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The Dot and Cross Products

April 14th, 2009 |

Author: Merrin

In this section we will focus on two fundamental operation between two vectors, the dot product and the cross product. The dot product can be used as a measure of how parallel two vectors are and the cross product can be used as a measure of how perpendicular two vectors are. While the dot product of two vectors is just a number the cross product of two vectors is a new vector perpendicular to the other two vectors. The dot product and cross products can be used to prove new identities amongst vectors such as the law of cosines or law of sines from trigonometry. Either product can be used to determine the angles between two vectors.

Definition 1. The dot product or scalar product between two vectors $latex \mathbf{A}$ and $latex \mathbf{B}$ is defined as

$\mathbf{A}\cdot \mathbf{B}=\mathbf{B}\cdot \mathbf{A}=A_{x} B_{x} +A_{y} B_{y} +A_{z} B_{z} =AB\cos \theta _{AB}$

$latex \theta _{AB} $ is the angle between the vectors.

Example 1. Find the dot product and angle between the following two vectors

$\mathbf{A}=\mathbf{i}+2\mathbf{j}+3\mathbf{k} \qquad \mathbf{B}=\mathbf{i}-\mathbf{j}$

Solution 1.

${\mathbf{A}\cdot \mathbf{B}=1-2+0=-1}$

${-1=AB\cos \theta _{AB} =\sqrt{1+4+9} \sqrt{1+1} \cos \theta _{AB} =2\sqrt{7} \cos \theta _{AB} }$

${\cos \theta _{AB} =\cfrac{-1}{2\sqrt{7} } }$

${\theta _{AB} =\arccos \left(-\cfrac{1}{2\sqrt{7} } \right)}=100.89 \,\text{deg}$

There can be unit vectors that point in some other direction besides one of the axis. To find a unit vector pointing in the direction of a vector, divide that vector by its magnitude. The magnitude can be found by taking the dot product of a vector with itself and then taking the square root. One special unit vector is called the normal vector which is perpendicular to a surface, $latex \mathbf{n}$.

Example 2. Construct the unit vector $latex \mathbf{u}$ pointing in the direction of a vector $latex \mathbf{V}$

Solution 2. To construct the unit vector from a vector, divide by its magnitude.

$\begin{array}{c} {\mathbf{V}=V_{x} \mathbf{i}+V_{y} \mathbf{j}+V_{z} \mathbf{k}} \\\\{\mathbf{v}=\cfrac{V_{x} }{\sqrt{V_{x}^{2} +V_{y}^{2} +V_{z}^{2} } } \mathbf{i}+\cfrac{V_{y} }{\sqrt{V_{x}^{2}+V_{y}^{2} +V_{z}^{2} } } \mathbf{j}+\cfrac{V_{z} }{\sqrt{V_{x}^{2} +V_{y}^{2}+V_{z}^{2}} } \mathbf{k}} \\\\ {\mathbf{v}=\cfrac{\mathbf{V}}{\left|\mathbf{V}\right|} }\end{array}$

To check that $latex \mathbf{u}$ is a unit vector just dot it into itself and find that the result is equal to one.

Example 3. An example of a vector field is the electric field produced by a point charge. The electric field is given as

$\mathbf{E}(x,y,z)=\cfrac{kq}{\left(x^{2} +y^{2} +z^{2} \right)^{3/2} } (x\mathbf{i}+y\mathbf{j}+z\mathbf{k})$

Find the unit vector $\mathbf{r}$ in the direction of the electric field and express the electric field in terms of it.

Solution 3. The electric field can be written in a different form with the following notation

$\begin{array}{c} {\mathbf{R}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}} \\ \\ R=\sqrt{x^{2} +y^{2} +z^{2}} \\\\ {\mathbf{r}=\cfrac{x}{\sqrt{x^{2} +y^{2} +z^{2} } } \mathbf{i}+\cfrac{y}{\sqrt{x^{2} +y^{2} +z^{2} } } \mathbf{j}+\cfrac{z}{\sqrt{x^{2} +y^{2} +z^{2} } } \mathbf{k}=\cfrac{\mathbf{R}}{R} } \\ \\{\mathbf{E}=\cfrac{kq}{R^3}\mathbf{R}=\cfrac{kq}{R^2}}\mathbf{r} \end{array}$

Example 4. Find the magnitude of $latex \mathbf{C}$ where $latex \mathbf{A}$ and $\mathbf{B}$ are given as

${\mathbf{A}+\mathbf{B}=\mathbf{C}} \\ {\mathbf{A}=2\mathbf{i}+4\mathbf{j}-3\mathbf{k}} \\ {\mathbf{B}=-3\mathbf{i}+\mathbf{j}+2\mathbf{k}}$

Solution 4. First add $\mathbf{A}$ and $\mathbf{B}$ together and get

${\mathbf{A}+\mathbf{B}=(2-3)\mathbf{i}+(4+1)\mathbf{j}+(-3+2)\mathbf{k}}$

${\mathbf{C}=-\mathbf{i}+5\mathbf{j}-\mathbf{k}}$

To calculate the magnitude of $\mathbf{C}$ we take the dot product of $\mathbf{C}$ with itself.

${\mathbf{C}\cdot \mathbf{C}=\left|\mathbf{C}\right|^{2} =1+25+1=27}$

${\left|\mathbf{C}\right|=3\sqrt{3}}$

Example 5. A hiker travels 5 km east, 2 km north-east, and then 1 km west. Find the net displacement of the hiker.

Solution 5. We first convert each distance the hiker travels into a vector

${\mathbf{A}=5\mathbf{i}\, {\rm km}} \\ {\mathbf{C}=-\mathbf{i}\, {\rm km}} \\ {\mathbf{B}=?}$

$latex \mathbf{B}$ is northwest so we write the unit vector in the direction of north-west and then multiply by the magnitude

$\left(\cfrac{1}{\sqrt{2} } \mathbf{i}+\cfrac{1}{\sqrt{2} } \mathbf{j}\right)\cdot \left(\cfrac{1}{\sqrt{2} } \mathbf{i}+\cfrac{1}{\sqrt{2} } \mathbf{j}\right)=\cfrac{1}{2} +\cfrac{1}{2} =1\quad unit\,\, vector$

$\mathbf{B}=2\left(\cfrac{1}{\sqrt{2} } \mathbf{i}+\cfrac{1}{\sqrt{2} } \mathbf{j}\right){\rm km}=\left(\sqrt{2} \mathbf{i}+\sqrt{2} \mathbf{j}\right){\rm km}$

To find the net displacement we add all the vectors together

${\mathbf{A}+\mathbf{B}+\mathbf{C}} = {[(5-1+\sqrt{2} )\mathbf{i}+\sqrt{2} \mathbf{j}]\, {\rm km}} \\ {} = {[(4+\sqrt{2} )\mathbf{i}+\sqrt{2} \mathbf{j}]\; {\rm km}}$

Definition 2. The Cross Product or Vector Product. The cross product is defined as

$\mathbf{A}\times \mathbf{B} = (A_{y} B_{z} -A_{z} B_{y})\mathbf{i} + (A_{z} B_{x} -A_{x} B_{z})\mathbf{j} + (A_{x} B_{y} -A_{y} B_{x})\mathbf{k}$

Another equivalent method to calculate the cross product, is to calculate the determinant in matrix notation

$\mathbf{A}\times \mathbf{B}=\left|\begin{array}{ccc} {\mathbf{i}} & {\mathbf{j}} & {\mathbf{k}} \\ {A_{x} } & {A_{y} } & {A_{z} } \\ {B_{x} } & {B_{y} } & {B_{z} } \end{array}\right|=\mathbf{i}\left|\begin{array}{cc} {A_{y} } & {A_{z} } \\ {B_{y} } & {B_{z} } \end{array}\right|-\mathbf{j}\left|\begin{array}{cc} {A_{x} } & {A_{z} } \\ {B_{x} } & {B_{z} } \end{array}\right|+\mathbf{k}\left|\begin{array}{cc} {A_{x} } & {A_{y} } \\ {B_{x} } & {B_{y} } \end{array}\right|$

Multiplying out the two by two determinants gives the form of the first definition.

The cross product in one sense is a measure of how perpendicular two vectors are. Another definition of the cross product is as

$\mathbf{A}\times\mathbf{B}=\left|\mathbf{A}\right|\left|\mathbf{B}\right|\sin \theta_{AB}\mathbf{n}$

The direction of $latex \mathbf{n}$ is perpendicular to both $latex \mathbf{A}$ and $latex \mathbf{B}$. This formula provides a means to calculate the angle between the two vectors. When the two vectors are more perpendicular the magnitude of the cross product is larger compared to the lengths of the two input vectors. If the cross product is zero then the two vectors are collinear.In another sense the cross product is the area of a parallelogram formed by the two vectors. There are six possible cross products of unit vectors, which can be used to calculate the cross products of general vectors. These relations are useful when “multiplying the cross product out” with the distributive law.

${\begin{array}{ccc} {\mathbf{i}\times \mathbf{j}=+\mathbf{k}} & {\mathbf{j}\times \mathbf{k}=+\mathbf{i}} & {\mathbf{k}\times \mathbf{i}=+\mathbf{j}} \end{array}} \\ {\begin{array}{ccc} {\mathbf{j}\times \mathbf{i}=-\mathbf{k}} & {\mathbf{k}\times \mathbf{j}=-\mathbf{i}} & {\mathbf{i}\times \mathbf{k}=-\mathbf{j}} \end{array}}$

The area of a parallelogram is the base times the height. This can be expressed as

$\left|\mathbf{A}\times \mathbf{B}\right|=AB\sin \theta _{AB}$

The magnitude of the cross product between two vectors is equal to the area of the parallelogram they form.

Example 6. Take the cross product of the following two vectors; find the angle between them, and the direction of the unit vector in the direction of the new vector.

$\mathbf{A}=(1,0,1)\quad \mathbf{B}=(2,-1,1)$

Solution 6. The determinant form of the cross product is generally the quickest way to do the calculation so we write.

${\mathbf{A}\times \mathbf{B}=\mathbf{C}=\left|\begin{array}{ccc} {\mathbf{i}} & {\mathbf{j}} & {\mathbf{k}} \\ {1} & {0} & {1} \\ {2} & {-1} & {1} \end{array}\right|=\mathbf{i}+\mathbf{j}-\mathbf{k}} \\ {C=AB\sin \theta _{AB} =\sqrt{3} =\sqrt{2} \sqrt{6} \sin \theta _{AB} } \\ {\sin \theta _{AB} =\cfrac{1}{2} \qquad \theta _{AB} =\cfrac{\pi }{6} } \\ {\mathbf{n}=\mathbf{C}=\cfrac{\mathbf{C}}{\left|\mathbf{C}\right|} =\cfrac{1}{\sqrt{3} } (\mathbf{i}+\mathbf{j}-\mathbf{k})}$

Example 7. Given the two vectors $\mathbf{A}$ and $\mathbf{B}$ calculate a unit vector normal to the plane A and B form

${\mathbf{A}} = {\mathbf{i}+\mathbf{j}+\mathbf{k}} \\ {\mathbf{B}} = {\mathbf{i}-\mathbf{j}}$

Solution 7. Given two vectors in a plane, a unit vector normal to the plane can be calculated by taking the cross product of the two vectors and converting the result into a unit vector. This only works if the two vectors are not collinear in which case the cross product is zero, and there infinitely many perpendicular vectors.

${\mathbf{C}} = {\mathbf{A}\times \mathbf{B}=(\mathbf{i}+\mathbf{j}+\mathbf{k})\times \mathbf{i}-(\mathbf{i}+\mathbf{j}+\mathbf{k})\times \mathbf{j}} \\ {} = {(\mathbf{j}+\mathbf{k})\times \mathbf{i}-(\mathbf{i}+\mathbf{k})\times \mathbf{j}} \\ {} = {-\mathbf{k}+\mathbf{j}-\mathbf{k}+\mathbf{i}=\mathbf{i}+\mathbf{j}-2\mathbf{k}} \\ {\mathbf{C}} = {\cfrac{1}{\sqrt{6} } (\mathbf{i}+\mathbf{k}-2\mathbf{k})}$

Example 8. Prove the law of cosines using vectors.

Solution 8. The law of cosines relates the lengths of the sides of a triangle and the angle between two of the sides. There are three possible equations relating the sides and an angle. The law of cosines can be easily derived using vectors. Take the vectors $latex \mathbf{A},\mathbf{B},\mathbf{C}$ and form the vector relation

${\mathbf{A}-\mathbf{B}=\mathbf{C}} \\ {(\mathbf{A}-\mathbf{B})\cdot (\mathbf{A}-\mathbf{B})=\mathbf{C}\cdot \mathbf{C}} \\ {\mathbf{A}\cdot \mathbf{A}+\mathbf{B}\cdot \mathbf{B}-2\mathbf{A}\cdot \mathbf{B}=\mathbf{C}\cdot \mathbf{C}} \\ {C^{2} =A^{2} +B^{2} -2AB\cos \theta _{AB} }$

Example 9. Prove the law of sines using vectors.

Solution 9. The law of sines can be easily derived using vectors. We know from the cross product that we get a quantity proportional to the sin of the angle between two vectors.

$\mathbf{A}\times \mathbf{B}=AB\sin \theta \mathbf{n}$

Form a relationship between vectors such that they form a loop

$\mathbf{A}+\mathbf{B}+\mathbf{C}=0$

We can now take the cross product of both sides with the vector A.

${\mathbf{A}\times \mathbf{A}+\mathbf{A}\times \mathbf{B}+\mathbf{A}\times \mathbf{C}=0} \\ {\mathbf{A}\times \mathbf{B}=\mathbf{A}\times (-\mathbf{C})} \\ {AB\sin \theta _{AB} =AC\sin \theta _{AC} } \\ {\cfrac{\sin \theta _{AB} }{C} =\cfrac{\sin \theta _{AC} }{B} }$

The last relation of the law of sines is derived by taking the cross product with $\mathbf{B}$.

${\mathbf{B}\times \mathbf{A}+\mathbf{B}\times \mathbf{B}+\mathbf{B}\times \mathbf{C}=0} \\ {\mathbf{A}\times \mathbf{B}=\mathbf{B}\times \mathbf{C}} \\ {AB\sin \theta _{AB} =BC\sin \theta _{BC} } \\ {\cfrac{\sin \theta _{AB} }{C} =\cfrac{\sin \theta _{BC} }{A} }$

Putting the results together we have the law of sines.

$\cfrac{\sin \theta _{BC} }{A} =\cfrac{\sin \theta _{AC} }{B} =\cfrac{\sin \theta _{AB} }{C}$

There is a useful calculation which involves both the dot product and the cross product. The triple product corresponds to the volume of the parallelpiped specified by the three vectors $latex \mathbf{A},\mathbf{B},\, {\rm and}\, \mathbf{C}$. The triple product is defined as $latex \mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})$

${Volume=\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})=\mathbf{C}\cdot (\mathbf{A}\times \mathbf{B})=\mathbf{B}\cdot (\mathbf{C}\times \mathbf{A})}$

${\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})=\left|\begin{array}{ccc} {A_{x} } & {A_{y} } & {A_{z} } \\ {B_{x} } & {B_{y} } & {B_{z} } \\ {C_{x} } & {C_{y} } & {C_{z} } \end{array}\right|}$

Due to the law for permuting rows of a determinant the cyclical property follows. Interchanging any two rows in a determinant changes the sign of the determinant by -1. Also, if the triple product is zero, this means that the three vectors are coplanar because the volume of the parallel piped they form is zero.

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