# Arc Length and Reparametrization

We have already discussed the length of a plane curve represented by the parametric equation ${\bf r}(t)=(x(t),y(t))$, $a\leq t\leq b$ here. The same goes for a space curve. Namely, if ${\bf r}(t)=(x(t),y(t),z(t))$, $a\leq t\leq b$, then its arc length $L$ is given by \begin{aligned}L&=\int_a^b|{\bf r}'(t)|dt\\&=\int_a^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2}dt\end{aligned}\label{eq:spacearclength}

Example. Find the length of the arc of the circular helix $${\bf r}(t)=\cos t{\bf i}+\sin t{\bf j}+t{\bf k}$$ from the point $(1,0,0)$ to the point $(1,0,2\pi)$.

Solution. ${\bf r}'(t)=-\sin t{\bf i}+\cos t{\bf j}+{\bf k}$ so we have $$|{\bf r}'(t)|=\sqrt{(-\sin t)^2+(\cos t)^2+1^2}=\sqrt{2}$$ The arc is going from $(1,0,0)$ to $(1,0,2\pi)$ and the $z$-component of ${\bf r}(t)$ is $t$, so $0\leq t\leq 2\pi$. Now, using \eqref{eq:spacearclength}, we obtain $$L=\int_0^{2\pi}|{\bf r}'(t)|dt=\int_0^{2\pi}\sqrt{2}dt=2\sqrt{2}\pi$$ Figure 1 shows the circular helix from $t=0$ to $t=2\pi$.

Given a curve ${\bf r}(t)$, $a\leq t\leq b$, sometimes we need to reparametrize it by another parameter $s$ for various reasons. Imagine that the curve represents the path of a particle moving in space. A reparametrization does not change the path of the particle (hence nor the distance it traveled) but it changes the particle’s speed! To see this, let $t=t(s)$, $\alpha\leq s\leq\beta$, $a=t(\alpha)$, $b=t(\beta)$ be an increasing and differentiable function. Since $t=t(s)$ is one-to-one and onto, ${\bf r}(t)$ and ${\bf r}(t(s))$, its reparametrization by the parameter $s$, represent the same path. By the chain rule, $$\label{eq:reparametrization}\frac{d}{ds}{\bf r}(t(s))=\frac{d}{dt}{\bf r}(t)\frac{dt}{ds}$$ Thus we see that the speed of the reparametrization $\left|{\bf r}(t(s))\right|$ differs from that of ${\bf r}(t)$ by a factor of $\left|\frac{dt}{ds}\right|=\frac{dt}{ds}$ (since $\frac{dt}{ds}>0$). However, the arc length of the reparametrization is \begin{align*}\int_{\alpha}^{\beta}\left|\frac{d}{ds}{\bf r}(t(s))\right|ds&=\int_{\alpha}^{\beta}\left|\frac{d}{dt}{\bf r}(t)\right|\frac{dt}{ds}ds\\&=\int_a^b\left|\frac{d}{dt}{\bf r}(t)\right|dt=L\end{align*} That is, no change of the distance.

There is a particular reparametrization we are interested. To discuss that, suppose ${\bf r}(t)$, $a\leq t\leq b$ be a differentiable curve in space such that ${\bf r}'(t)\ne 0$ for all $t$. Such a curve is said to be regular or smooth. Let us now define the arc length function $$\label{eq:arclengthfunction}s(t)=\int_a^t|{\bf r}'(u)|du$$ By the Fundamental Theorem of Calculus, we have $$\label{eq:arclengthfunction2}\frac{ds}{dt}=|{\bf r}'(t)|>0$$ and so the arc length function $s=s(t)$ is increasing. This means that $s(t)$ is one-to-one and onto, so it is invertible. It’s inverse function can be written as $t=t(s)$ and ${\bf r}(t(s))$ is called the reparamerization by arc length. The reason we are interested in this particular reparametrization is that it results in the unit speed: From \eqref{eq:reparametrization} and \eqref{eq:arclengthfunction2}, $$\left|\frac{d}{ds}{\bf r}(t(s))\right|=|{\bf r}'(t)|\left|\frac{dt}{ds}\right|=|{\bf r}'(t)|\frac{1}{|{\bf r}'(t)|}=1$$ So it is also called the unit-speed reparametrization. The reparametrization by arc length plays an important role in defining the curvature of a curve. This will be discussed elsewhere.

Example. Reparametrize the helix ${\bf r}(t)=\cos t{\bf i}+\sin t{\bf j}+t{\bf k}$ by arc length measured from $(1,0,0)$in the direction of increasing $t$.

Solution. The initial point $(1,0,0)$ corresponds to $t=0$. From the previous example, we know that the helix has the constant speed $|{\bf r}'(t)|=\sqrt{2}$. Thus, $$s(t)=\int_0^t|{\bf r}'(u)|du=\sqrt{2}t$$ Hence, we obtain $t=\frac{s}{\sqrt{2}}$. The reparametrization is then given by $${\bf r}(t(s))=\cos\left(\frac{s}{\sqrt{2}}\right){\bf i}+\sin\left(\frac{s}{\sqrt{2}}\right){\bf j}+\frac{s}{\sqrt{2}}{\bf k}$$

Examples in this note have been taken from [1].

References.

[1] Calculus, Early Transcendentals, James Stewart, 6th Edition, Thompson Brooks/Cole

# Derivatives and Integrals of Vector-Valued Functions

The derivative $\frac{d{\bf r}}{dt}={\bf r}'(t)$ of a vector-valued function ${\bf r}(t)=(x(t),y(t),z(t))$ is defined by $$\label{eq:vectorderivative}\frac{d{\bf r}}{dt}=\lim_{\Delta t\to 0}\frac{{\bf r}(t+\Delta t)-{\bf r}(t)}{\Delta t}$$ In case of a scalar-valued function or a real-valued function, the geometric meaning of derivative is that it is the slope of tangent line. In case of a vector-valued function, the geometric meaning of derivative is that it is the tangent vector. It can be easily seen from Figure 1. As $\Delta t$ gets smaller and smaller, $\frac{{\bf r}(t+\Delta t)-{\bf r}(t)}{\Delta t}$ gets closer to a line tangent to ${\bf r}(t)$.

From the definition \eqref{eq:vectorderivative}, it is straightforward to show $$\label{eq:vectorderivative2}{\bf r}'(t)=(x'(t),y'(t),z'(t))$$

If ${\bf r}'(t)\ne 0$, the unit tangent vector ${\bf T}(t)$ of ${\bf r}(t)$ is given by $$\label{eq:unittangent}{\bf T}(t)=\frac{{\bf r}'(t)}{|{\bf r}'(t)|}$$

Example.

1. Find the derivative of ${\bf r}(t)=(1+t^3){\bf i}+te^{-t}{\bf j}+\sin 2t{\bf k}$.
2. Find the unit tangent vector when $t=0$.

Solution.

1. Using \eqref{eq:vectorderivative2}, we have $${\bf r}'(t)=3t^2{\bf i}+(1-t)e^{-t}{\bf j}+2\cos 2t{\bf k}$$
2. ${\bf r}'(0)={\bf j}+2{\bf k}$ and $|{\bf r}'(0)|=\sqrt{5}$. So by \eqref{eq:unittangent}, we have $${\bf T}(0)=\frac{{\bf r}'(0)}{|{\bf r}'(0)|}=\frac{1}{\sqrt{5}}{\bf j}+\frac{2}{\sqrt{5}}{\bf k}$$

The following theorem is a summary of differentiation rules for vector-valued functions. We omit the proofs of these rules. They can be proved straightforwardly from differentiation rules for real-valued functions. Note that there are three different types of the product rule or the Leibniz rule for vector-valued functions (rules 3, 4, and 5).

Theorem. Let ${\bf u}(t)$ and ${\bf v}(t)$ be differentiable vector-valued functions, $f(t)$ a scalar function, and $c$ a scalar. Then

1. $\frac{d}{t}[{\bf u}(t)+{\bf v}(t)]={\bf u}'(t)+{\bf v}'(t)$
2. $\frac{d}{dt}[c{\bf u}(t)]=c{\bf u}'(t)$
3. $\frac{d}{dt}[f(t){\bf u}(t)]=f'(t){\bf u}(t)+f(t){\bf u}'(t)$
4. $\frac{d}{dt}[{\bf u}(t)\cdot{\bf v}(t)]={\bf u}'(t)\cdot{\bf v}(t)+{\bf u}(t)\cdot{\bf v}'(t)$
5. $\frac{d}{dt}[{\bf u}(t)\times{\bf v}(t)]={\bf u}'(t)\times{\bf v}(t)+{\bf u}(t)\times{\bf v}'(t)$
6. $\frac{d}{dt}[{\bf u}(f(t))]=f'(t){\bf u}'(f(t))$ (Chain Rule)

Example. Show that if $|{\bf r}(t)|=c$ (a constant), then ${\bf r}'(t)$ is orthogonal to ${\bf r}(t)$ for all $t$.

Proof. Differentiating ${\bf r}(t)\cdot{\bf r}(t)=|{\bf r}(t)|^2=c^2$ using the Leibniz rule 5, we obtain $$0=\frac{d}{dt}[{\bf r}(t)\cdot{\bf r}(t)]={\bf r}'(t){\bf r}(t)+{\bf r}(t){\bf r}'(t)=2{\bf r}'(t)\cdot{\bf r}(t)$$ This implies that ${\bf r}'(t)$ is orthogonal to ${\bf r}(t)$.

As seen in \eqref{eq:vectorderivative2}, the derivative of a vector-valued functions is obtained by differentiating component-wise. The indefinite and definite integral of a vector-valued function are done similarly by integrating component-wise, namely$$\label{eq:vectorintegral}\int{\bf r}(t)dt=\left(\int x(t)dt\right){\bf i}+\left(\int y(t)dt\right){\bf j}+\left(\int z(t)dt\right){\bf k}$$ and $$\label{eq:vectorintegral2}\int_a^b{\bf r}(t)dt=\left(\int_a^b x(t)dt\right){\bf i}+\left(\int_a^b y(t)dt\right){\bf j}+\left(\int_a^b z(t)dt\right){\bf k}$$, respectively. When evaluate the definite integral of a vector-valued function, one can use \eqref{eq:vectorintegral2} but it would be easier to first find the indefinite integral using \eqref{eq:vectorintegral} and then evaluate the definite integral the Fundamental Theorem of Calculus (and yes, the Fundamental Theorem of Calculus still works for vector-valued functions).

Example. Find $\int_0^{\frac{\pi}{2}}{\bf r}(t)dt$ if ${\bf r}(t)=2\cos t{\bf i}+\sin t{\bf j}+2t{\bf k}$.

Solution. \begin{align*}\int{\bf r}(t)dt&=\left(\int 2\cos tdt\right){\bf i}+\left(\int \sin tdt\right){\bf j}+\left(\int 2tdt\right){\bf k}\\&=2\sin t{\bf i}-\cos t{\bf j}+t^2{\bf k}+{\bf C}\end{align*} where ${\bf C}$ is a vector-valued constant of integration. Now, $$\int_0^{\frac{\pi}{2}}{\bf r}(t)dt=[2\sin t{\bf i}-\cos t{\bf j}+t^2{\bf k}]_0^{\frac{\pi}{2}}=2{\bf i}+{\bf j}+\frac{\pi^2}{4}{\bf k}$$

Examples in this note have been taken from [1].

References.

[1] Calculus, Early Transcendentals, James Stewart, 6th Edition, Thompson Brooks/Cole

# Lines and Planes in Space

You remember from algebra that a line in the plane can be determined by its slope and a point on the line or two points on the line (in which case those two points determine the slope). For a space line, slope is not a suitable ingredient. It’s alternative ingredient is a vector parallel to the line. As shown in Figure 1, with a point ${\bf r}_0=(x_0,y_0,z_0)$ on the line $L$ and a vector ${\bf v}=(a,b,c)$ parallel to $L$, we can determine any point ${\bf r}=(x,y,z)$ on $L$ by vector addition of ${\bf r}_0$ and $t{\bf v}$, a dilation of ${\bf v}$: $$\label{eq:spaceline}{\bf r}={\bf r}_0+t{\bf v}$$ where $-\infty<t<\infty$. The equation \eqref{eq:spaceline} is called a vector equation of $L$.

In terms of the components, \eqref{eq:spaceline} can be written as \begin{aligned}x&=x_0+at\\y&=y_0+bt\\z&=z_0+ct\end{aligned}\label{eq:spaceline2} where $-\infty<t<\infty$. The equations in \eqref{eq:spaceline2} are called parametric equations of $L$. Solving the parametric equations in \eqref{eq:spaceline2} for $t$, we also obtain $$\label{eq:spaceline3}\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$$ The equations in \eqref{eq:spaceline3} are called symmetric equations of $L$. Often we need to work with a line segment. The vector equation \eqref{eq:spaceline} can be used to figure out an equation of a line segment. Consider the line segment from ${\bf r}_0$ to ${\bf r}_1$. Then the difference ${\bf r}_1-{\bf r}_0$ is a vector parallel to the line segment and by \eqref{eq:spaceline}, we have \begin{align*}{\bf r}(t)&={\bf r}_0+t({\bf r}_1-{\bf r}_0)\\&=(1-t){\bf r}_0+t{\bf r}_1\end{align*} This still represents an infinite line through ${\bf r}_0$ and ${\bf r}_1$. By limiting the range of $t$ to represent only the line segment from ${\bf r}_0$ to ${\bf r}_1$, we obtain an equation of the line segment $$\label{eq:linesegment}{\bf r}(t)=(1-t){\bf r}_0+t{\bf r}_1$$ where $0\leq t\leq 1$.

Example.

1. Find a vector equation and parametric equations for the line that passes through the point $(5,1,3)$ and is parallel to the vector ${\bf i}+4{\bf j}-2{\bf k}$.
2. Find two other points on the line.

Solution.

1. ${\bf r}_0=(5,1,3)$ and ${\bf v}=(1,4,-2)$. Hence by \eqref{eq:spaceline}, we have $${\bf r}(t)=(5,1,3)+t(1,4,-2)=(5+t,1+4t,3-2t)$$ Parametric equations are $$x(t)=5+t,\ y(t)=1+4t,\ z(t)=3-2t$$
2. From the parametric equations, for example, $t=1$ gives $(6,5,1)$ and $t=-1$ gives $(4,-3,5)$.

Example.

1. Find parametric equations and symmetric equations of the line that passes through the points $A(2,4,-3)$ and $B(3,-1,1)$.
2. At what point does this line intersect the $xy$-plane?

Solution.

1. Note that the vector ${\bf v}=\overrightarrow{AB}=(1, -5, 4)$ is parallel to the line. So $a=1$, $b=-5$, and $c=4$. By taking ${\bf r}_0=(x_0,y_0,z_0)=(2,4,-3)$, we have the parametric equations $$x=2+t,\ y=4-5t,\ z=-3+4t$$ Symmetric equations are then $$\frac{x-2}{1}=\frac{y-4}{-5}=\frac{z+3}{4}$$
2. The line intersects the $xy$-plane when $z=0$. By setting $z=0$ in the symmetric equations from part 1, we get $$\frac{x-2}{1}=\frac{y-4}{-5}=\frac{3}{4}$$ Solving these equations for $x$ and $y$ respectively, we find $x=\frac{11}{4}$ and $y=\frac{1}{4}$.

As we have seen, a line is determined by a point on the line and a vector parallel to the line. A plane, on the other hand, can be determined by a point ${\bf r}_0$ on the plane and a vector ${\bf n}$ perpendicular to the plane (such a vector is called a normal vector to the plane). See Figure 2.

From Figure 2, we see that $$\label{eq:plane}{\bf n}\cdot({\bf r}-{\bf r}_0)=0$$ The equation \eqref{eq:plane} is called a vector equation of the plane. If ${\bf n}=(a,b,c)$, ${\bf r}=(x,y,z)$, and ${\bf r}_0=(x_0,y_0,z_0)$, then the equation \eqref{eq:plane} can be written as $$\label{eq:plane2}a(x-x_0)+b(y-y_0)+c(z-z_0)=0$$

Example. Find an equation of the plane through the point $(2,4,-1)$ with normal vector ${\bf n}=(2,3,4)$. Find the intercepts and sketch the plane.

Solution. $a=2$, $b=3$, $c=4$, $x_0=2$, $y_0=4$, and $z_0=-1$. So by the equation \eqref{eq:plane2}, we have $$2(x-2)+3(y-4)+4(z+1)=0$$ or $$2x+3y+4z=12$$ To find the $x$-intercept, set $y=z=0$ in the equation and we get $x=6$. Similarly, we find the $y$- and $z$-intercepts $y=4$ and $z=3$, respectively. Figure 3 shows the plane.

Example. Find an equation of the plane that passes through the points $P(1,3,2)$, $Q(3,-1,6)$, and $R(5,2,0)$.

Solution. The vectors $\overrightarrow{PQ}=(2,-4,4)$ and $\overrightarrow{PR}=(4,-1,-2)$ are on the plane, so the cross product $$\overrightarrow{PQ}\times\overrightarrow{PR}=\begin{vmatrix}{\bf i} & {\bf j} & {\bf k}\\2 & -4 & 4\\4 & -1 & -2\end{vmatrix}=12{\bf i}+20{\bf j}+14{\bf k}$$ is a normal vector to the plane. With $(x_0,y_0,z_0)=(1,3,2)$ and $(a,b,c)=(12,20,14)$, we find an equation of the plane $$12(x-1)+20(y-3)+14(z-2)=0$$ or $$6x+10y+7z=50$$

Using basic geometry, we see that the angle between two planes $P_1$ and $P_2$ is the same as the angle between their respective normal vectors ${\bf n}_1$ and ${\bf n}_2$. See Figure 4 where cross sections of two planes $P_1$ and $P_2$ are shown.

Example.

1. Find the angle between the planes $x+y+z=1$ and $x-2y+3z=1$.
2. Find the symmetric equations for the line of intersection $L$ of these two planes

Solution.

1. The normal vectors of these planes are ${\bf n}_1=(1,1,1)$ and ${\bf n}_2=(1,-2,3)$. Let $\theta$ be the angle between ${\bf n}_1$ and ${\bf n}_2$ . Then $$\cos\theta=\frac{{\bf n}_1\cdot{\bf n }_2}{|{\bf n}_1||{\bf n}_2|}=\frac{1(1)+1(-2)+1(3)}{\sqrt{1^2+1^2+1^2}\sqrt{1^2+(-2)^2+3^2}}=\frac{2}{\sqrt{42}}$$ Thus, $$\theta=\cos^{-1}\left(\frac{2}{\sqrt{42}}\right)\approx 72^\circ$$
2. First, let us find a point on $L$. Set $z=0$. Then we have $x+y=1$ and $x-2y=1$. Solving these two equations simultaneously we find $x=1$ and $y=0$. So $(1,1,0)$ is on the line $L$. Now we need a vector parallel to the line $L$. The cross product $${\bf n}_1\times{\bf n}_2=\begin{vmatrix}{\bf i} & {\bf j} & {\bf k}\\1 & 1 & 1\\1 & -2 & 3\end{vmatrix}=5{\bf i}-2{\bf j}-3{\bf k}$$ is perpendicular to both ${\bf n}_1$ and ${\bf n}_2$, hence it is parallel to $L$. Therefore, the symmetric equations of $L$ are $$\frac{x-1}{5}=\frac{y}{-5}=\frac{z}{-3}$$

Let us find the distance $D$ from a point $Q(x_1,y_1,z_1)$ to the plane $ax+by+cz+d=0$. See Figure 5.

Let $P(x_0,y_0,z_0)$ be a point in the plane and let ${\bf b}=\overrightarrow{PQ}=(x_1-x_0,y_1-y_0,z_1-z_0)$. Then from Figure 5, we see that the distance from $Q$ to the plane is given by the scalar projection of ${\bf b}$ onto the normal vector ${\bf n}$: \begin{align*}D&=|\mathrm{comp}_{\bf n}{\bf b}|=\frac{|{\bf b}\cdot{\bf n}|}{|{\bf n}|}\\&=\frac{|a(x_1-x_0)+b(y_1-y_0)+c(z_1-z_0)|}{\sqrt{a^2+b^2+c^2}}\\&=\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}\end{align*} The last expression is obtained by the fact that $ax_0+by_0+cz_0=-d$. Therefore, the distance $D$ from a point $Q(x_1,y_1,z_1)$ to the plane $ax+by+cz+d=0$ is $$\label{eq:distance2plane}D=\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$$

Example. Find the distance between the parallel planes $10x+2y-2z=5$ and $5x+y-z=1$.

Solution. One can easily see that the two planes are parallel because their respective normal vectors $(10,2,-2)$ and $(5,1,-1)$ are parallel. To find the distance between the planes, first one will have to find a point in one plane and then use the formula \eqref{eq:distance2plane} to find the distance. Let us find a point in the plane $10x+2y-2z=5$. One can, for instance, use the $x$-intercept, so let $y=z=0$. Then $10x=5$ i.e. $x=\frac{1}{2}$. The distance from $\left(\frac{1}{2},0,0\right)$ to the plane $5x+y-z=1$ is $$D=\frac{\left|5\left(\frac{1}{2}\right)+1(0)-(0)\right|}{\sqrt{5^2+1^2+(-1)^2}}=\frac{\sqrt{3}}{6}$$

Examples in this note have been taken from [1].

References.

[1] Calculus, Early Transcendentals, James Stewart, 6th Edition, Thompson Brooks/Cole

# The Cross Product

Definition. Let ${\bf u}=(u_1,u_2,u_3)$ and ${\bf v}=(v_1,v_2,v_3)$. Then the cross product ${\bf u}\times {\bf v}$ is defined by $$\label{eq:crossprod}{\bf u}\times{\bf v}=(u_2v_3-u_3v_2,u_3v_1-u_1v_3,u_1v_2-u_2v_1)$$ The cross product can be also written as the determinant $$\label{eq:crossprod2}{\bf u}\times{\bf v}=\begin{vmatrix}{\bf i} & {\bf j} & {\bf k}\\u_1 & u_2 & u_3\\v_1 & v_2 & v_3\end{vmatrix}$$ One can calculate the determinant as shown in Figure 1. You multiply three entries along each indicated arrow. When you multiply three entries along each red arrow, you also multiply by −1. This is called the Rule of Sarrus named after a French mathematician Pierre Frédéric Sarrus.

Unlike the dot product, the outcome of the dot product is a vector. Also unlike the dot product, the cross product is anticommutative i.e. $${\bf u}\times{\bf v}=-{\bf v}\times{\bf u}$$ Furthermore, ${\bf u}\times{\bf v}$ is orthogonal to both ${\bf u}$ and ${\bf v}$. This can be seen by showing that $$({\bf u}\times{\bf v})\cdot{\bf u}=({\bf u}\times{\bf v})\cdot{\bf v}=0$$ The cross product tells us about the orientation of the plane containing two vectors ${\bf u}$ and ${\bf v}$ as shown in Figure 2.

Theorem. If $\theta$ is the angle between ${\bf u}$ and ${\bf v}$ ($0\leq\theta\leq\pi$), then $$\label{eq:crossprod3}|{\bf u}\times{\bf v}|=|{\bf u}||{\bf v}|\sin\theta$$

Proof. It would require some work with algebra but one can show that $$|{\bf u}\times{\bf v}|^2=|{\bf u}|^2|{\bf v}|^2-({\bf u}\cdot{\bf v})^2$$ This, along with ${\bf u}\cdot{\bf v}=|{\bf u}||{\bf v}|\cos\theta$, will lead to \eqref{eq:crossprod3}.

From \eqref{eq:crossprod3}, we can easily see that two nonzero vectors ${\bf u}$ and ${\bf v}$ are parallel if and only if ${\bf u}\times{\bf v}=0$.

The standard basis vectors ${\bf i}$, ${\bf j}$, ${\bf k}$ satisfy the following cross products: $${\bf i}\times{\bf j}={\bf k},\ {\bf j}\times{\bf k}={\bf i},\ {\bf k}\times{\bf i}={\bf j}$$

The following theorem summarizes the properties of the cross product.

Theorem. Let ${\bf u}$, ${\bf v}$, and ${\bf w}$ be vectors and $c$ a scalar. Then

1. ${\bf u}\times{\bf v}=-{\bf v}\times{\bf u}$
2. $(c{\bf u})\times{\bf v}=c({\bf u}\times{\bf v})={\bf u}\times(c{\bf v})$
3. ${\bf u}\times({\bf v}+{\bf w})={\bf u}\times{\bf v}+{\bf u}\times{\bf w}$
4. $({\bf u}+{\bf v})\times{\bf w}={\bf u}\times{\bf w}+{\bf v}\times{\bf w}$
5. ${\bf u}\cdot({\bf v}\times{\bf w})=({\bf u}\times{\bf v})\cdot{\bf w}$
6. ${\bf u}\times({\bf v}\times{\bf w})=({\bf u}\cdot{\bf w}){\bf v}-({\bf u}\cdot{\bf v}){\bf w}$

The products in 5 and 6 are called, respectively, a scalar triple product and a vector triple product.

From Figure 3, we see that $$\label{eq:areaparallelogram}|{\bf u}\times{\bf v}|$$ is equal to the area of the parallelogram determined by ${\bf u}$ and ${\bf v}$.

Example. Find a vector perpendicular to the plane that passes through the points $P(1,4,6)$, $Q(-2,5,-1)$, and $R(1,-1,1)$.

Solution. The vectors $\overrightarrow{PQ}=(-3,1,-7)$ and $\overrightarrow{PR}=(0,-5,-5)$ lie in the plane through $P,Q,R$. So the cross product $$\overrightarrow{PQ}\times\overrightarrow{PR}=(-40,-15,15)$$ is perpendicular to the plane.

Example. Find the area of the triangle with vertices $P(1,4,6)$, $Q(-2,5,-1)$, and $R(1,-1,1)$.

Solution. In the previous example, we found $\overrightarrow{PQ}\times\overrightarrow{PR}=(-40,-15,15)$ and by \eqref{eq:areaparallelogram} we know that $|\overrightarrow{PQ}\times\overrightarrow{PR}|=\sqrt{(-40)^2+(-15)^2+{15}^2}=5\sqrt{82}$ is the area of the parallelogram determined by the two vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$. The area of the triangle with vertices $P$, $Q$, and $R$ is just the half of the area of the parallelogram i.e. $\frac{5}{2}\sqrt{82}$.

From Figure 4, the volume of the parallelepiped determined by ${\bf u}$, ${\bf v}$, and ${\bf w}$ is $$V=|{\bf v}\times{\bf w}||{\bf u}|\cos\theta={\bf u}\cdot({\bf v}\times{\bf w})$$ In Figure 4, the vectors ${\bf u}$, ${\bf v}$, and ${\bf w}$ are positioned well enough so that the triple scalar product ${\bf u}\cdot({\bf v}\times{\bf w})$ is positive but depending on how they are positioned, it could be negative. Since the volume always has to be positive, it is given by $$\label{eq:volumeparallelepiped}V=|{\bf u}\cdot({\bf v}\times{\bf w})|$$

The scalar triple product ${\bf u}\cdot({\bf v}\times{\bf w})$ can be written nicely by the determinant $$\label{eq:scalartripleprod}{\bf u}\cdot({\bf v}\times{\bf w})=\begin{vmatrix}u_1 & u_2 & u_3\\v_1 & v_2 & v_3\\w_1 & w_2 & w_3\end{vmatrix}$$ The calculation of the determinant can be done by the rule of Sarrus shown in Firgure 1.

Example. Determine if ${\bf u}=(1,4,-7)$, ${\bf v}=(2,-1,4)$, and ${\bf w}=(0,-9,18)$ are coplanar.

Solution. From Figure 4 above, one can easily see that the three vectors ${\bf u}$, ${\bf v}$ and ${\bf w}$ are coplanar (i.e. they are in the same plane) if and only if $\theta=\frac{\pi}{2}$ if and only if ${\bf u}\cdot ({\bf v}\times{\bf w})=0$. \begin{align*}{\bf u}\cdot ({\bf v}\times{\bf w})&=\begin{vmatrix}1 & 4 & -7\\2 & -1 & 4\\0 & -9 & 18\end{vmatrix}\\&=0\end{align*} Therefore, ${\bf u}$, ${\bf v}$ and ${\bf w}$ are coplanar.

The notion of the cross product can be used to describe physical effects involving rotations such as the circulation of electric/magnetic fields or fluids. Here we discuss the torque as a physical application of the cross product. Look at Figure 5.

Assume that a force ${\bf F}$ is acting on a rigid body at a point given by a position vector ${\bf r}$. The resulting turning effect ${\bf\tau}$, called the torque, can be measured by $$\label{eq:torque}{\bf\tau}={\bf r}\times{\bf F}$$

Example. A bolt is tightened by applying a 40 N force to a 0.25 m wrench as shown in Figure 6. Find the magnitude of the torque about the center of the bolt.

Solution. The magnitude of the torque is \begin{align*}|{\bf\tau}|&=|{\bf r}\times{\bf F}|=|{\bf r}||{\bf F}|\sin 75^\circ=(0.25)(40)\sin 75^\circ\\&=10\sin 75^\circ\approx 9.66\ \mathrm{Nm}\end{align*}

Examples in this note have been taken from [1].

References.

[1] Calculus, Early Transcendentals, James Stewart, 6th Edition, Thompson Brooks/Cole

# The Dot Product

Let us begin with the following definition.

Definition. Let ${\bf u}=(u_1,u_2,u_3)$ and ${\bf v}=(v_1,v_2,v_3)$. Then the dot product ${\bf u}\cdot{\bf v}$ is defined by $${\bf u}\cdot{\bf v}=u_1v_1+u_2v_2+u_3v_3$$

The name “product” is misleading as it is not really an operation. The reason is simple because the outcome of a dot product is a scalar, not a vector. So what is a big deal about this dot product? The dot product defines the length of a vector. Let ${\bf u}=(u_1,u_2,u_3)$. Then $$|{\bf u}|=\sqrt{{\bf u}\cdot {\bf u}}=\sqrt{u_1^2+u_2^2+u_3^3}$$ Furthermore it can also define the distance between two points in space as shown in Figure 1:

Let two position vectors ${\bf v}=(v_1,v_2,v_3)$ and ${\bf w}=(w_1,w_2.w_3)$ respectively represent points $P$ and $Q$ in space. The the distance $\overline{PQ}$ between the two points $P$ and $Q$ is the length of the vector ${\bf v}-{\bf w}$ $$\overline{PQ}=|{\bf v}-{\bf w}|=\sqrt{({\bf v}-{\bf w})\cdot({\bf v}-{\bf w})}=\sqrt{(v_1-w_1)^2+(v_2-w_2)^2+(v_3-w_3)^2}$$

Example. \begin{align*}(2,4)\cdot(3,-1)&=2(3)+4(-1)=2\\(-1,7,4)\cdot\left(6,2,-\frac{1}{2}\right)&=-1(6)+7(2)+4\left(-\frac{1}{2}\right)=6\\({\bf i}+2{\bf j}-3{\bf k})\cdot(2{\bf j}-{\bf k})&=1(0)+2(2)+(-3)(-1)=7\end{align*}

The dot product satisfies the following properties. These properties can be easily verified from its definition.

Theorem. Let ${\bf u}$, ${\bf v}$ and ${\bf w}$ be vectors in space and $c$ a scalar. Then

1. ${\bf u}\cdot{\bf v}={\bf v}\cdot{\bf u}$
2. ${\bf u}\cdot ({\bf v}+{\bf w})={\bf u}\cdot{\bf v}+{\bf u}\cdot{\bf w}$
3. $(c{\bf u})\cdot{\bf v}=c({\bf u}\cdot{\bf v})={\bf u}\cdot(c{\bf v})$
4. ${\bf 0}\cdot {\bf u}=0$

Although this is beyond the scope of our discussion here, I would like to mention that the notion of the dot product can be generalized so that it can give rise to a different kind of length. Such generalization is called a scalar product or an inner product. You can read more about it here in case you are interested. A scalar product would satisfy the properties 1-4 in the above theorem. The dot product is associated with the length we are most familiar with, called the Euclidean length but that is not the only kind of length out there. For example, in the vector space of continuous functions on the closed interval $[0,1]$ (which was mentioned here), the scalar product of two functions $f$ and $g$, denoted by $\langle f,g\rangle$ is defined by $$\langle f,g\rangle=\int_0^1 f(x)g(x)dx$$ and the length of $f$ is defined by $$|f|=\sqrt{\langle f,f\rangle}=\sqrt{\int_0^1|f(x)|^2dx}$$ This type of a scalar product plays a very important role in quantum mechanics. It is used to measure the probability of a particle (such as an electron) to be in a particular quantum mechanical state.

There is an alternative description of the dot product.

Theorem. If $\theta$ is the angle between the vectors ${\bf u}$ and ${\bf v}$, where $0\leq\theta\leq\pi$, then $$\label{eq:dotproduct}{\bf u}\cdot{\bf v}=|{\bf u}||{\bf v}|\cos\theta$$

Proof. By applying the Law of Cosines to triangle $\triangle OPQ$ in Figure 1, we obtain $$\label{eq:lawcosine}|{\bf u}-{\bf v}|^2=|{\bf u}|^2+|{\bf v}|^2-2|{\bf u}||{\bf v}|\cos\theta$$ \begin{align*}|{\bf u}-{\bf v}|^2&=({\bf u}-{\bf v})\cdot({\bf u}-{\bf v})\\&={\bf u}\cdot{\bf u}-{\bf u}\cdot{\bf v}-{\bf v}\cdot{\bf u}+{\bf v}\cdot{\bf v}\\&=|{\bf u}|^2-2{\bf u}\cdot{\bf v}+|{\bf v}|^2\end{align*}Replacing $|{\bf u}-{\bf v}|^2$ in \eqref{eq:lawcosine} by this last expression results in $$|{\bf u}|^2-2{\bf u}\cdot{\bf v}+|{\bf v}|^2=|{\bf u}|^2+|{\bf v}|^2-2|{\bf u}||{\bf v}|\cos\theta$$ and this simplifies to $${\bf u}\cdot{\bf v}=|{\bf u}||{\bf v}|\cos\theta$$

Example. Find the angle between the vector ${\bf u}=(2,2,-1)$ and ${\bf v}=(5,-3,2)$.

Solution. From \eqref{eq:dotproduct}, \begin{align*}\cos\theta&=\frac{{\bf u}\cdot{\bf v}}{|{\bf u}||{\bf v}|}\\&=\frac{2(5)+2(-3)+(-1)(2)}{\sqrt{2^2+2^2+(-1)^2}\sqrt{5^2+(-3)^2+2^2}}\\&=\frac{2}{3\sqrt{38}}\end{align*} Hence, $$\theta=\cos^{-1}\left(\frac{2}{3\sqrt{38}}\right)\approx 1.46\ \mathrm{rad}\ (84^\circ)$$

The alternative description of the dot product in \eqref{eq:dotproduct} is usually introduced as the definition of the dot product in high school/freshmen physics course.

From \eqref{eq:dotproduct}, we see that two vectors ${\bf u}$ and ${\bf v}$ are perpendicular or orthogonal (i.e. the angle $\theta$ between ${\bf u}$ and ${\bf v}$ is $\frac{\pi}{2}$) if and only if ${\bf u}\cdot{\bf v}=0$.

Example. $2{\bf i}+2{\bf j}-{\bf k}$ is perpendicular to $5{\bf i}-4{\bf j}+2{\bf k}$ because $$(2{\bf i}+2{\bf j}-{\bf k})\cdot(5{\bf i}-4{\bf j}+2{\bf k})=2(5)+2(-4)+(-1)(2)=0$$

This is beyond the scope of our discussion here but the notion of orthogonality of two vectors can be extended to higher dimensional spaces or more abstract vector spaces by defining that: two vectors ${\bf u}$ and ${\bf v}$ are said to be orthogonal if $\langle{\bf u},{\bf v}\rangle=0$, where $\langle\ ,\ \rangle$ denotes a scalar product. In our case, $\langle{\bf u},{\bf v}\rangle={\bf u}\cdot{\bf v}$. For example, Let $V$ be the set of all continuous functions on the closed interval $[-1,1]$. Then $V$ is a vector space with addition and scalar multiplication defined in the usual way that I discussed here. Also $\langle\ ,\ \rangle$ defined by $$\langle f,g\rangle=\int_{-1}^1f(x)g(x)dx$$ for $f,g\in V$ is a scalar product. The two functions $\sin(2n\pi x)$ and $\cos(2n\pi x)$ are continuous on $[-1,1]$ so they belong to $V$ i.e. they are vectors. They are also orthogonal because $$\langle\sin(2n\pi x),\cos(2n\pi x)\rangle=\int_{-1}^1\sin(2n\pi x)\cos(2n\pi x)dx=0$$

Let us take a look at Figure 2.

Imagine that light rays coming down on the vector ${\bf u}$ at the direction perpendicular to the vector ${\bf v}$. The the shadow of ${\bf u}$ will be cast on ${\bf v}$ (the red line segment in Figure 2). Mathematically, this shadow is called the orthographic projection of ${\bf u}$ onto ${\bf v}$. In fact, the red line segment is the orthographic projection of the length of the vector ${\bf u}$ onto ${\bf v}$. We denote it by $\mathrm{comp}_{\bf v}{\bf u}$ and call it the scalar projection of ${\bf u}$ onto ${\bf v}$. Using basic trigonometry, we can easily find that $$\mathrm{comp}_{\bf v}{\bf u}=|{\bf u}|\cos\theta$$ However, we prefer to express the scalar projection free of the angle $\theta$ i.e. in terms of only ${\bf u}$ and ${\bf v}$. This can be done using \eqref{eq:dotproduct}: \begin{align*}\mathrm{comp}_{\bf v}{\bf u}&=|{\bf u}|\cos\theta\\&=|{\bf u}|\frac{{\bf u}\cdot{\bf v}}{|{\bf u}||{\bf v}|}\\&=\frac{{\bf u}\cdot{\bf v}}{|{\bf v}|}\end{align*} Hence we obtained our preferred form of the scalar projection $$\label{eq:scalarprojection}\mathrm{comp}_{\bf v}{\bf u}=\frac{{\bf u}\cdot{\bf v}}{|{\bf v}|}$$ One can also consider the vector projection of ${\bf u}$ onto ${\bf v}$. All you have to do is to multiplying the scalar projection \eqref{eq:scalarprojection} by the direction of ${\bf v}$: $$\label{eq:vectorprojection}\mathrm{proj}_{\bf v}{\bf u}=\mathrm{comp}_{\bf v}{\bf u}\frac{{\bf v}}{|{\bf v}|}=\frac{{\bf u}\cdot{\bf v}}{|{\bf v}|^2}{\bf v}$$

Example. Find the scalar projection and the vector projection of ${\bf u}=(1,1,2)$ onto ${\bf v}=(-2,3,1)$.

Solution. The scalar projection is $$\mathrm{comp}_{\bf v}{\bf u}=\frac{{\bf u}\cdot{\bf v}}{|{\bf v}|}=\frac{1(-2)+1(3)+2(1)}{\sqrt{(-2)^2+3^2+1^2}}=\frac{3}{\sqrt{14}}$$ The direction of ${\bf v}$ is $\frac{1}{\sqrt{14}}(-2,3,1)$. Hence, the vector projection is $$\mathrm{proj}_{\bf v}{\bf u}=\mathrm{comp}_{\bf v}{\bf u}\frac{1}{\sqrt{14}}(-2,3,1)=\frac{3}{14}(-2,3,1)=\left(-\frac{3}{7},\frac{9}{14},\frac{3}{14}\right)$$

Consider a vector ${\bf v}=(v_1,v_2,v_3)$ in space. The angle $\alpha$ between ${\bf v}$ and ${\bf i}$, the angle $\beta$ between ${\bf v}$ and ${\bf j}$, and the angle $\gamma$ between ${\bf v}$ and ${\bf k}$ are called the direction angles of ${\bf v}$. (See Figure 3.)

Now, \begin{aligned}\cos\alpha&=\frac{{\bf v}\cdot{\bf i}}{|{\bf v}||{\bf i}|}=\frac{v_1}{|{\bf v}|}\\\cos\beta&=\frac{{\bf v}\cdot{\bf j}}{|{\bf v}||{\bf j}|}=\frac{v_2}{|{\bf v}|}\\\cos\gamma&=\frac{{\bf v}\cdot{\bf k}}{|{\bf v}||{\bf k}|}=\frac{v_3}{|{\bf v}|}\end{aligned}\label{eq:directioncosine} $\cos\alpha$, $\cos\beta$ and $\cos\gamma$ are called the direction cosines of vector ${\bf v}$. It follows from \eqref{eq:directioncosine} that $(\cos\alpha,\cos\beta,\cos\gamma)$ is the direction of ${\bf v}$, hence the name directions cosines.

Example. Find the direction angles of the vector ${\bf v}=(1,2,3)$.

Solution. $|{\bf v}|=\sqrt{1^2+2^2+3^2}=\sqrt{14}$. Using \eqref{eq:directioncosine}, we have \begin{align*}\alpha&=\cos^{-1}\left(\frac{1}{\sqrt{14}}\right)\approx 74^\circ\\\beta&=\cos^{-1}\left(\frac{2}{\sqrt{14}}\right)\approx 58^\circ\\\gamma&=\cos^{-1}\left(\frac{3}{\sqrt{14}}\right)\approx 37^\circ\end{align*}

Work

Consider a linear motion i.e. a motion of an object along a straight line. See Figure 4.

Suppose that an object is moved by a force ${\bf F}$. If the displacement is ${\bf D}$, then the work $W$ done by this force ${\bf F}$ is defined by the scalar projection of ${\bf F}$ onto ${\bf D}$, $|{\bf F}|\cos\theta$ (this is the component of ${\bf F}$ that actually moved the object) times the distance moved $|{\bf D}|$: $$\label{eq:work}W={\bf F}\cdot{\bf D}$$

Example. A wagon is pulled a distance of 100 m along a horizontal path by a constant force of 70 N. The handle of the wagon is held at an angle of $35^\circ$ above the horizontal. Find the work done by the force.

Solution. The force ${\bf F}$ and the displacement ${\bf D}$ are as depicted in Figure 5.

Thus the work $W$ is \begin{align*}W&={\bf F}\cdot{\bf D}=|{\bf F}||{\bf D}|\cos 35^\circ\\&=70(100)\cos 35^\circ=5734\ \mathrm{J}\end{align*} where J, called Joule, is a unit for work which stands for Newton times meter.

Example. A force given by the vector ${\bf F}=3{\bf i}+4{\bf j}+5{\bf k}$ moves a particle from the point $P(2,1,0)$ to the point $Q(4,6,2)$. Find the work done by the force.

Solution. Here, there is no mention of a particular path the particle is taking. We assume that the motion is again linear. The displacement is ${\bf D}=\overrightarrow{PQ}=(4-2,6-1,2-0)=(2,5,2)$. Hence the work is \begin{align*}W&={\bf F}\cdot{\bf D}=(3,4,5)\cdot(2,5,2)\\&=6+20+10=36\end{align*}

Examples in this note have been taken from [1].

References.

[1] Calculus, Early Transcendentals, James Stewart, 6th Edition, Thompson Brooks/Cole