)
2. \[\begin{align*} \left\vert \begin{array}{r@{\quad}r@{\quad}r} A & B & C \\ 2 & 3 & 1 \\ -1 & 2 & -1 \end{array} \right\vert &=\left\vert \begin{array}{r@{\quad}r} 3 & 1 \\ 2 & -1 \end{array} \right\vert A-\left\vert \begin{array}{r@{\quad}r} 2 & 1 \\ -1 & -1 \end{array} \right\vert B+\left\vert \begin{array}{r@{\quad}r} 2 & 3 \\ -1 & 2 \end{array} \right\vert C\\ &=(-3-2)A-(-2+1)B+(4+3)C=-5A+B+7C \end{align*}\]

NOW WORK

Problem 11.

Finding a Cross Product Using Determinants

If \(\mathbf{v}=2\mathbf{i}-\mathbf{j}+\mathbf{k}\) and \(\mathbf{w}=4\mathbf{i} +2\mathbf{j}-\mathbf{k}\), find

1. \(\mathbf{v}\times \mathbf{w}\)
2. \(\mathbf{w} \times \mathbf{v}\)
3. \(\mathbf{v\times v}\)

Solution (a) \(\mathbf{v}\times \mathbf{w}=\left\vert \begin{array}{r@{\quad}r@{\quad}r} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 4 & 2 & -1 \end{array} \right\vert =(1-2)\mathbf{i}-(-2-4)\mathbf{j}+(4+4)\mathbf{k}=-\mathbf{i}+6 \mathbf{j}+8\mathbf{k}\)

(b) \(\mathbf{w}\times \mathbf{v}=\left\vert \begin{array}{r@{\quad}r@{\quad}r} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 2 & -1 \\ 2 & -1 & 1 \end{array} \right\vert =(2-1) \,\mathbf{i}-( 4+2) \,\mathbf{ j\,+\,}( -4-4) \,\mathbf{k=i}-6\mathbf{j}-8\mathbf{k=-}( \mathbf{v}\times \mathbf{w}) \)

(c) \(\mathbf{v}\times \mathbf{v}=\left\vert \begin{array}{r@{\quad}r@{\quad}r} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 2 & -1 & 1 \end{array} \right\vert =0\,\mathbf{i}-0\,\mathbf{j}+0\mathbf{\,k}=\mathbf{0}\)

NOTE

The cross product is not a commutative operation. That is, \(\mathbf{v}\times \mathbf{w}\neq \mathbf{w}\times \mathbf{v}\).

NOW WORK

Problem 15(a).

THEOREM Algebraic Properties of the Cross Product

If \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) are vectors in space and if \(a\) is a scalar, then

Proof

Let \(\mathbf{v}=v_{1}\mathbf{i}+v_{2}\mathbf{j}+v_{3} \mathbf{k}\) and \(\mathbf{w}=w_{1}\mathbf{i}+w_{2}\mathbf{j}+w_{3}\mathbf{k}\).

(1) \(\mathbf{v}\times \mathbf{v}=(v_{2}v_{3}-v_{2}v_{3})\mathbf{i} -(v_{1}v_{3}-v_{1}v_{3})\mathbf{j}+(v_{1}v_{2}-v_{1}v_{2})\mathbf{k}=\mathbf{ 0}\)

(2) \( \begin{array}{@{\hspace*{-6.2pc}}rcl} \mathbf{v}\times \mathbf{w}&=&(v_{2}w_{3}-w_{2}v_{3})\mathbf{i} -(v_{1}w_{3}-w_{1}v_{3})\mathbf{j}+(v_{1}w_{2}-w_{1}v_{2})\mathbf{k}\\[2pt] \mathbf{w}\times \mathbf{v}&=&(w_{2}v_{3}-v_{2}w_{3}) \mathbf{i}-(w_{1}v_{3}-v_{1}w_{3})\,\mathbf{j}+(w_{1}v_{2}-v_{1}w_{2})\, \mathbf{k} \end{array} \)

Consequently, \(\mathbf{v}\times \mathbf{w}=-(\mathbf{w}\times \mathbf{v})\).

(5) First we find \(\left\Vert \mathbf{v}\times \mathbf{w}\right\Vert ^{2}\). \[ \begin{array}{@{\hspace*{-2.2pc}}rcl} \mathbf{v}\times \mathbf{w} &=&(v_{2}w_{3}-w_{2}v_{3})\mathbf{i}-(v_{1}w_{3}-w_{1}v_{3})\mathbf{j}+(v_{1}w_{2}-w_{1}v_{2})\mathbf{k} \\ \Vert \mathbf{v}\times \mathbf{w}\Vert ^{2} &=&(v_{2}w_{3}-w_{2}v_{3})^{2}+(v_{1}w_{3}-w_{1}v_{3})^{2}+(v_{1}w_{2}-w_{1}v_{2})^{2}\\ &=&[ (v_{2}\,w_{3})^{2}-2v_{2}\,w_{2}\,v_{3}\,w_{3}+(w_{2}\,v_{3})^{2}] +[ (v_{1}w_{3})^{2}-2v_{1}w_{1}v_{3}w_{3}+(w_{1}v_{3})^{2}] \\ &&+\,[ (v_{1}\,w_{2})^{2}-2\,v_{1}\,w_{1}\,v_{2}\,w_{2}+(w_{1}v_{2})^{2}] \end{array} \]

Then we find \(\left\Vert \mathbf{v}\right\Vert ^{2}\,\left\Vert \mathbf{w} \right\Vert ^{2}-(\mathbf{v}\,{\cdot}\, \mathbf{w})^{2}\). \[ \begin{eqnarray*} \Vert \mathbf{v}\Vert ^{2}\Vert \mathbf{w}\Vert ^{2}-(\mathbf{v}\,{\cdot}\, \mathbf{w})^{2}& =&\big[ \big( v_{1}^{2}+v_{2}^{2}+v_{3}^{2}\big) \big( w_{1}^{2}+w_{2}^{2}+w_{3}^{2}\big) \big] -(v_{1}w_{1}+v_{2}w_{2}+v_{3}w_{3})^{2} \\[4pt] &=& (v_{1}w_{2})^{2}+(v_{1}w_{3})^{2}+(v_{2}w_{1})^{2}+(v_{2}w_{3})^{2}+( v_{3}w_{1}) ^{2}+\left( v_{3}w_{2}\right) ^{2} \\[4pt] && -2v_{1}v_{2}w_{1}w_{2}-2v_{1}v_{3}w_{1}w_{3}-2v_{2}v_{3}w_{2}w_{3} \end{eqnarray*} \]

Reordering the terms in each equation proves the property.

THEOREM Geometric Properties of the Cross Product

If \(\mathbf{v}\) and \(\mathbf{w}\) are two nonzero vectors in space, then

Proof of (6)

Let \(\mathbf{v}=v_{1}\mathbf{i}+v_{2}\mathbf{j}+v_{3} \mathbf{k}\) and \(\mathbf{w}=w_{1}\mathbf{i}+w_{2}\mathbf{j}+w_{3}\mathbf{k}\). Then \[ \mathbf{v}\times \mathbf{w}=(v_{2}w_{3}-w_{2}v_{3})\mathbf{i} -(v_{1}w_{3}-w_{1}v_{3})\mathbf{j}+(v_{1}w_{2}-w_{1}v_{2})\mathbf{k} \]

We show that \(\mathbf{v}\times \mathbf{w}\) is orthogonal to \(\mathbf{v}\) by computing the dot product \(\mathbf{v}\,{\cdot}\, (\mathbf{v}\times \mathbf{w}).\) \[ \begin{eqnarray*} \mathbf{v}\,{\cdot}\, (\mathbf{v}\times \mathbf{w}) &=&v_{1}( v_{2}w_{3}-w_{2}v_{3}) -v_{2}( v_{1}w_{3}-w_{1}v_{3}) +v_{3}( v_{1}w_{2}-w_{1}v_{2}) \\[3pt] &=&v_{1}v_{2}w_{3}-v_{1}w_{2}v_{3}-v_{2}v_{1}w_{3}+v_{2}w_{1}v_{3}+v_{3}v_{1}w_{2}-v_{3}w_{1}v_{2}=0 \end{eqnarray*} \]

So, \(\mathbf{v}\times \mathbf{w}\) is orthogonal to \(\mathbf{v}\).

Now we show \(\mathbf{v}\times \mathbf{w}\) is orthogonal to \(\mathbf{w}\) by computing the dot product \(\mathbf{w}\,{\cdot}\, (\mathbf{v}\times \mathbf{w}).\) \[ \begin{eqnarray*} \mathbf{w}\,{\cdot}\, (\mathbf{v}\times \mathbf{w}) &=&w_{1} ( v_{2}w_{3}-w_{2}v_{3}) -w_{2} ( v_{1}w_{3}-w_{1}v_{3}) +w_{3} ( v_{1}w_{2}-w_{1}v_{2}) \\[3pt] &=&w_{1}v_{2}w_{3}-w_{1}w_{2}v_{3}-w_{2}v_{1}w_{3}+w_{2}w_{1}v_{3}+w_{3}v_{1}w_{2}-w_{3}w_{1}v_{2}=0 \end{eqnarray*} \]

So, \(\mathbf{v}\times \mathbf{w}\) is orthogonal to \(\mathbf{w}\).

Finding a Vector Orthogonal to Two Given Vectors

Find a vector orthogonal to the vectors \(\mathbf{v}=2\mathbf{i}-\mathbf{j}+ \mathbf{k}\) and \(\mathbf{w}=2\mathbf{i}+2\mathbf{j}-\mathbf{k}\).

Solution Since \(\mathbf{v\times w}\) is orthogonal to both the vector \(\mathbf{v}\) and the vector \(\mathbf{w}\), a vector orthogonal to \( \mathbf{v}\) and \(\mathbf{w}\) is \[ \mathbf{v}\times \mathbf{w}= \left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \hphantom{-}\mathbf{j} & \hphantom{-}\mathbf{k} \\[3pt] 2 & -1 & \hphantom{-}1 \\[3pt] 2 & \hphantom{-}2 & -1 \end{array}\right| =-\mathbf{i}+4\mathbf{j}+6\mathbf{k} \]

NOW WORK

Problem 29.

Proof of (7)

Let \(\theta \) be the angle between the nonzero vectors \(\mathbf{v}\) and \(\mathbf{w}\). Then from algebraic property (5), we have \[ \begin{equation*} \Vert \mathbf{v}\times \mathbf{w}\Vert ^{2}=\Vert \mathbf{v}\Vert ^{2}\Vert \mathbf{w}\Vert ^{2}-(\mathbf{v}\,{\cdot}\, \mathbf{w})^{2}\tag {5} \end{equation*} \]

Since \(\cos \theta =\dfrac{\mathbf{v}\,{\cdot}\, \mathbf{w}}{\left\Vert \mathbf{v} \right\Vert \left\Vert \mathbf{w}\right\Vert },\) we have \(\mathbf{v}\,{\cdot}\, \mathbf{w=}\left\Vert \mathbf{v}\right\Vert \left\Vert \mathbf{w}\right\Vert \cos \theta \) and \[ \begin{align*} \Vert \mathbf{v}\times \mathbf{w}\Vert ^{2}&=\Vert \mathbf{v}\Vert^{2}\Vert \mathbf{w}\Vert ^{2}-\Vert \mathbf{v}\Vert ^{2}\Vert \mathbf{w} \Vert ^{2}\cos ^{2}\theta & {\color{#0066A7}{\hbox{\({(\mathbf{v}\,{\cdot}\, \mathbf{w})}^{2}{=\Vert \mathbf{v}\Vert }^{2}{\Vert \mathbf{w}\Vert }^{2}\cos ^{2}{\theta }\)}}} \\ &=\Vert \mathbf{v}\Vert ^{2}\Vert \mathbf{w}\Vert ^{2}(1-\cos ^{2}\theta )&{\color{#0066A7}{\hbox{Factor out \(\Vert \mathbf{v}\Vert ^{2} \Vert \mathbf{w}\Vert ^{2}\)}}}\\ &=\Vert \mathbf{v}\Vert ^{2}\Vert \mathbf{w}\Vert ^{2}\sin ^{2}\theta & {\color{#0066A7}{\hbox{\(1-\cos ^{2}\theta =\sin ^{2}\theta\)}}} \end{align*} \]

Since \(0\leq \theta \leq \pi ,\) we know \(\sin \theta \geq 0\). Now take the square root of both sides to obtain \(\Vert \mathbf{v}\times \mathbf{w} \Vert =\Vert \mathbf{v}\Vert \Vert \mathbf{w}\Vert \sin \theta .\)

Proof of (8)

If \(\mathbf{v}\) and \(\mathbf{w}\) are nonzero vectors and \(\theta \) is the angle between \(\mathbf{v}\) and \(\mathbf{w}\), then \( \Vert \mathbf{v}\Vert \) and \(\Vert \mathbf{w}\Vert \) represent the lengths of the sides of a parallelogram. As shown in Figure 48, the base of the parallelogram is \(\left\Vert \mathbf{v}\right\Vert \) and its altitude is \( \left\Vert \mathbf{w}\right\Vert \sin \theta ,\) so its area \(A\) is \[ \begin{equation*} A=\hbox{ (base)(altitude)}=\left\Vert \mathbf{v}\right\Vert \left\Vert \mathbf{w}\right\Vert \sin \theta \end{equation*} \]

Since \(\Vert \mathbf{v}\times \mathbf{w}\Vert =\Vert \mathbf{v}\Vert \Vert \mathbf{w}\Vert \sin \theta ,\) the magnitude of \(\mathbf{v}\times \mathbf{w}\) is the area of the parallelogram whose sides are the vectors \(\mathbf{v}\) and \(\mathbf{w}\).

Finding the Area of a Parallelogram

Find the area \(A\) of the parallelogram shown in Figure 49, which has vertices at the points \(P_{1}=(0, 0, 0)\), \(P_{2}=(-1, 2, 4)\), \(P_{3}=(1, 1, 8)\), and \(P_{4}=(2, -1, 4)\).

Figure 49

Solution The area \(A\) of a parallelogram is \(\Vert \mathbf{v} \times \mathbf{w}\Vert \), where \(\mathbf{v}\) and \(\mathbf{w}\) are two adjacent sides of the parallelogram. Be careful! Not all pairs of vertices give a side. For example, \(\overrightarrow{P_{1}~P_{3}}\) is not a side; it is a diagonal of the parallelogram. It is helpful to sketch the parallelogram before choosing \(\mathbf{v}\) and \(\mathbf{w}\). Choose \[ \begin{equation*} \mathbf{v}=\overrightarrow{P_{1}~P_{2}}= \left\langle -1,\,2,4\right\rangle =-\mathbf{i}+2\mathbf{j}+4\mathbf{k} \end{equation*} \]

and \[ \begin{equation*} \hbox{}\mathbf{w}=\overrightarrow{P_{1}~P_{4}} =\left\langle 2,-1,4\right\rangle =2\mathbf{i}-\mathbf{j}+4\mathbf{k} \end{equation*} \]

Then \[ \begin{equation*} \mathbf{v}\times \mathbf{w}= \left|\begin{array}{r@{\quad}r@{\quad}r} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] -1 & 2 & 4 \\[3pt] 2 & -1 & 4 \end{array}\right| =12\mathbf{i}+12\mathbf{j}-3\mathbf{k} \end{equation*} \]

The area of the parallelogram is \[ \left\Vert \mathbf{v}\times \mathbf{w}\right\Vert = \sqrt{144+144+9}= \sqrt{297} \approx 17.234~\hbox{square units} \]

NOW WORK

Problem 37 and 45.

Proof of (9)

If \(\mathbf{v}\) and \(\mathbf{w}\) are parallel nonzero vectors, then the angle \(\theta\) between the vectors \(\mathbf{v}\) and \(\mathbf{w}\) is either \(\theta =0\) or \(\theta =\pi .\) For either angle, \( \sin \theta =0\), and \[ \begin{equation*} \left\Vert \mathbf{v}\times \mathbf{w}\right\Vert =\Vert \mathbf{v}\Vert \Vert \mathbf{w}\Vert \sin \theta =0\qquad \hbox{so}\quad \mathbf{v}\times \mathbf{w }=\mathbf{0} \end{equation*} \]

Now suppose \(\mathbf{v}\times \mathbf{w}=\mathbf{0}\). Since \(\left\Vert \mathbf{v}\times \mathbf{w}\right\Vert =\Vert \mathbf{v}\Vert \Vert \mathbf{w }\Vert \sin \theta ,\) we have \[ \begin{eqnarray*} \Vert \mathbf{v}\Vert \left\Vert \mathbf{w}\right\Vert \sin \theta &=&0 \\ \sin \theta &=&0 & {\color{#0066A7}{\hbox{\(\mathbf{v} \neq \mathbf{0},\quad \mathbf{w}\neq \mathbf{0}\)}}} \\ \theta &=&0\quad\hbox{or}\quad \theta =\pi \end{eqnarray*} \]

That is, the vectors \(\mathbf{v}\) and \(\mathbf{w}\) are parallel.

Verifying Two Vectors Are Parallel

Show that the vectors \(\mathbf{v}=2\mathbf{i}+\mathbf{j}-\mathbf{k}\) and \( \mathbf{w}=-4\mathbf{i}-2\mathbf{j}+2\mathbf{k}\) are parallel.

Solution We show that \(\mathbf{v}\times \mathbf{w}=\mathbf{0}\). \[ \begin{eqnarray*} \mathbf{v}\times \mathbf{w}&=&\left\vert \begin{array}{r@{\quad}r@{\quad}r} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -1 \\ -4 & -2 & 2 \end{array} \right\vert =\left\vert \begin{array}{r@{\quad}r} 1 & -1 \\ -2 & 2 \end{array} \right\vert \,\mathbf{i}-\left\vert \begin{array}{r@{\quad}r} 2 & -1 \\ -4 & 2 \end{array} \right\vert \,\mathbf{j}+\left\vert \begin{array}{r@{\quad}r} 2 & 1 \\ -4 & -2 \end{array} \right\vert \,\mathbf{k}\\[5pt] &=&0\mathbf{i}+0\mathbf{j}+0\mathbf{k}=\mathbf{0} \end{eqnarray*} \]

NEED TO REVIEW?

Circular motion is discussed in Appendix A.4, p. A-27.

Finding the Speed of an Object in Circular Motion

A rigid body rotates with a constant angular speed of \(\omega \) radians per second about a line through the origin in the direction of the vector \(2 \mathbf{i}+\mathbf{j}+2\mathbf{k}\). Find the speed of an object on this body at the instant the object passes through the point \((1,3,5)\). Assume the distance is in meters.

Solution The vector \( {\omega }\) is parallel to the axis of rotation of the rigid body, and so it is in the direction of the vector \(2 \mathbf{i}+\mathbf{j}+2\mathbf{k}\). Since \(\dfrac{2\mathbf{i}+\mathbf{j}+2 \mathbf{k}}{ \sqrt{2^{2}+1^{2}+2^{2}}}=\dfrac{2}{3}\mathbf{i}+\dfrac{1}{3} \mathbf{j}+\dfrac{2}{3}\mathbf{k}\) is a unit vector in the direction of \(2 \mathbf{i}+\mathbf{j}+2\mathbf{k}\), the angular velocity \( {\omega }\) is \[ \begin{equation*} {\omega }=\omega \left( \frac{2}{3}\,\mathbf{i}+\frac{1}{3}\,\mathbf{j }+\frac{2}{3}\,\mathbf{k}\right) \end{equation*} \]

The position of the object is \(\mathbf{r}=\mathbf{i}+3\mathbf{j}+5\mathbf{k}\). So, the velocity \(\mathbf{v}\) of the object is \[ \begin{eqnarray*} \mathbf{v}&=& {\omega }\times \mathbf{r}=\left[ \omega \left( \frac{2}{ 3}\,\mathbf{i}+\frac{1}{3}\,\mathbf{j}+\frac{2}{3}\,\mathbf{k}\right) \right] \times \left[ \mathbf{i}+3\mathbf{j}+5\mathbf{k}\right] =\,\omega \left\vert \begin{array}{l@{\quad}l@{\quad}l} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \dfrac{2}{3} & \dfrac{1}{3} & \dfrac{2}{3} \\ 1 & 3 & 5 \end{array} \right\vert\\[5pt] &=&\omega \left( -\dfrac{1}{3}\mathbf{i}-\dfrac{8}{3}\mathbf{j}+ \dfrac{5}{3}\mathbf{k}\right) \end{eqnarray*} \]

The speed \(v\) of the object is \(v=\left\Vert \mathbf{v}\right\Vert =\omega \sqrt{\dfrac{1}{9}+\dfrac{64}{9}+\dfrac{25}{9}}= \sqrt{10}\,\omega\) meters per second.

NOW WORK

Problem 47.