[b]Dot Products[/b][br][br]Given any two vectors [math]\vec{u}[/math] and [math]\vec{v}[/math] in [math]\mathbb{R}^2[/math] or [math]\mathbb{R}^3[/math], we define their [b]dot product[/b] as follows:[br][br]In [math]\mathbb{R}^2[/math], [math]\vec{u}=\langle u_1,u_2\rangle[/math] and [math]\vec{v}=\langle v_1,v_2\rangle[/math],[br][br][math]\vec{u}\cdot\vec{v}=u_1v_1+u_2v_2[/math][br][br]In [math]\mathbb{R}^3[/math], [math]\vec{u}=\langle u_1,u_2,u_3\rangle[/math] and [math]\vec{v}=\langle v_1,v_2,v_3\rangle[/math],[br][br][math]\vec{u}\cdot\vec{v}=u_1v_1+u_2v_2+u_3v_3[/math][br][br][u]Remark[/u]: The dot product of any two vectors is a scalar i.e. a real number.[br][br][br][br]We can easily derive the following properties of dot product from its definition:[br][br]Let [math]\vec{u},\vec{v},\vec{w}[/math] be vectors in [math]\mathbb{R}^2[/math] or [math]\mathbb{R}[br]^3[/math] and [math]k[/math] be a real number.[br][br]1. [math]\vec{u}\cdot\vec{v}=\vec{v}\cdot\vec{u}[/math][br]2. [math]\vec{u}\cdot(\vec{v}+\vec{w})=\vec{u}\cdot\vec{v}+\vec{u}\cdot\vec{w}[/math][br]3. [math]k(\vec{u}\cdot\vec{v})=(k\vec{u})\cdot\vec{v}=\vec{u}\cdot(k\vec{v})[/math][br]4. [math]\vec{u}\cdot\vec{u}=|\vec{u}|^2[/math][br]5. [math]\vec{0}\cdot\vec{u}=0[/math][br]

It turns out that for any two non-zero vectors [math]\vec{u}[/math] and [math]\vec{v}[/math] in [math]\mathbb{R}^2[/math] or [math]\mathbb{R}^3[/math], it can be shown that their dot product can also be expressed as follows:[br][br][math]\vec{u}\cdot \vec{v}=|\vec{u}| |\vec{v}| \cos\theta [/math][br][br]when both vectors are non-zero and [math]\theta[/math] is the angle in the range from [math]0[/math] to [math]\pi[/math] formed by translating two non-zero vectors such that their tails meet at a point.[br][br]In the applet below, bring the two vectors together in the plane to find out their dot product.[br]

[u]Definition[/u]: Two vectors are [b]orthogonal[/b] if their dot product is zero i.e. if two vectors are orthogonal, they are either both non-zero and perpendicular to each other, or at least one of the vectors is a zero vector. [br][br][u]Remark[/u]: Zero vector is orthogonal to any vector.

[u]Theorem[/u]: Given any two vectors [math]\vec{u}[/math] and [math]\vec{v}[/math] in [math]\mathbb{R}^2[/math] or [math]\mathbb{R}^3[/math], [br][br][math]\vec{u}\cdot\vec{v}=|\vec{u}||\vec{v}|\cos(\theta)[/math][br][br]The proof is as follows:[br]

[u]Orthogonal Projections[/u][br][br]Given non-zero vectors [math]\vec{u}[/math] and [math]\vec{v}[/math], let [math]L[/math] be the line through the tail of [math]\vec{u}[/math] in the direction of [math]\vec{v}[/math]. We define the [b](orthogonal) projection[/b] of [math]\vec{u}[/math] onto [math]\vec{v}[/math], denoted by [math]\text{Proj}_{\vec{v}}\vec{u}[/math], to be the vector pointing from the tail of [math]\vec{u}[/math] such that its head is the foot of the perpendicular from the head of [math]\vec{u}[/math] to the the line [math]L[/math].[br][br][br]In the applet below, the red vector is the projection of [math]\vec{u}[/math] onto [math]\vec{v}[/math]. [br][br]If [math]\theta< \frac{\pi}2[/math], its magnitude is [math]|\vec{u}|\cos\theta[/math] and it is in the same direction as [math]\vec{v}[/math]. Hence, we have[br][br][math]\text{Proj}_{\vec{v}}\vec{u}=|\vec{u}|\cos\theta \frac{\vec{v}}{|\vec{v}|}[/math] ([math]\star[/math])[br][br]If [math]\frac{\pi}2 <\theta\leq\pi[/math], its magnitude is [math]-|\vec{u}|\cos\theta[/math] and it is in opposite direction to [math]\vec{v}[/math]. Hence, we have[br][br][math]\text{Proj}_{\vec{v}}\vec{u}=-|\vec{u}|\cos\theta \left(-\frac{\vec{v}}{|\vec{v}|}\right)=|\vec{u}|\cos\theta \frac{\vec{v}}{|\vec{v}|}[/math][br][br](Note: When [math]\theta=\frac{\pi}2[/math], the tail of [math]\vec{u}[/math] is the foot of the perpendicular and hence [math]\text{Proj}_{\vec{v}}\vec{u}=\vec{0}[/math].)[br][br]Combining above, we have the formula for the projection:[br][br][math]\text{Proj}_{\vec{v}}\vec{u}=|\vec{u}|\cos\theta \frac{\vec{v}}{|\vec{v}|}[/math][br][br]Using the dot product, we can also rewrite the formula as follows:[br][br][math]\text{Proj}_{\vec{v}}\vec{u}=\frac{|\vec{u}||\vec{v}|\cos\theta}{|\vec{v}|} \frac{\vec{v}}{|\vec{v}|}=\left(\frac{\vec{u}\cdot\vec{v}}{\vec{v}\cdot\vec{v}}\right)\vec{v}[/math].[br][br]In ([math]\star[/math]), the factor [math]|\vec{u}|\cos\theta[/math] multiplying to the unit vector in the direction of [math]\vec{v}[/math] is called the [b]scalar projection[/b] of [math]\vec{u}[/math] onto [math]\vec{v}[/math], denoted by [math]\text{Comp}_{\vec{v}}\vec{u}[/math] i.e.[br][br][math]\text{Comp}_{\vec{v}}\vec{u}=|\vec{u}|\cos\theta=\frac{\vec{u}\cdot\vec{v}}{|\vec{v}|}[/math][br][br]It is also called the [b]scalar component[/b] of [math]\vec{u}[/math] in the direction of [math]\vec{v}[/math].[br][br][br][br][br]

[u]An Application - Work[/u][br][br]A constant force is applied to an object. We define the [b]work done (W)[/b] by the force over the displacement of the object as follows:

Therefore, we have [math] W=\vec{F}\cdot \vec{d}[/math].[br][br][u]Example[/u]: Given a force [math]\vec{F}=\langle 3,2,1\rangle[/math] that moves an object from [math]P(1,-1,0)[/math] to [math]Q(6,5,1)[/math]. Find the work done by [math]\vec{F}[/math].[br][br][math]\vec{d}=\overrightarrow{PQ}=\langle 5,6,1\rangle[/math]. Therefore, we have[br][math]W=\vec{F}\cdot\vec{d}=\langle 3,2,1\rangle \cdot \langle 5,6,1\rangle=15+12+1=28 J[/math] (J stands for "joule", the unit for energy i.e. work done).[br][br][br]

[size=150][b]Cross Product[/b][/size][br][br]Given two vectors [math]\vec{u}=\langle u_1,u_2,u_3\rangle[/math] and [math]\vec{v}=\langle v_1,v_2,v_3\rangle[/math] in [math]\mathbb{R}^3[/math], we are going to define the [b]cross product[/b] of these two vectors. But first of all, we need to recall the definition of the [b]determinant[/b] of a matrix:[br][br]Suppose [math]\begin{pmatrix} a & b \\ c & d\end{pmatrix}[/math] is a [math]2\times 2[/math] matrix. Its determinant [math]\begin{vmatrix} a & b \\ c & d\end{vmatrix}=ad-bc[/math][br][br]Then we can define the determinant of a [math]3\times 3[/math] matrix [math]\begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{pmatrix}[/math] as follows:[br][br] [math]\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix}=a_1\begin{vmatrix} b_2 & b_3 \\ c_2 & c_3\end{vmatrix}-a_2\begin{vmatrix} b_1 & b_3 \\ c_1 & c_3\end{vmatrix}+a_3\begin{vmatrix} b_1 & b_2 \\ c_1 & c_2\end{vmatrix}[/math][br][br][u]Example[/u]:[br][br][math]\begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ -1 & 2 & 0\end{vmatrix}=1\begin{vmatrix} 1 & 1 \\ 2 & 0\end{vmatrix}-2\begin{vmatrix} 0 & 1 \\ -1 & 0\end{vmatrix}+3\begin{vmatrix} 0 & 1 \\ -1 & 2\end{vmatrix}=1\cdot(-2)-2\cdot 1+3\cdot 1=-1[/math][br][br]Write [math]\vec{u}=u_1\vec{i}+u_2\vec{j}+u_3\vec{k}[/math] and [math]\vec{v}=v_1\vec{i}+v_2\vec{j}+v_3\vec{k}[/math]. The [b]cross product[/b], [math]\vec{u}\times\vec{v}[/math] is defined as follows:[br][br][math]\vec{u}\times\vec{v}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3\end{vmatrix}=\vec{i}\begin{vmatrix} u_2 & u_3 \\ v_2 & v_3\end{vmatrix}-\vec{j}\begin{vmatrix} u_1 & u_3 \\ v_1 & v_3\end{vmatrix}+\vec{k}\begin{vmatrix} u_1 & u_2 \\ v_1 & v_2\end{vmatrix}[/math][br][br][u]Remark[/u]: [br][list][*]Cross product of two vectors in [math]\mathbb{R}^3[/math] is a vector in [math]\mathbb{R}^3[/math].[br][/*][*]The order of taking cross product is important: In general, [math]\vec{u}\times\vec{v}\ne \vec{v}\times\vec{u}[/math] [/*][/list][br][br][u]Example[/u]:[br][br]Given [math]\vec{u}=\vec{i}+2\vec{j}+3\vec{k}[/math] and [math]\vec{v}=4\vec{i}+5\vec{j}+6\vec{k}[/math]. We have[br][br][math]\vec{u}\times\vec{v}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 2 & 3 \\ 4 & 5 & 6\end{vmatrix}=-3\vec{i}-(-6)\vec{j}+(-3)\vec{k}=-3\vec{i}+6\vec{j}-3\vec{k}[/math].[br][br]

[u]Properties of Cross Product[/u][br][br]Recall that if we interchange any two rows of a [math]3\times 3[/math] matrix, its determinant will be changed as follows:[br][br][math]\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix}=-\begin{vmatrix} b_1 & b_2 & b_3 \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3\end{vmatrix}[/math][br][br][math]\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix}=-[br]\begin{vmatrix} a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 \\ b_1 & b_2 & b_3 \end{vmatrix}[/math][br][br][math]\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix}=-[br]\begin{vmatrix} c_1 & c_2 & c_3 \\ b_1 & b_2 & b_3 \\ a_1 & a_2 & a_3 \end{vmatrix}[/math][br][br]Therefore, by the definition of cross product, we have the following properties:[br][br][list=1][*][math]\vec{u}\times\vec{v}=-\vec{v}\times\vec{u}[/math] and in particular, [math]\vec{u}\times\vec{u}=\vec{0}[/math].[/*][*]Suppose [math]\vec{w}=w_1\vec{i}+w_2\vec{j}+w_3\vec{k}[/math]. [math]\vec{w}\cdot(\vec{u}\times\vec{v})=\begin{vmatrix} w_1 & w_2 & w_3 \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3\end{vmatrix} [/math] (Note: In some textbooks, this is called the [b]scalar triple product[/b]). In particular, [math]\vec{u}\cdot(\vec{u}\times\vec{v})=\vec{v}\cdot(\vec{u}\times\vec{v})=0[/math] i.e. [math]\vec{u}\times\vec{v}[/math] is orthogonal to both [math]\vec{u}[/math] and [math]\vec{v}[/math]. [/*][*]Let [math]\lambda[/math] and [math]\mu[/math] be real numbers, then [math](\lambda\vec{u})\times(\mu\vec{v})=\lambda\mu(\vec{u}\times\vec{v})[/math]. [/*][*]For any vectors [math]\vec{u},\vec{v},\vec{w}[/math], [math]\vec{u}\times (\vec{v}+\vec{w})=\vec{u}\times\vec{v}+\vec{u}\times\vec{w}[/math] and [math] (\vec{u}+\vec{v})\times\vec{w}=\vec{u}\times\vec{w}+\vec{v}\times\vec{w}[/math]. (Distributive laws).[/*][/list][br][br]For any two non-parallel vectors [math]\vec{u}[/math] and [math]\vec{v}[/math], there are exactly two possible directions that are orthogonal to both vectors. We can use the following [b]right-hand rule[/b] to determine the direction of the cross product:[br][br]

In the applet below, you can do the following: [br][br][list=1][*]First construct two vectors [math]\vec{u}[/math] and [math]\vec{v}[/math] by typing "u=vector((1,2,3))" and "v=vector((4,5,6))".[/*][*]Type "cross(u,v)" to get the cross product [math]\vec{u}\times\vec{v}[/math].[/*][*]Draw the plane containing the two vectors through the origin by typing "plane((0,0,0),u,v)". Then you will see that the cross product is perpendicular to the plane.[/*][*]Verify the right-hand rule using some other vectors.[/*][*]You can also find [math]\vec{i}\times\vec{j}, \vec{j}\times\vec{k}[/math], and [math]\vec{k}\times\vec{i}[/math]. [/*][/list]

We have the following nice theorem about the norm of a cross product:[br][br][u]Theorem[/u]: [math]|\vec{u}\times\vec{v}|=|\vec{u}||\vec{v}|\sin\theta[/math], where [math]\theta[/math] is the angle between the two vectors.[br][br][u]Proof[/u]: [br][br][math]|\vec{u}||\vec{v}|\sin\theta=|\vec{u}||\vec{v}|\sqrt{1-\cos^2\theta}[/math] (Note: positive square root here because [math]\sin\theta\geq 0[/math] for [math]0\leq\theta\leq\pi[/math]).[br][br][math]=|\vec{u}||\vec{v}|\sqrt{1-\left(\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}\right)^2}=\sqrt{|\vec{u}|^2|\vec{v}|^2-(\vec{u}\cdot\vec{v})^2}[/math][br][math]=\sqrt{(u_1^2+u_2^2+u_3^2)(v_1^2+v_2^2+v_3^2)-(u_1v_1+u_2v_2+u_3v_3)^2}[/math][br][math]=\sqrt{(u_2v_3-u_3v_2)^2+(u_1v_3-u_3v_1)^2+(u_1v_2-u_2v_1)^2}=|\vec{u}\times\vec{v}|[/math][br][br][br][u]Remark[/u]: [math]|\vec{u}\times\vec{v}|[/math] is exactly the area of the parallelogram spanned by [math]\vec{u}[/math] and [math]\vec{v}[/math].[br][br]

[u]The Volume of a Parallelepiped[/u][br][br]Given three vectors [math]\vec{u}, \vec{v}, \vec{w}[/math] in [math]\mathbb{R}^3[/math]. We can compute its scalar triple product [math]\vec{w}\cdot(\vec{u}\times\vec{v})[/math]. It turns out that it is related to the volume of the parallelepiped formed by these three vectors, as shown in the applet below. [br][br]In general, the volume of a parallelepiped is the product of the area of its base parallelogram and its height (the distance between the two planes that contain the base and top respectively). By the previous theorem, the area of the base parallelogram is [math]|\vec{u}\times\vec{v}|[/math]. Moreover, the height of the parallelepiped is [math]|\text{Proj}_{\vec{u}\times\vec{v}}\vec{w}|[/math]. Therefore, we have[br][br]Volume = [math]|\text{Proj}_{\vec{u}\times\vec{v}}\vec{w}||\vec{u}\times\vec{v}|=\left|\left(\frac{\vec{w}\cdot(\vec{u}\times\vec{v})}{|\vec{u}\times\vec{v}|^2}\right)\vec{u}\times\vec{v}\right||\vec{u}\times\vec{v}|=|\vec{w}\cdot(\vec{u}\times\vec{v})|[/math][br][br]In the applet below, the dotted line contains [math]\vec{u}\times\vec{v}[/math], which is perpendicular to both planes containing the top and base of the parallelepiped. The red vector is [math]\text{Proj}_{\vec{u}\times\vec{v}}\vec{w}[/math] and its length is exactly the height of the parallelepiped.[br][br]In this example, [math]\vec{u}=\langle 2,-2,1\rangle, \vec{v}=\langle 1,2,1 \rangle, \vec{w}=\langle -2,-2,2\rangle[/math]. By the right-hand rule, the red vector [math]\text{Proj}_{\vec{u}\times\vec{v}}\vec{w}[/math] is in the same direction as [math]\vec{u}\times\vec{v}[/math]. However, if you interchange [math]\vec{u}[/math] and [math]\vec{v}[/math], the red vector will point to the opposite direction. Its length is still the height of the parallelepiped though.[br][br]Moreover, you can check that the sign of [math]\vec{w}\cdot(\vec{u}\times\vec{v})[/math] to see whether [math]\vec{w}[/math] and [math]\vec{u}\times\vec{v}[/math] are in the same side of the plane containing the base.[br][br]