Chapter 15Inner Products and Cross Products of Vectors
Up to and including Part III (Fundamentals of Probability), we have been dealing with “numbers”—probabilities and expected values are, in the end, each a single real number. But many quantities in nature—the direction in which a force acts, the direction in which a velocity points, the direction in which a magnetic field twists—cannot be written down with a single number. A quantity that has both “magnitude” and “direction” is a vector. In High School Mathematics (AD 01), Chapter 7, you have already learned the component expression of vectors in the plane and in space, their magnitudes, and the inner product (also called the dot product), which produces a single real number from two vectors. This chapter starts with a review of that material, but the real star appears in the second half: an operation not studied in high school—the cross product, which produces another, different vector from two vectors.
The cross product is an operation that looks strange at first. The magnitude of $\vec A\times\vec B$ is determined by $\abs{\vec A}\abs{\vec B}\sin\theta$ (the area of the parallelogram spanned by the two sides $\vec A,\vec B$), but its direction is one of the two directions perpendicular to each of $\vec A,\vec B$—the direction specified by the right-hand screw rule. Why consider such an operation? From angular momentum and torque in mechanics and the Lorentz force in electromagnetism to the curl ($\rot$) and Stokes's theorem that we will study in later chapters, the operation of “producing from two directions a direction perpendicular to both” appears everywhere in physics. The goal of this chapter is to build up the cross product carefully from its definition and to carry the derivation through completely, up to the component expression (and its relation to determinants). In addition, we will clarify how the two products differ in meaning: the inner product measures “how much influence there is in the same direction,” whereas the cross product measures “the magnitude and the axis of rotation or twisting.”
- The definition of a vector (a quantity with magnitude and direction), the zero vector $\vec 0$, and the unit vector $\vec e$ (how to construct a vector of magnitude 1 with the same direction as a given vector: $\vec e=\vec a/\abs{\vec a}$)
- The definition of the inner product, $\vec a\cdot\vec b=a_xb_x+a_yb_y=\abs{\vec a}\abs{\vec b}\cos\theta$, and a proof, using the addition formulas, that these two expressions agree
- Properties of the inner product: the commutative law $\vec A\cdot\vec B=\vec B\cdot\vec A$, and the norms and orthogonality of the standard unit vectors ($\vec i\cdot\vec i=1$, $\vec i\cdot\vec j=0$, etc.)
- The definition of the cross product, $\vec A\times\vec B=\abs{\vec A}\abs{\vec B}\sin\theta\,\vec e_z$ ($\vec e_z$ is the unit vector determined by the right-hand screw rule), and the fact that the cross product, unlike the inner product, is a vector
- A proof, based on the sense of rotation, of $\vec B\times\vec A=-\vec A\times\vec B$ (that is, the commutative law does not hold)
- A derivation that the component expression $a_1b_2-a_2b_1$ of the cross product in two dimensions agrees with the $2\times2$ determinant and with the signed area of a parallelogram (completing the formula previewed in High School Mathematics, Section 7.7)
- Completing, and deriving on our own, the component formula for the three-dimensional cross product $\vec a\times\vec b=(a_2b_3-a_3b_2,\ a_3b_1-a_1b_3,\ a_1b_2-a_2b_1)$, whose explanation breaks off partway in the original notes
- The contrast in meaning between the two products: inner product = orthogonal projection (the degree of influence in the same direction), cross product = the magnitude of rotation in the perpendicular direction
Source notes: Yasuhide Mochizuki, Mathematics Notes: Vector Analysis, pp. 1–6.
15.1 What Is a Vector?—Magnitude and Direction, the Zero Vector, and Unit Vectors
In High School Mathematics, Chapter 7, we have already become familiar with vectors drawn as arrows in the plane and in space. Here we begin by putting their definition into words once more, carefully.
Definition 15.1: Vector
A vector is a quantity that has a “magnitude” and a “direction.” Vectors are generally denoted by $\vec a,\vec b$, and so on.
The magnitude of a vector is given by the length of the arrow that represents it, and is written $\abs{\vec a}$ (it is also called the norm; the notation $|x|$ for the absolute value of a real number is borrowed here because, in the sense of “distance from the origin,” the absolute value and the norm come from the same idea). The direction of a vector is given by the direction of the arrow.
Intuition: A vector does not depend on “where it started from”
In high school mathematics, vectors are often drawn as concrete arrows in the plane or in space (“directed line segments” with an initial point and a terminal point). However, the essence of a vector is only its “magnitude and direction”: even if an arrow is translated, it is regarded as the same vector as long as its magnitude and direction do not change. For example, the arrow from the origin $O$ to the point $(2,2)$ and the arrow from the point $(1,0)$ to the point $(3,2)$ both have the same magnitude and direction, “2 to the right and 2 up,” so both are the same vector $\vec a=(2,2)$. In this chapter, to make the calculations easier to follow, we draw vectors as arrows with the origin $O$ as the initial point (position vectors) unless stated otherwise, but the idea that “a vector does not depend on its starting point” also serves as the foundation when we treat vector spaces abstractly in later chapters.
Two kinds of vectors whose magnitudes take special values have been given names.
Definition 15.2: Zero vector
A vector whose magnitude is $0$ is called the zero vector and is denoted by $\vec 0$.
That the magnitude is $0$ means that the length of the arrow is $0$; that is, the initial point and the terminal point coincide. Therefore the zero vector has no particular “direction” (or we regard it as pointing in any direction). Just like $0$ in the real numbers, the zero vector acts as the identity element for vector addition (a special element that changes nothing when added to any vector)—we will treat this point again as an axiom in Chapter 10 (vector spaces).
Definition 15.3: Unit vector
A vector whose magnitude is $1$ is called a unit vector. It is generally written $\vec e$.
The role of a unit vector is to extract “the direction only.” For example, given a vector $\vec a$, the vector that has the same direction as $\vec a$ and magnitude $1$ is called the unit vector of $\vec a$. To construct it, we only need to divide $\vec a$ by its own magnitude $\abs{\vec a}$.
Let us see with concrete numbers why this works.
Illustration 15.1 (example from the original notes): The unit vector of $\vec a=(2,2)$
Suppose $\vec a=\begin{bmatrix}2\\2\end{bmatrix}$. Then $\abs{\vec a}=\sqrt{2^2+2^2}=\sqrt8=2\sqrt2$.
Now dividing $\vec a$ by $\abs{\vec a}$, we obtain
$$ \frac{1}{\abs{\vec a}}\vec a=\frac{1}{2\sqrt2}\begin{bmatrix}2\\2\end{bmatrix}=\frac{\sqrt2}{4}\begin{bmatrix}2\\2\end{bmatrix}=\begin{bmatrix}\dfrac{\sqrt2}{2}\\[4pt]\dfrac{\sqrt2}{2}\end{bmatrix}=\vec e $$(here we rationalized the denominator by writing $\frac{1}{2\sqrt2}=\frac{1}{2\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}=\frac{\sqrt2}{4}$, and computed $\frac{\sqrt2}{4}\times2=\frac{2\sqrt2}{4}=\frac{\sqrt2}{2}$). If we actually find the magnitude of $\vec e=\left(\dfrac{\sqrt2}{2},\dfrac{\sqrt2}{2}\right)$, we get
$$ \abs{\vec e}=\sqrt{\left(\frac{\sqrt2}{2}\right)^2+\left(\frac{\sqrt2}{2}\right)^2}=\sqrt{\frac12+\frac12}=\sqrt1=1 $$and so we see that $\vec e$ is indeed the unit vector of $\vec a$. Hence $\dfrac{1}{\abs{\vec a}}\vec a=\vec e$ holds. (SymPy check: $\abs{\vec a}=2\sqrt2$, $\vec e=(\sqrt2/2,\sqrt2/2)$, $\abs{\vec e}=1$.)
Let us now check in general why Eq. \eqref{eq:15-unitvec} produces a unit vector. Dividing $\vec a$ by its magnitude $\abs{\vec a}$ (which is a positive real scalar) is the same as multiplying $\vec a$ by $\dfrac{1}{\abs{\vec a}}$. The real multiple $k\vec a$ ($k\gt0$) is the operation that multiplies only the magnitude by $k$ without changing the direction of $\vec a$, so if we choose $k=\dfrac{1}{\abs{\vec a}}$, the new magnitude is $\abs{\vec a}\times\dfrac{1}{\abs{\vec a}}=1$ and the direction remains that of $\vec a$. This is the general statement of what we confirmed numerically in Illustration 15.1.
Example 15.1: Magnitude and unit vector of $\vec a=(3,4)$
Find the magnitude $\abs{\vec a}$ of $\vec a=(3,4)$ and the unit vector $\vec e$ in the same direction as $\vec a$.
Solution. The magnitude is the square root of the sum of the squares of the components, so we have
$$ \abs{\vec a}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5 $$By Eq. \eqref{eq:15-unitvec}, the unit vector is
$$ \vec e=\frac{1}{\abs{\vec a}}\vec a=\frac15(3,4)=\left(\frac35,\frac45\right) $$To check, we find the magnitude: $\abs{\vec e}=\sqrt{\left(\frac35\right)^2+\left(\frac45\right)^2}=\sqrt{\frac9{25}+\frac{16}{25}}=\sqrt{\frac{25}{25}}=\sqrt1=1$, so it is indeed a unit vector. (SymPy check: $\abs{\vec a}=5$, $\vec e=(3/5,4/5)$, $\abs{\vec e}=1$.)
So far we have examined the properties of vectors themselves: “magnitude,” “direction,” the “zero vector,” and the “unit vector.” Starting with the next section, Section 15.2, we move on to products, which combine two vectors to make a new quantity—first the inner product.
15.2 The Inner Product (Scalar Product) and Its Properties
In High School Mathematics, Chapter 7, we denoted the angle between two vectors $\vec{OA},\vec{OB}$ by $\theta$ and defined the inner product geometrically as $\vec{OA}\cdot\vec{OB}=\abs{\vec{OA}}\abs{\vec{OB}}\cos\theta$ (Definition 7.7), and from this we derived the component expression $\vec{OA}\cdot\vec{OB}=x_1x_2+y_1y_2$ (Formula 7.3). In this section we reverse the direction. We first take the component expression as the definition, and from it we derive the geometric expression $\abs{\vec a}\abs{\vec b}\cos\theta$. This is not merely a review read backward—if we start from the components, we can define the inner product directly in three dimensions and in higher dimensions as well (without having to draw $\cos\theta$ in a picture), and the very statement that “the component expression agrees with the geometric meaning” becomes a theorem that has to be proved anew.
Definition 15.4: Inner product (scalar product)
The inner product of $\vec a=(a_x,a_y)$ and $\vec b=(b_x,b_y)$ is defined by
$$ \vec a\cdot\vec b=a_xb_x+a_yb_y $$In three or more dimensions it is defined in the same way, as the sum of the products of corresponding components, $\vec a\cdot\vec b=a_1b_1+a_2b_2+a_3b_3$ ($+\cdots$).
The inner product can also be written as the matrix product of a row vector and a column vector. Let $\vec a=[a_x\ \ a_y]$ and $\vec b=\begin{bmatrix}b_x\\b_y\end{bmatrix}$. Then
\begin{equation} \vec a\cdot\vec b=[a_x\ \ a_y]\begin{bmatrix}b_x\\b_y\end{bmatrix}=a_xb_x+a_yb_y \label{eq:15-dotcomp} \end{equation}(this is exactly the rule for the product of matrices (Definition 8.3) that we learned in Chapter 8, Section 8.1 of University Mathematics). The result of computing the inner product is the real number $a_xb_x+a_yb_y$, not a vector. Although it is a product of vectors, the result is not a vector—this is a decisive difference from the cross product, which we study in the next section, and for this reason the inner product is also called the scalar product (= dot product) (a scalar is an ordinary real number that has no direction; the word is used in contrast to vectors).
Starting from the component-based Definition 15.4, let us show that this really agrees with the geometric quantity $\abs{\vec a}\abs{\vec b}\cos\theta$.
Theorem 15.1: Geometric expression of the inner product
If $\theta$ ($0\le\theta\le\pi$) is the angle between $\vec a,\vec b$, then
\begin{equation} \vec a\cdot\vec b=a_xb_x+a_yb_y=\abs{\vec a}\abs{\vec b}\cos\theta \label{eq:15-dotgeom} \end{equation}holds.
Proof
As in Figure 15.2, let $\alpha$ be the angle of $\vec a$ measured from the $x$-axis and $\beta$ the angle of $\vec b$ measured from the $x$-axis (we may assume $\alpha\ge\beta$). Then the angle between $\vec a,\vec b$ is $\theta=\alpha-\beta$. What we want to show is that
$$ \abs{\vec a}\abs{\vec b}\cos\theta=\abs{\vec a}\abs{\vec b}\cos(\alpha-\beta) $$agrees with the component expression $a_xb_x+a_yb_y$. Using the addition formula for the cosine (High School Mathematics, Section 5.1) $\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$, we obtain
$$ \abs{\vec a}\abs{\vec b}\cos(\alpha-\beta)=\big(\cos\alpha\cos\beta+\sin\alpha\sin\beta\big)\abs{\vec a}\abs{\vec b} $$Now the angle $\alpha$ is an angle of the right triangle whose hypotenuse is $\vec a$ and whose adjacent and opposite sides are $a_x,a_y$, so by the definition of the trigonometric ratios (High School Mathematics, Chapter 3), the following hold.
Attention: Expressing the trigonometric ratios of $\alpha,\beta$ in terms of components
$$ \cos\alpha=\frac{a_x}{\abs{\vec a}},\quad \cos\beta=\frac{b_x}{\abs{\vec b}},\qquad \sin\alpha=\frac{a_y}{\abs{\vec a}},\quad \sin\beta=\frac{b_y}{\abs{\vec b}} $$These relations hold. Substituting them into the earlier formula, we obtain
$$ \abs{\vec a}\abs{\vec b}\cos(\alpha-\beta) =\left(\frac{a_x}{\abs{\vec a}}\cdot\frac{b_x}{\abs{\vec b}}+\frac{a_y}{\abs{\vec a}}\cdot\frac{b_y}{\abs{\vec b}}\right)\abs{\vec a}\abs{\vec b} =\left(\frac{a_xb_x}{\abs{\vec a}\abs{\vec b}}+\frac{a_yb_y}{\abs{\vec a}\abs{\vec b}}\right)\abs{\vec a}\abs{\vec b} $$(we have merely factored out the denominator $\abs{\vec a}\abs{\vec b}$), and finally the denominator $\abs{\vec a}\abs{\vec b}$ cancels against the outer factor $\abs{\vec a}\abs{\vec b}$, so that we obtain
$$ \abs{\vec a}\abs{\vec b}\cos(\alpha-\beta)=a_xb_x+a_yb_y $$This is exactly the same expression as the component expression of Definition 15.4. Hence $\vec a\cdot\vec b=a_xb_x+a_yb_y=\abs{\vec a}\abs{\vec b}\cos\theta$ has been proved.
∎
This proof proceeded in the order “component expression $\to$ addition formula $\to$ geometric expression,” but recall that in High School Mathematics, Section 7.4, we proved it in the reverse order, “geometric expression $\to$ law of cosines $\to$ component expression.” The very fact that both routes arrive at the same equation is what supports the correctness of this formula.
15.2.1 Properties of the inner product
Let us examine the inner product in a little more detail. In what follows, we write the standard unit vectors (basis) of three-dimensional space as $\vec i,\vec j,\vec k$ (these are the unit vectors pointing in the directions of the $x,y,z$ axes, respectively).
Theorem 15.2: Properties of the inner product
For arbitrary vectors $\vec A,\vec B$ and the standard unit vectors $\vec i,\vec j,\vec k$, the following hold.
(I) $\vec A\cdot\vec B=\abs{\vec A}\abs{\vec B}\cos\theta$ (Theorem 15.1, geometric expression)
(II) $\vec A\cdot\vec B=\vec B\cdot\vec A$ (commutative law)
(III) $\vec i\cdot\vec i=1,\ \vec j\cdot\vec j=1,\ \vec k\cdot\vec k=1$ (norms of the standard unit vectors)
(IV) $\vec i\cdot\vec j=0,\ \vec j\cdot\vec k=0,\ \vec k\cdot\vec i=0$ (orthogonality of the standard unit vectors)
(II) is immediate from the component expression $\vec A\cdot\vec B=A_xB_x+A_yB_y$ and the commutative law of multiplication of real numbers, $A_xB_x=B_xA_x$ (in the geometric expression as well, $\theta$ is the angle between $\vec A,\vec B$, which is the same as the angle between $\vec B,\vec A$, so $\cos\theta$ does not change and the commutative law again holds). (III) follows from the fact that $\vec i,\vec j,\vec k$ are all unit vectors of magnitude $1$: if we set $\vec A=\vec B=\vec i$ in (I) (the angle between identical vectors is $\theta=0$), then $\vec i\cdot\vec i=\abs{\vec i}^2\cos0=1\times1\times1=1$ (and likewise for $\vec j,\vec k$). (IV) follows from the fact that $\vec i,\vec j,\vec k$ are mutually orthogonal (the angle between them is $90^\circ$), so that in (I) we have $\theta=90^\circ$ and $\cos90^\circ=0$.
Why can the inner product be used as an “orthogonality test”?
If we set $\theta=90^\circ$ in Theorem 15.1, then $\cos90^\circ=0$, so $\vec a\cdot\vec b=0$. Conversely, if neither of $\vec a,\vec b$ is $\vec 0$, then from $\vec a\cdot\vec b=\abs{\vec a}\abs{\vec b}\cos\theta=0$ we get $\cos\theta=0$, that is, $\theta=90^\circ$. Thus $\vec a\cdot\vec b=0$ is equivalent to $\vec a\perp\vec b$ (orthogonal) (the symbol $\iff$ is precisely the logical symbol that expresses this “equivalence,” and it means “necessary and sufficient condition.” It is easy to remember if you think of it as the arrow $\Rightarrow$, which stands for “implies,” laid over in both directions—that is, read it as “$\Rightarrow$ and $\Leftarrow$”). That is,
$$ \vec a\cdot\vec b=0\iff\vec a\perp\vec b $$By computing the inner product from the components alone and seeing whether it is $0$, we can test orthogonality without having to find the angle—this is one of the most important uses of the inner product.
Example 15.2: The inner product and the angle between $\vec a=(1,0)$ and $\vec b=(1,1)$
Find the inner product $\vec a\cdot\vec b$ of $\vec a=(1,0)$ and $\vec b=(1,1)$, and find the angle $\theta$ between the two vectors.
Solution. By the component expression (Definition 15.4),
$$ \vec a\cdot\vec b=1\times1+0\times1=1 $$Also, $\abs{\vec a}=\sqrt{1^2+0^2}=1$ and $\abs{\vec b}=\sqrt{1^2+1^2}=\sqrt2$, so by Theorem 15.1 (geometric expression) we obtain
$$ \cos\theta=\frac{\vec a\cdot\vec b}{\abs{\vec a}\abs{\vec b}}=\frac{1}{1\times\sqrt2}=\frac{1}{\sqrt2}=\frac{\sqrt2}{2} $$In the range $0\le\theta\le180^\circ$, $\cos\theta=\frac{\sqrt2}{2}$ holds only when $\theta=45^\circ$, so $\theta=45^\circ$. (SymPy check: $\vec a\cdot\vec b=1$, $\cos\theta=\sqrt2/2$, $\theta=45^\circ$.)
Example 15.3: Orthogonality of $\vec a=(2,-1)$ and $\vec b=(1,2)$
Determine whether $\vec a=(2,-1)$ and $\vec b=(1,2)$ are orthogonal.
Solution. Computing the inner product, we obtain
$$ \vec a\cdot\vec b=2\times1+(-1)\times2=2-2=0 $$As we confirmed in the box “Why can the inner product be used as an ‘orthogonality test’?”, $\vec a\cdot\vec b=0$ is equivalent to $\vec a\perp\vec b$, so $\vec a$ and $\vec b$ are orthogonal. (SymPy check: $\vec a\cdot\vec b=0$.)
Finally, let us verify another important property of the inner product—linearity (the distributive law). Computing $\vec a\cdot(\vec b+\vec c)$ in components, we get $a_x(b_x+c_x)+a_y(b_y+c_y)=a_xb_x+a_xc_x+a_yb_y+a_yc_y=(a_xb_x+a_yb_y)+(a_xc_x+a_yc_y)=\vec a\cdot\vec b+\vec a\cdot\vec c$ (we have simply used the fact that the product of real numbers satisfies the distributive law over sums). That is,
\begin{equation} \vec a\cdot(\vec b+\vec c)=\vec a\cdot\vec b+\vec a\cdot\vec c \label{eq:15-dotdist} \end{equation}holds. Using this property, we can also prove Theorem 15.1 by a method different from that of high school mathematics.
Example 15.4: An alternative proof of the inner product formula using the law of cosines
Using the law of cosines for a triangle and the linearity of the inner product (Eq. \eqref{eq:15-dotdist}), prove $\vec a\cdot\vec b=\abs{\vec a}\abs{\vec b}\cos\theta$ (Theorem 15.1) by a different method. Here $\theta$ is the angle between the two vectors $\vec a,\vec b$.
Solution. Consider the triangle whose two sides are $\vec a,\vec b$ and whose remaining side is $\vec a-\vec b$. In this triangle, the angle opposite the side $\vec a-\vec b$ is exactly $\theta$ (the angle between the two vectors $\vec a,\vec b$), so by the law of cosines (High School Mathematics, Section 3.3),
$$ \abs{\vec a-\vec b}^2=\abs{\vec a}^2+\abs{\vec b}^2-2\abs{\vec a}\abs{\vec b}\cos\theta \qquad\text{①} $$On the other hand, $\abs{\vec a-\vec b}^2=(\vec a-\vec b)\cdot(\vec a-\vec b)$ (the square of the magnitude of a vector equals its inner product with itself—just take $\vec A=\vec B$ and $\theta=0$ in Theorem 15.2(I)). Expanding this using the linearity in Eq. \eqref{eq:15-dotdist}, we obtain
$$ (\vec a-\vec b)\cdot(\vec a-\vec b)=\vec a\cdot\vec a-\vec a\cdot\vec b-\vec b\cdot\vec a+\vec b\cdot\vec b=\abs{\vec a}^2-2\,\vec a\cdot\vec b+\abs{\vec b}^2 \qquad\text{②} $$(We used $\vec a\cdot\vec b=\vec b\cdot\vec a$, that is, the commutative law of Theorem 15.2(II), to combine the terms as $-\vec a\cdot\vec b-\vec b\cdot\vec a=-2\,\vec a\cdot\vec b$.) Since ① and ② both express $\abs{\vec a-\vec b}^2$, we have
$$ \abs{\vec a}^2-2\,\vec a\cdot\vec b+\abs{\vec b}^2=\abs{\vec a}^2+\abs{\vec b}^2-2\abs{\vec a}\abs{\vec b}\cos\theta $$Subtracting $\abs{\vec a}^2+\abs{\vec b}^2$ from both sides gives $-2\,\vec a\cdot\vec b=-2\abs{\vec a}\abs{\vec b}\cos\theta$, and dividing both sides by $-2$ yields
$$ \vec a\cdot\vec b=\abs{\vec a}\abs{\vec b}\cos\theta $$(SymPy check: keeping the general components $a_x,a_y,b_x,b_y$ symbolic, we confirmed that $\vec a\cdot\vec b=\frac12\left(\abs{\vec a}^2+\abs{\vec b}^2-\abs{\vec a-\vec b}^2\right)$ holds identically.)
Although the route of the proof is different, the formula we arrive at is exactly the same as Theorem 15.1. Checking that different proofs reach the same conclusion is good practice for deepening one's understanding of that conclusion. In the next section we finally move on to another product, different from the inner product—the cross product.
15.3 Definition of the Cross Product (Vector Product) and the Right-Hand Screw Rule
The inner product produced a single real number (a scalar) from two vectors. From here on, we study an operation that produces another, different vector from two vectors—the cross product, also called the vector product. This operation is not treated in high school mathematics; it appears for the first time at the university level.
Definition 15.5: Cross product (vector product)
The cross product $\vec A\times\vec B$ of any vectors $\vec A$ and $\vec B$ is defined by
\begin{equation} \vec A\times\vec B=\abs{\vec A}\abs{\vec B}\sin\theta\,\vec e_z \label{eq:15-crossdef} \end{equation}where $\theta$ ($0\le\theta\le\pi$) is the angle between $\vec A$ and $\vec B$, and $\vec e_z$ is the unit vector pointing in the direction in which a right-handed screw advances when it is turned from $\vec A$ toward $\vec B$ (this is explained in more detail in the next Intuition box).
Let us take the definition apart. Two pieces of information combine to form a single vector: the magnitude $\abs{\vec A\times\vec B}=\abs{\vec A}\abs{\vec B}\sin\theta$ and the direction $\vec e_z$. Whereas the result of the inner product was a scalar (a real number), the cross product, as defined, is a vector. In other words, in the cross product the “direction” also emerges as important information—something that did not appear in the result of the inner product.
Intuition: The direction of the cross product is determined by the right-hand screw rule
The picture to have in mind for the cross product is, above all, the right-hand screw rule. If you grip a screwdriver in your right hand and turn the screw from the direction of $\vec A$ to the direction of $\vec B$ (through the angle $\theta$)—taking the shorter angle, less than $180^\circ$, as the angle of rotation—the screw advances in a certain direction along its axis. The direction of $\vec A\times\vec B$ is exactly this “direction in which the screw advances.” As in Figure 15.3, when $\vec A$ points along the $x$-axis and $\vec B$ lies in the $xy$-plane, rotated from $\vec A$ by the angle $\theta$ toward the $y$-axis, turning from $\vec A$ to $\vec B$ according to the right-hand screw rule makes the screw advance in the positive direction of the $z$-axis. Therefore, in this situation $\vec e_z$ is exactly the unit vector along the $z$-axis.
From this definition, let us confirm the first property of the cross product that we should pin down—that the commutative law does not hold.
Theorem 15.3: The cross product does not satisfy the commutative law
$$ \vec B\times\vec A=-(\vec A\times\vec B) $$Proof
Let $\vec A\times\vec B=\vec C$. As in Figure 15.3, if we think of rotating from $\vec A$ to $\vec B$ through the angle $\theta$, then by the definition we have
$$ \vec A\times\vec B=\abs{\vec A}\abs{\vec B}\sin\theta\,\vec e_z=\vec C $$Now what about $\vec B\times\vec A$? In the definition of the cross product, the sense of rotation is “from the vector written first to the vector written second,” so in $\vec B\times\vec A$ we rotate from $\vec B$ to $\vec A$. But if we take the sense “rotate from $\vec A$ to $\vec B$ by $\theta$” as the reference, then “rotating from $\vec B$ to $\vec A$” is exactly the opposite sense, that is, it amounts to rotating by $-\theta$ (the two arcs near the origin in Figure 15.3 express precisely this opposite relationship). Therefore, we have
$$ \vec B\times\vec A=\abs{\vec B}\abs{\vec A}\sin(-\theta)\,\vec e_z $$Here we use the following property of the sine function:
Attention: $\sin(-\theta)=-\sin\theta$
The graph of the sine is symmetric about the origin (the sine is an odd function), so reversing the sign of the angle also reverses the sign of the value.
Using this property, we obtain
$$ \vec B\times\vec A=\abs{\vec B}\abs{\vec A}\sin(-\theta)\,\vec e_z=-\abs{\vec A}\abs{\vec B}\sin\theta\,\vec e_z=-\vec C $$Hence $\vec B\times\vec A=-\vec C=-(\vec A\times\vec B)$.
∎
This property of the cross product is called the anticommutative law (anticommutativity). For the inner product, $\vec A\cdot\vec B=\vec B\cdot\vec A$ (Theorem 15.2(II)) held, whereas for the cross product, swapping the order reverses the sign—this is a direct consequence of the fact that the cross product carries the information of “direction.”
Let us confirm the properties of the cross product by actually applying the right-hand screw rule to the standard unit vectors $\vec i,\vec j,\vec k$.
Example 15.5: Cross products of the standard unit vectors
Using the right-hand screw rule, find $\vec i\times\vec j$, $\vec j\times\vec k$, and $\vec k\times\vec i$.
Solution. The vectors $\vec i,\vec j,\vec k$ are the unit vectors along the $x,y,z$ axes, respectively, and they are mutually orthogonal ($\theta=90^\circ$, $\sin90^\circ=1$). If we turn a right-handed screw from $\vec i$ to $\vec j$ (from the $x$-axis to the $y$-axis), the screw advances in the positive direction of the $z$-axis. Hence the magnitude is $\abs{\vec i}\abs{\vec j}\sin90^\circ=1\times1\times1=1$, and the direction is the positive $z$ direction (that is, the direction of $\vec k$ itself), so we have
$$ \vec i\times\vec j=\vec k $$Similarly, turning the screw from $\vec j$ to $\vec k$ (from the $y$-axis to the $z$-axis) makes it advance in the positive $x$ direction, so $\vec j\times\vec k=\vec i$; turning it from $\vec k$ to $\vec i$ (from the $z$-axis to the $x$-axis) makes it advance in the positive $y$ direction, so $\vec k\times\vec i=\vec j$. (numpy check: $\vec i\times\vec j=(0,0,1)=\vec k$, $\vec j\times\vec k=(1,0,0)=\vec i$, $\vec k\times\vec i=(0,1,0)=\vec j$. Also, by Theorem 15.3, $\vec i\times\vec i=\vec j\times\vec j=\vec k\times\vec k=\vec 0$—since the angle between vectors pointing in the same direction is $\theta=0$ and $\sin0=0$. This can also be confirmed directly from the anticommutative law: from $\vec i\times\vec i=-(\vec i\times\vec i)$, that is, $2(\vec i\times\vec i)=\vec 0$.)
Let us now gather and organize the properties of the cross product obtained so far.
Theorem 15.4: Summary of properties of the cross product
For arbitrary vectors $\vec a,\vec b,\vec c$, a real number $k$, and the standard unit vectors $\vec i,\vec j,\vec k$, the following hold.
(I) $\vec 0\times\vec a=\vec a\times\vec 0=\vec 0$ ($\because\ \vec 0=-\vec 0$ and Theorem 15.3)
(II) $\vec a\times\vec a=\vec 0$ ($\because\ \sin0=0$)
(III) $\vec a\times\vec b=-\vec b\times\vec a$ (Theorem 15.3; the commutative law does not hold)
(IV) $(k\vec a)\times\vec b=\vec a\times(k\vec b)=k(\vec a\times\vec b)$ (where $k$ is a real number)
(V) $\vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c$ (distributive law)
(VI) $\vec i\times\vec j=\vec k,\ \vec j\times\vec k=\vec i,\ \vec k\times\vec i=\vec j$, and $\vec i\times\vec i=\vec j\times\vec j=\vec k\times\vec k=\vec 0$
Property (I) follows from setting $\vec b=\vec 0$ in (III), which gives $\vec 0\times\vec a=-(\vec a\times\vec 0)$, together with the fact that $\vec 0=-\vec 0$ (a vector of magnitude $0$ has no definite direction, so whichever way we rotate, the magnitude stays $0$). Property (II) holds because the angle between a vector and itself is $\theta=0$ and $\sin0=0$, so the magnitude is $0$; that is, $\vec a\times\vec a=\vec 0$. For (IV) and (V), a proof by computation with components is the natural one, and we will prove them together in the next section, where we actually obtain the component expression. Property (VI) is as we confirmed in Example 15.5.
15.4 Component Expression of the Cross Product—Determinants and the Area of a Parallelogram
The definition of the cross product (Definition 15.5) was given in geometric terms: “magnitude $\abs{\vec A}\abs{\vec B}\sin\theta$, direction given by the right-hand screw rule.” As we did for the inner product, we want a way to compute it from the components alone. Let us begin with the simple case in which both vectors lie in the $xy$-plane (that is, their $z$ components are $0$).
15.4.1 The cross product of two vectors in a plane—component expression as a determinant
Derivation
Let $\vec A\times\vec B=\vec C$, and let $\vec A=(a_1,a_2,0)$ and $\vec B=(b_1,b_2,0)$. Since $\vec A,\vec B$ both lie in the $xy$-plane, Definition 15.5 shows that $\vec C$ points along the $z$-axis. As in Figure 15.4, let $\alpha,\beta$ be the angles of $\vec A,\vec B$, respectively, measured from the $x$-axis (suppose $\beta\gt\alpha$); then the angle from $\vec A$ to $\vec B$ is $\theta=\beta-\alpha$. The magnitude of the cross product is
$$ \abs{\vec A}\cdot\abs{\vec B}\cdot\sin\theta=\abs{\vec A}\abs{\vec B}\sin(\beta-\alpha) $$Using the addition formula for sine (High School Mathematics, Section 5.2), $\sin(\beta-\alpha)=\sin\beta\cos\alpha-\cos\beta\sin\alpha$, this becomes
$$ =\abs{\vec A}\abs{\vec B}\big(\sin\beta\cos\alpha-\cos\beta\sin\alpha\big) $$Now, by exactly the same idea as in the proof in Section 15.2, we express the angles $\alpha,\beta$ in terms of the components.
Attention: Expressing the trigonometric ratios of $\alpha,\beta$ in terms of components
$$ \sin\beta=\frac{b_2}{\abs{\vec B}},\qquad \sin\alpha=\frac{a_2}{\abs{\vec A}},\qquad \cos\beta=\frac{b_1}{\abs{\vec B}},\qquad \cos\alpha=\frac{a_1}{\abs{\vec A}} $$Substituting these gives
$$ \abs{\vec A}\abs{\vec B}\left(\frac{b_2}{\abs{\vec B}}\cdot\frac{a_1}{\abs{\vec A}}-\frac{b_1}{\abs{\vec B}}\cdot\frac{a_2}{\abs{\vec A}}\right) =\abs{\vec A}\abs{\vec B}\cdot\frac{a_1b_2-a_2b_1}{\abs{\vec A}\abs{\vec B}} =a_1b_2-a_2b_1 $$Here the factor $\abs{\vec A}\abs{\vec B}$ in the denominator cancels and disappears (the computation has exactly the same form as the proof for the inner product in Section 15.2). Therefore we obtain
$$ \abs{\vec A}\abs{\vec B}\sin\theta=a_1b_2-a_2b_1 $$The combination $a_1b_2-a_2b_1$ on the right-hand side is none other than the $2\times2$ determinant (Chapter 8, Section 8.3 of University Mathematics, or the very combination foreshadowed in the “Math Note” of High School Mathematics, Section 7.7),
$$ \begin{vmatrix}a_1&a_2\\b_1&b_2\end{vmatrix}=a_1b_2-a_2b_1 $$Writing $S$ for this quantity, we have found that $S=\abs{\vec C}$ (the magnitude of $\vec C=\vec A\times\vec B$).
Theorem 15.5: Component expression of the cross product (the case of vectors in a plane)
If $\vec A=(a_1,a_2,0)$ and $\vec B=(b_1,b_2,0)$ (vectors in the $xy$-plane), then
$$ \vec A\times\vec B=(a_1b_2-a_2b_1)\,\vec k=\begin{vmatrix}a_1&a_2\\b_1&b_2\end{vmatrix}\vec k $$Moreover, $\abs{\vec A\times\vec B}=\abs{a_1b_2-a_2b_1}$ equals the area $S$ of the parallelogram having $\vec A,\vec B$ as two of its sides.
Why is the absolute value needed?—Signed area
The quantity $a_1b_2-a_2b_1$ itself can take a positive value if $\vec B$ lies on the counterclockwise side of $\vec A$ ($\beta\gt\alpha$) and a negative value if it lies on the clockwise side ($\beta\lt\alpha$) (since an area itself is never negative, in general we attach the absolute value and write $S=\abs{a_1b_2-a_2b_1}$). This very sign corresponds to the “direction” of the cross product—whether $\vec C$ points in the $+z$ direction or in the $-z$ direction. When $\vec B$ lies on the clockwise side of $\vec A$, turning a right-handed screw from $\vec A$ toward $\vec B$ makes the screw advance in the $-z$ direction, so $\vec C$ points in the $-z$ direction. The sign thus tells us exactly this difference in direction.
15.4.2 The cross product of general three-dimensional vectors—completing the continuation that breaks off in the original notes
Here we must make an honest admission about the original notes. The notes set out to find the component formula for the cross product in the continuation of this argument, that is, for the case where $\vec A,\vec B$ are general three-dimensional vectors not restricted to the $xy$-plane (“Let $\vec a=[a_1,a_2,a_3]$ and $\vec b=[b_1,b_2,b_3]$ be arbitrary position vectors. Also, the plane through the origin $O$ spanned by $\vec a,\vec b$ is”), but the text breaks off midway. Since this book is written under the policy of “not omitting anything from the source notes,” we will not omit this continuation or gloss over it; instead, we will step in here and complete it to the end with our own hands.
Math Note: Accepting the linearity of the cross product and using it
To move on to general three-dimensional vectors, we use (IV) and (V) of Theorem 15.4—$(k\vec a)\times\vec b=\vec a\times(k\vec b)=k(\vec a\times\vec b)$ and $\vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c$. As for (IV), if $k\gt0$ then $k\vec a$ is merely a vector in the same direction as $\vec a$ with $k$ times the magnitude, so it follows almost directly from the definition that the cross product is multiplied by $k$ in both magnitude and direction (when $k\lt0$, the direction of $k\vec a$ is reversed, so the sense of rotation is also reversed; one more sign thus enters, and in the end the result is again multiplied by $k$). A complete geometric proof of (V) (that the cross product satisfies the distributive law over sums) requires a somewhat intricate argument—orthogonally project $\vec b,\vec c$ onto the plane perpendicular to $\vec a$, rotate the projected vectors by 90 degrees, and multiply by $\abs{\vec a}$—which goes beyond the scope of this book, so we omit the proof. In what follows, we accept that (IV) and (V) hold and derive the component formula; then, as a confirmation in the opposite direction, we verify with Python in Example 15.7 and Example 15.8 that the derived formula actually satisfies (IV) and (V).
Derivation
Write $\vec a=(a_1,a_2,a_3)=a_1\vec i+a_2\vec j+a_3\vec k$ and $\vec b=(b_1,b_2,b_3)=b_1\vec i+b_2\vec j+b_3\vec k$ as linear combinations of the standard unit vectors. Using the distributive law (V) twice (expanding on both the $\vec a$ side and the $\vec b$ side), we get
$$ \vec a\times\vec b=(a_1\vec i+a_2\vec j+a_3\vec k)\times(b_1\vec i+b_2\vec j+b_3\vec k) $$ $$ =a_1b_1(\vec i\times\vec i)+a_1b_2(\vec i\times\vec j)+a_1b_3(\vec i\times\vec k) +a_2b_1(\vec j\times\vec i)+a_2b_2(\vec j\times\vec j)+a_2b_3(\vec j\times\vec k) +a_3b_1(\vec k\times\vec i)+a_3b_2(\vec k\times\vec j)+a_3b_3(\vec k\times\vec k) $$That is, the product expands into $3\times3=9$ terms in all (we have written out every one of them without skipping any). Substituting here Theorem 15.4(VI) ($\vec i\times\vec j=\vec k,\ \vec j\times\vec k=\vec i,\ \vec k\times\vec i=\vec j$, and $\vec 0$ for a product of identical letters) and Theorem 15.3 (the anticommutative law, $\vec j\times\vec i=-\vec k,\ \vec k\times\vec j=-\vec i,\ \vec i\times\vec k=-\vec j$), we obtain
$$ \vec a\times\vec b=a_1b_1\vec 0+a_1b_2\vec k+a_1b_3(-\vec j)+a_2b_1(-\vec k)+a_2b_2\vec 0+a_2b_3\vec i+a_3b_1\vec j+a_3b_2(-\vec i)+a_3b_3\vec 0 $$Removing the terms containing $\vec 0$ and collecting the coefficients of each of $\vec i,\vec j,\vec k$, we find that
$$ \vec a\times\vec b=(a_2b_3-a_3b_2)\vec i+(a_3b_1-a_1b_3)\vec j+(a_1b_2-a_2b_1)\vec k $$holds.
Theorem 15.6: Component expression of the cross product (general three-dimensional case)
If $\vec a=(a_1,a_2,a_3)$ and $\vec b=(b_1,b_2,b_3)$, then
\begin{equation} \vec a\times\vec b=(a_2b_3-a_3b_2,\ a_3b_1-a_1b_3,\ a_1b_2-a_2b_1) \label{eq:15-cross3d} \end{equation}As a way of remembering it, the formula can be written in exactly the same form as the cofactor expansion (Chapter 8, Section 8.4 of University Mathematics) along the first row of the formal $3\times3$ determinant
$$ \vec a\times\vec b=\begin{vmatrix}\vec i&\vec j&\vec k\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix} $$(This is only a device for remembering how to compute the components: we are merely arranging $\vec i,\vec j,\vec k$ formally—as if they were numbers—in the positions of the entries of the determinant where numbers (scalars) would properly stand, and expanding. A determinant is originally a quantity defined for a square array of numbers (Section 8.3), and a determinant whose entries are vectors has not been defined, so the single word “formal” is needed.)
Let us also check that this formula does not contradict Theorem 15.5. Substituting $a_3=b_3=0$ (vectors in the $xy$-plane) into Eq. \eqref{eq:15-cross3d} gives $\vec a\times\vec b=(a_2\cdot0-0\cdot b_2,\ 0\cdot b_1-a_1\cdot0,\ a_1b_2-a_2b_1)=(0,0,a_1b_2-a_2b_1)$, which agrees exactly with Theorem 15.5.
Example 15.6: Finding the area of a triangle formed by three points using the cross product
Using the cross product, find the area of the triangle $PQR$ whose vertices are the three points $P(0,0)$, $Q(3,0)$, $R(1,4)$ in the plane.
Solution. Let $\vec u=\vec{PQ}=(3,0)$ and $\vec v=\vec{PR}=(1,4)$. By Theorem 15.5, the area of the parallelogram (spanned by the two sides $\vec u,\vec v$) is
$$ S_{\text{parallelogram}}=\abs{u_1v_2-u_2v_1}=\abs{3\times4-0\times1}=\abs{12}=12 $$Triangle $PQR$ is exactly half of this parallelogram, cut in two by a diagonal, so
$$ S_{\triangle PQR}=\frac12S_{\text{parallelogram}}=\frac12\times12=6 $$(SymPy check: $u_1v_2-u_2v_1=12$, area $=6$. Note that this formula is the same as $S=\frac12\abs{x_1y_2-x_2y_1}$ in the “Math Note” in High School Mathematics, Section 7.7, and the previews given there—“a quantity called the 2×2 determinant in university linear algebra” and “the planar version of the cross product, which you will study in vector analysis in the future”—have now been fulfilled here.)
Example 15.7: Verifying the distributive law with general components
Using the component formula of Theorem 15.6, verify that $\vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c$ (Theorem 15.4(V)) holds for general components $a_1,a_2,a_3,b_1,b_2,b_3,c_1,c_2,c_3$.
Solution. Since $\vec b+\vec c=(b_1+c_1,b_2+c_2,b_3+c_3)$, Eq. \eqref{eq:15-cross3d} shows that the first component of $\vec a\times(\vec b+\vec c)$ is
$$ a_2(b_3+c_3)-a_3(b_2+c_2)=a_2b_3+a_2c_3-a_3b_2-a_3c_2=(a_2b_3-a_3b_2)+(a_2c_3-a_3c_2) $$which is exactly the sum of “the first component of $\vec a\times\vec b$” and “the first component of $\vec a\times\vec c$.” For the second and third components, we see that they agree by exactly the same computation (we only use the distributive law of real numbers component by component). Hence $\vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c$. (SymPy check: we expanded both sides keeping the symbols $a_1,\ldots,c_3$ as they are, and confirmed that the difference is identically $\vec 0$.)
Example 15.8: Verifying scalar multiples with general components
Using the component formula of Theorem 15.6, verify that $(k\vec A)\times\vec B=\vec A\times(k\vec B)=k(\vec A\times\vec B)$ (Theorem 15.4(IV)).
Solution. Since $k\vec A=(ka_1,ka_2,ka_3)$, Eq. \eqref{eq:15-cross3d} shows that the first component of $(k\vec A)\times\vec B$ is
$$ (ka_2)b_3-(ka_3)b_2=k(a_2b_3-a_3b_2) $$which agrees with the first component $k(a_2b_3-a_3b_2)$ of $k(\vec A\times\vec B)$. The second and third components are similar. For $\vec A\times(k\vec B)$ as well, substituting $k\vec B=(kb_1,kb_2,kb_3)$ gives $a_2(kb_3)-a_3(kb_2)=k(a_2b_3-a_3b_2)$, and we obtain the same conclusion. (SymPy check: keeping the symbols as they are, we confirmed that the difference of the two sides is identically $\vec 0$.)
Now that the component expression (Theorem 15.6) is in hand, the cross product has become not only something defined geometrically (by magnitude and direction) but also a tool for concrete numerical computation. In the next section, we place the inner product and the cross product we have obtained side by side and consider the difference in their meanings—what each of them “measures.”
15.5 How the Meanings of the Inner Product and the Cross Product Differ
The inner product $\vec a\cdot\vec b=\abs{\vec a}\abs{\vec b}\cos\theta$ and the cross product $\vec a\times\vec b=\abs{\vec a}\abs{\vec b}\sin\theta\,\vec e_z$ are both built from the same two vectors $\vec a,\vec b$ and the angle $\theta$ between them. One uses $\cos\theta$ and the other uses $\sin\theta$—so what does each of them measure? Let us think about their meanings with the help of figures.
15.5.1 Meaning of the inner product—orthogonal projection onto the same direction
First, let us consider the meaning of the inner product formula $\vec a\cdot\vec b=\abs{\vec a}\abs{\vec b}\cos\theta$.
As Figure 15.6 shows, $\abs{\vec b}\cos\theta$ is the length of the red segment in the figure. This is nothing other than the distance from the origin to the foot of the perpendicular when a perpendicular is dropped from the tip of $\vec b$ to the line containing $\vec a$ (in the right triangle, the hypotenuse is $\abs{\vec b}$ and the adjacent side along the direction of $\vec a$ is the length we want, so from $\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}$ we get adjacent side $=\abs{\vec b}\cos\theta$). This quantity is called the length of the orthogonal projection of $\vec b$ onto the direction of $\vec a$. Moreover, $\abs{\vec b}\cos\theta$ is always parallel to $\vec a$ and overlaps it (lies on the same line).
Attention: Why $\vec b$ may be taken as the reference
Here we explained this as “projecting $\vec b$ onto the direction of $\vec a$,” but even if we swap the roles of $\vec a$ and $\vec b$ and think of “projecting $\vec a$ onto the direction of $\vec b$,” the value of the inner product itself, $\vec a\cdot\vec b=\abs{\vec a}\abs{\vec b}\cos\theta$, does not change. This is because the inner product satisfies the commutative law $\vec a\cdot\vec b=\vec b\cdot\vec a$ (Theorem 15.2(II)).
In other words, the inner product can be regarded as showing the degree (magnitude) to which $\vec b$ exerts influence in the same direction as $\vec a$, with $\vec a$ as the reference. The closer $\theta$ is to $0$ ($\vec a,\vec b$ pointing in almost the same direction), the closer $\cos\theta$ is to $1$, and the projection approaches its maximum value $\abs{\vec b}$. Conversely, as $\theta$ approaches $90^\circ$ the projection approaches $0$, and when $\theta$ exceeds $90^\circ$ and becomes obtuse, the projection becomes negative (opposite in direction to $\vec a$)—this is what the word “influence” means.
15.5.2 Meaning of the cross product—magnitude of rotation in the perpendicular direction
On the other hand, what about the cross product? Let us first consider the meaning of the cross product formula $\vec a\times\vec b=\abs{\vec a}\abs{\vec b}\sin\theta\cdot\vec e_z=\vec c$.
Attention: $\vec e_z$ is the standard unit vector in the direction of the $z$-axis
Here $\vec e_z$ is the unit vector perpendicular to the plane spanned by $\vec a,\vec b$ whose direction is determined by the right-hand screw rule (when $\vec a,\vec b$ are taken in the $xy$-plane as in Section 15.3, this coincides exactly with the standard unit vector $\vec k$ in the direction of the $z$-axis).
As Figure 15.7 shows, $\abs{\vec a}\abs{\vec b}\sin\theta$ is the area of the parallelogram with $\vec a,\vec b$ as two sides (this is the same content as Theorem 15.5 in Section 15.4, grasped anew from the figure alone, without using components). Considering the perpendicular direction, this area can also be interpreted as follows—it suffices to make a right triangle with $\vec b$ as the hypotenuse and measure the height perpendicular to the direction of $\vec a$.
The quantity $\abs{\vec b}\cos\theta$ in Figure 15.6 (the component in the direction parallel to $\vec a$) and the quantity $\abs{\vec b}\sin\theta$ in Figure 15.8 (the component in the direction perpendicular to $\vec a$) are related exactly as the adjacent side and the opposite side of a right triangle—when $\vec b$ is decomposed into a part parallel to $\vec a$ and a part perpendicular to it, the inner product extracts the magnitude of the parallel part, and the cross product extracts the magnitude of the perpendicular part (multiplied by the magnitude of $\vec a$). To sum up: whereas the inner product measures the “degree of influence in the same direction,” the cross product is a quantity that carries, at the same time, the “direction perpendicular to the plane spanned by $\vec a,\vec b$”—the axis of rotation—and the “strength of the twist (rotation).”
Example 15.9: Checking the orthogonal projection and the area of a parallelogram with concrete vectors
Let $\vec a=(4,0)$ and $\vec b=(3,3)$. (a) Find the angle $\theta$ between $\vec a,\vec b$ and the orthogonal projection $\abs{\vec b}\cos\theta$, and verify that this agrees with $\dfrac{\vec a\cdot\vec b}{\abs{\vec a}}$. (b) Find $\abs{\vec a}\abs{\vec b}\sin\theta$ (the area of the parallelogram), and verify that this agrees with the magnitude of the cross product, $\abs{a_1b_2-a_2b_1}$.
Solution. (a) Since $\vec a\cdot\vec b=4\times3+0\times3=12$, $\abs{\vec a}=4$, and $\abs{\vec b}=\sqrt{3^2+3^2}=3\sqrt2$, we have
$$ \cos\theta=\frac{\vec a\cdot\vec b}{\abs{\vec a}\abs{\vec b}}=\frac{12}{4\times3\sqrt2}=\frac{1}{\sqrt2}=\frac{\sqrt2}{2}\qquad(\theta=45^\circ) $$Therefore the orthogonal projection is
$$ \abs{\vec b}\cos\theta=3\sqrt2\times\frac{\sqrt2}{2}=3 $$On the other hand, $\dfrac{\vec a\cdot\vec b}{\abs{\vec a}}=\dfrac{12}{4}=3$, so the two indeed agree. (This is only to be expected—substituting $\cos\theta=\dfrac{\vec a\cdot\vec b}{\abs{\vec a}\abs{\vec b}}$ into $\abs{\vec b}\cos\theta$ gives $\abs{\vec b}\cdot\dfrac{\vec a\cdot\vec b}{\abs{\vec a}\abs{\vec b}}=\dfrac{\vec a\cdot\vec b}{\abs{\vec a}}$, which shows that the two agree identically.)
(b) Since $\sin\theta=\sin45^\circ=\dfrac{\sqrt2}{2}$, we have
$$ \abs{\vec a}\abs{\vec b}\sin\theta=4\times3\sqrt2\times\frac{\sqrt2}{2}=4\times3\times\frac{2}{2}=12 $$On the other hand, computing directly from the components gives $a_1b_2-a_2b_1=4\times3-0\times3=12$, so it indeed agrees with $\abs{a_1b_2-a_2b_1}=12$ (the area is $S=12$). (SymPy check: $\theta=45^\circ$, orthogonal projection $=3=\vec a\cdot\vec b/\abs{\vec a}$, $\abs{\vec a}\abs{\vec b}\sin\theta=12=\abs{a_1b_2-a_2b_1}$.)
Application: Torque $\vec\tau=\vec r\times\vec F$
As previewed in the lead at the beginning of this chapter, one of the most representative applications of the cross product is the torque (also called the moment of force) in mechanics. Let $\vec r$ be the position vector from a given point (the center of rotation) to the point where the force is applied, and let $\vec F$ be the force acting there. Then the torque $\vec\tau$ is defined by
$$ \vec\tau=\vec r\times\vec F $$For example, suppose that a force $\vec F=(0,2,0)$ [N] acts at the position $\vec r=(1,0,0)$ [m] from the origin (the center of rotation). Since $\vec r,\vec F$ both have $z$ component $0$, that is, both are vectors in the $xy$-plane, Theorem 15.5 (the component expression of the cross product in the plane) can be used as it is, and we obtain
$$ \vec\tau=\vec r\times\vec F=(r_1F_2-r_2F_1)\,\vec k=(1\times2-0\times0)\,\vec k=2\,\vec k\ [\text{N}\!\cdot\!\text{m}] $$The magnitude $\abs{\vec\tau}=2$ [N·m] equals the area of the parallelogram spanned by $\vec r,\vec F$ (the $S$ of Section 15.4.1), and the direction (the $+z$ direction) is determined by the right-hand screw rule itself—if we turn a right-handed screw from $\vec r$ (the direction of the $x$-axis) to $\vec F$ (the direction of the $y$-axis), the screw advances in the positive direction of the $z$-axis. Physically, this direction represents the “axis about which the body tends to rotate,” and the magnitude represents the “strength of the effect that tries to produce the rotation.” Whereas the inner product was a quantity measuring the “degree of influence in the same direction,” the torque, as an application of the cross product, is precisely a tool for quantifying the “rotation in the direction perpendicular to $\vec r,\vec F$” (the angular momentum $\vec L=\vec r\times\vec p$, which uses the momentum $\vec p$, is also a quantity with exactly the same structure). (SymPy check: $\vec\tau=(0,0,2)$, $\abs{\vec\tau}=2$.)
To close this chapter, let us borrow the words of the original notes just as they are—“Think about the perpendicular direction!” What the inner product taught us was information about the direction parallel to $\vec a$. What the cross product teaches us is the direction perpendicular to the plane spanned by $\vec a,\vec b$. In the next chapter, we move on to coordinate tools for treating such “perpendicular directions” systematically: the rectangular coordinate system whose axes are the three mutually orthogonal standard unit vectors $\vec e_x,\vec e_y,\vec e_z$, and the cylindrical and spherical (polar) coordinate systems.
15.6 Summary and Exercises
15.6.1 Summary
- Vector: a quantity with magnitude and direction. The zero vector $\vec 0$ (magnitude $0$); the unit vector $\vec e=\vec a/\abs{\vec a}$ (same direction as $\vec a$, with magnitude $1$).
- Inner product (scalar product): $\vec a\cdot\vec b=a_xb_x+a_yb_y=\abs{\vec a}\abs{\vec b}\cos\theta$ (Theorem 15.1, proved using the addition formulas). The result is a real number. The commutative law $\vec a\cdot\vec b=\vec b\cdot\vec a$ holds (Theorem 15.2). The standard unit vectors satisfy $\vec i\cdot\vec i=\vec j\cdot\vec j=\vec k\cdot\vec k=1$ and $\vec i\cdot\vec j=\vec j\cdot\vec k=\vec k\cdot\vec i=0$. $\vec a\cdot\vec b=0\iff\vec a\perp\vec b$.
- Cross product (vector product): $\vec A\times\vec B=\abs{\vec A}\abs{\vec B}\sin\theta\,\vec e_z$ ($\vec e_z$ is the unit vector determined by the right-hand screw rule; Definition 15.5). The result is a vector. The anticommutative law $\vec B\times\vec A=-(\vec A\times\vec B)$ holds (Theorem 15.3). The standard unit vectors satisfy $\vec i\times\vec j=\vec k,\ \vec j\times\vec k=\vec i,\ \vec k\times\vec i=\vec j$ (Theorem 15.4).
- Component expression of the cross product: in the plane, $\vec A\times\vec B=(a_1b_2-a_2b_1)\vec k$ (Theorem 15.5, a $2\times2$ determinant); in general three-dimensional space, $\vec a\times\vec b=(a_2b_3-a_3b_2,\ a_3b_1-a_1b_3,\ a_1b_2-a_2b_1)$ (Theorem 15.6, a formal $3\times3$ determinant). In either case, it corresponds to the (signed) area of the parallelogram spanned by $\vec a,\vec b$.
- The difference in meaning of the two products: the inner product $\abs{\vec b}\cos\theta$ is the orthogonal projection of $\vec b$ onto the direction of $\vec a$ (the degree of influence in the same direction). The cross product $\abs{\vec a}\abs{\vec b}\sin\theta$ is the strength of the rotation in the direction perpendicular to the plane spanned by $\vec a,\vec b$ (the area of the parallelogram).
In the next chapter (Chapter 16), starting from the standard unit vectors $\vec i,\vec j,\vec k$ (rectangular coordinates) that appeared in this chapter, we move on to other coordinate systems often used in physics—cylindrical and spherical (polar) coordinates—and verify the inner and cross products (orthogonality) of the basis vectors with one another in each coordinate system. The cross product is a tool used repeatedly throughout vector analysis as a whole, from the gradient ($\grad$) in Chapter 18 and the divergence and curl ($\mathrm{div},\ \rot$) in Chapter 19 all the way to the integral theorems (Stokes's theorem) in Chapter 20.
15.6.2 Exercises
Exercise 15.1: Unit vector
Find the unit vector of $\vec a=(1,2,2)$.
Hint: Use Eq. \eqref{eq:15-unitvec} ($\vec e=\vec a/\abs{\vec a}$). Even in three dimensions, the magnitude is the square root of the sum of the squares of the components.
Exercise 15.2: Inner product and the angle between vectors
Find the inner product of $\vec a=(2,1)$ and $\vec b=(-1,3)$, and the angle $\theta$ between the two vectors (the value of $\cos\theta$ suffices).
Hint: Compute the inner product using Definition 15.4, and use $\cos\theta=\dfrac{\vec a\cdot\vec b}{\abs{\vec a}\abs{\vec b}}$ from Theorem 15.1.
Exercise 15.3 (Advanced): The scalar triple product and volume
Let $\vec a=(1,0,0)$, $\vec b=(0,1,0)$, and $\vec c=(0,0,1)$. After finding $\vec b\times\vec c$, compute $\vec a\cdot(\vec b\times\vec c)$ (a quantity called the scalar triple product). Also verify that this value equals the volume of the parallelepiped (rectangular box) having $\vec a,\vec b,\vec c$ as three of its edges.
Hint: $\vec b\times\vec c$ can be computed with the component formula of Theorem 15.6. Compute the inner product (Definition 15.4) of the cross product $\vec b\times\vec c$ with $\vec a$. Note that $\vec a,\vec b,\vec c$ are three edges of the unit cube.
Exercise 15.4: Proof of the anticommutative law for general components
Using the component formula of Theorem 15.6, prove $\vec A\times\vec B=-(\vec B\times\vec A)$ for general components $a_1,a_2,a_3,b_1,b_2,b_3$.
Hint: Following the same approach as in Example 15.7 and Example 15.8, write down each component of $\vec A\times\vec B$ and $\vec B\times\vec A$ from Eq. \eqref{eq:15-cross3d}, and confirm that only the signs are reversed. (This result also supports, from the standpoint of component computation, Theorem 15.3, which was proved geometrically in Section 15.3 from the sense of rotation.)
Exercise 15.5: Area of a triangle formed by three points
Using the cross product, find the area of the triangle whose vertices are the three points $P(0,0)$, $Q(4,0)$, and $R(2,5)$ in the plane.
Hint: Following the same approach as in Example 15.6, form $\vec u=\vec{PQ}$ and $\vec v=\vec{PR}$, compute $\abs{u_1v_2-u_2v_1}$ from Theorem 15.5, and then multiply by $\frac12$.
Exercise 15.6: Does the cross product satisfy the associative law?
Using the standard unit vectors $\vec i,\vec j,\vec k$, compute $(\vec i\times\vec i)\times\vec j$ and $\vec i\times(\vec i\times\vec j)$ separately, and determine whether the two agree. From this result, answer whether one can say that the associative law $(\vec a\times\vec b)\times\vec c=\vec a\times(\vec b\times\vec c)$ holds in general for the cross product $\times$, or that it does not hold.
Hint: Use Theorem 15.4(VI) ($\vec i\times\vec i=\vec 0$, $\vec i\times\vec j=\vec k$). If one of the two turns out to be $\vec 0$ and the other a vector different from $\vec 0$, this is a counterexample showing that the associative law does not hold in general.
15.6.3 References
- Yasuhide Mochizuki, Mathematics Notes: Vector Analysis (handwritten lecture notes, in Japanese). The source of this chapter.
- Nagayoshi Iwahori, Bekutoru Kaiseki [Vector Analysis], Shōkabō, 1960 (in Japanese).