# Related Rates

Related rates problems often involve (context-wise) real-life applications of the chain rule/implicit differentiation. Here are some of the examples that are commonly seen in calculus textbooks.

Example. Car A is traveling west at 50mi/h and car B is traveling north at 60mi/h. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the intersection?

Solution.

Denote by $x$ and $y$ the distances from the intersection to car A and to car B, respectively. Then we have $\frac{dx}{dt}=-50$mi/h and $\frac{dy}{dt}=-60$mi/h. Let us denote $z$ the distance between $A$ and $B$. Then by Pythagorean law we have $$z^2=x^2+y^2$$ Differentiating this with respect to $t$, we obtain $$z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$$ and thus \begin{align*}\frac{dz}{dt}&=\frac{1}{z}\left[x\frac{dx}{dt}+y\frac{dy}{dt}\right]\\&=\frac{1}{0.5}[0.3(-50)+0.4(-60)]=-78\mathrm{mi/h}\end{align*}

Example. Air is being pumped into a spherical balloon so that its volume increases at a rate of $100\mathrm{cm}^3/\mathrm{s}$. How fast is the radius of the balloon increasing when the diameter is 50 cm?

Solution. Let $V$ and $r$ denote the volume and the radius of the spherical balloon. Then $V=\frac{4}{3}\pi r^3$. Differentiating this with respect to $t$, we obtain $$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$ So, \begin{align*}\frac{dr}{dt}&=\frac{1}{4\pi r^2}\frac{dV}{dt}\\&=\frac{1}{4\pi(25)^2}100\\&=\frac{1}{25\pi}\mathrm{cm/s}\end{align*}

Example. Gravel is being dumped from a conveyor belt at a rate of $30 \mathrm{ft}^3/\mathrm{min}$ and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are the same. How fast is the height of the pile increasing when the pile is 10 ft high?

Solution. The cross section of the gravel pile is shown in the figure below.

The amount of gravel dumped is the same as the volume of the cone. Let us denote the volume by $V$, its base radius by $r$, and its height by $h$. Then $V=\frac{1}{3}\pi r^2h$. Since $h=2r$, $V$ can be written as $$V=\frac{1}{12}\pi h^3$$ Differentiating this with respect to $t$, we obtain $$\frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}$$ So, we have \begin{align*}\frac{dh}{dt}&=\frac{4}{\pi h^2}\frac{dV}{dt}\\&=\frac{4}{\pi(10)^2}(30)=\frac{1.2}{\pi}\mathrm{ft/min}\approx 0.38\mathrm{ft/min}\end{align*}

Example. A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall?

Solution.

Let us denote by $x$ and $y$ the distance from the wall to the bottom of the ladder and the distance from the top of the ladder to the floor, respectively. By Pythagorean law, we have $x^2+y^2=100$. Differentiating this with respect to $t$, we obtain $$x\frac{dx}{dt}+y\frac{dy}{dt}=0$$ Hence, we have \begin{align*}\frac{dy}{dt}&=-\frac{x}{y}\frac{dx}{dt}\\&=-\frac{6}{8}(1)=-\frac{3}{4}\mathrm{ft/s}\end{align*}

Example. A water tank has the shape of an inverted circular cone with base radius 2m and heigh 4 m. If water is being pumped into the tank at a rate of $2 \mathrm{m}^3/\mathrm{min}$, find the rate at which the water level is rising when the water is 3 m deep.

Solution. The cross section of the water tank is shown in the figure below.

The amount of water $V$ when the water level is $h$ and the surface radius is $r$ is $V=\frac{1}{3}\pi r^2h$. From the above figure we have the following ratio holds $$\frac{2}{4}=\frac{r}{h}$$ i.e. $r=\frac{h}{2}$. SO $V$ can be written as $$V=\frac{1}{12}\pi h^3$$ Differentiating this with respect to $t$, we obtain $$\frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}$$ Hence, \begin{align*}\frac{dh}{dt}&=\frac{4}{\pi h^2}\frac{dV}{dt}\\&=\frac{4}{\pi(3)^2}(2)\\&=\frac{8}{9\pi}\mathrm{m/min}\approx 0.28\mathrm{m/min}\end{align*}

# 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