Chapter 7Vectors

7.1 Vector Operations — Equality, Addition, Subtraction, and Scalar Multiplication

In “walking 3 km east” and “walking 3 km north,” the distance (magnitude) walked is the same 3 km in both cases, but the direction walked is completely different. A quantity that has both this “magnitude” and a “direction” is called a vector, and it is written with an arrow over a letter, as in $\vec a,\ \vec b$. In figures, the direction is shown by the direction of the arrow and the magnitude by its length. The magnitude of a vector $\vec a$ is written $|\vec a|$. In contrast, a quantity that has no direction and is determined by its magnitude alone (length, mass, temperature, and so on) is called a scalar (that is, a real number).

A vector whose arrow has initial point $A$ and terminal point $B$ is also written $\vec{AB}$. In this case the magnitude $|\vec{AB}|$ of $\vec{AB}$ equals the distance between the two points $A,B$. What matters is that a vector is determined by its magnitude and direction alone, so it does not matter where we start drawing the arrow. Arrows with the same magnitude and the same direction represent the same single vector, wherever their initial points are.

7.1.1 Addition of vectors

The sum $\vec a+\vec b$ of two vectors $\vec a,\vec b$ is constructed as follows (the triangle rule). Draw an arrow so that $\vec a=\vec{OA}$, and then, from its terminal point $A$, draw an arrow $\vec{AB}$ with the same direction and magnitude as $\vec b$. Then the arrow $\vec{OB}$ going from the initial point $O$ to the terminal point $B$ is the sum $\vec a+\vec b$.

7.1.2 The zero vector and opposite vectors

It is convenient to consider a vector whose initial point and terminal point coincide (a vector that does not move). This is called the zero vector.

(I) follows from the fact that if, by the triangle rule, we take $\vec a=\vec{OA}$ and then join $-\vec a=\vec{AO}$ (the arrow returning from $A$ to $O$), the initial point and the terminal point are both $O$—that is, we obtain a vector of magnitude $0$. (II) follows from the fact that attaching the motionless arrow $\vec 0$ changes neither the initial point nor the terminal point.

Using opposite vectors, we can define the subtraction of vectors in terms of addition alone: $\vec a-\vec b:=\vec a+(-\vec b)$. In a figure, when $\vec a=\vec{OA}$ and $\vec b=\vec{OB}$ are drawn from the same initial point $O$, we get $\vec a-\vec b=\vec{BA}$ (the arrow from $B$ to $A$). Indeed, reading the triangle rule as a route that passes through $B$, we have $\vec{OB}+\vec{BA}=\vec{OA}$, that is, $\vec b+\vec{BA}=\vec a$, and adding the opposite vector of $\vec b$ to both sides gives $\vec{BA}=\vec a-\vec b$.

7.1.3 Scalar multiplication of vectors

Multiplying a vector $\vec a$ by $k$ (where $k$ is a real number) is defined by the following convention: it denotes the vector with the same direction as $\vec a$ and $k$ times the magnitude if $k\gt0$, the vector with the direction opposite to $\vec a$ and $|k|$ times the magnitude if $k\lt0$, and the zero vector $\vec 0$ if $k=0$. This is written $k\vec a$.

(I) follows from the fact that “multiplying $\vec a$ by $l$ and then by $k$ as well” and “multiplying $\vec a$ by $kl$ from the start” agree, whether we look at the direction or at the magnitude. (II) and (III) are, when the directions are the same, exactly the distributive law for lengths $(k+l)|\vec a|=k|\vec a|+l|\vec a|$, and when the direction is reversed, they can be checked in the same way by keeping track of the signs.

7.2 Conditions for Parallelism, Unit Vectors, and Linear Independence

7.2.1 The condition for parallelism of vectors

From the definition of scalar multiplication of a vector (Section 7.1.3), the following necessary and sufficient condition follows immediately. This is because $k\vec a$ is parallel to $\vec a$ and has the same direction if $k\gt0$, and has the opposite direction if $k\lt0$.

7.2.2 Unit vectors

The unit vectors (vectors of magnitude $1$) parallel to an arbitrary vector $\vec a\ (\ne\vec 0)$ are $\pm\dfrac{\vec a}{|\vec a|}$. Indeed, $\dfrac{\vec a}{|\vec a|}$ is the vector obtained by multiplying $\vec a$ by the real number $\dfrac1{|\vec a|}\ (\gt0)$, so it has the same direction as $\vec a$, and its magnitude is $|\vec a|\times\dfrac1{|\vec a|}=1$. The vector $-\dfrac{\vec a}{|\vec a|}$, obtained by reversing the sign, is another unit vector that differs only in that its direction is opposite.

7.2.3 Decomposition of vectors and linear independence

In the plane, if we have two arrows pointing in different directions, then any arrow can be built by “scaling those two arrows and adding them together”—this is the main theorem of this section. First, we prepare a term that states the condition “pointing in different directions” precisely.

7.3 Component Representation of Vectors and Internal and External Division Points

7.3.1 Components of a vector (in the plane)

Let $O$ be the origin of the coordinate plane, and for a vector $\vec a$ take the point $A$ so that $\vec a=\vec{OA}$, and write $A(x_1,y_1)$. Now take the point $E(1,0)$ on the $x$-axis and the point $F(0,1)$ on the $y$-axis, and let $\vec{OE}=\vec e_x$ and $\vec{OF}=\vec e_y$. Since $\vec e_x,\vec e_y$ are both vectors of magnitude $1$ (unit vectors) and are not parallel, they form a basis to which Theorem 7.5 of Section 7.2.3 (decomposition in terms of linearly independent vectors) applies.

Dropping perpendiculars $AX,AY$ from the point $A$ to the $x$-axis and the $y$-axis, respectively, we obtain $\vec{OA}=\vec{OX}+\vec{OY}$ by the triangle rule (Section 7.1.1). Here, $X$ is the point on the $x$-axis at a distance of $x_1$ (the coordinate of $X$) from $O$, so $\vec{OX}=x_1\vec e_x$, and similarly $\vec{OY}=y_1\vec e_y$. Therefore, this vector can be written in the form

in exactly one way (the uniqueness of the decomposition in terms of the linearly independent $\vec e_x,\vec e_y$ is precisely Theorem 7.5 itself).

7.3.2 Position vectors of internal and external division points

Let $\vec a=\vec{OA}$ and $\vec b=\vec{OB}$ be the position vectors of the points $A,B$, respectively (the position of a point expressed as a vector with respect to the origin $O$ is called its position vector). If $P$ is the point that divides segment $AB$ internally in the ratio $m:n$, then $\vec{AP}=\dfrac{m}{m+n}\vec{AB}$ holds (this merely restates, in the language of vectors, the idea of an internal division point that we have used since junior high school). Therefore,

(the last step merely regroups the numerator as $(m+n)\vec a-m\vec a+m\vec b=(m+n-m)\vec a+m\vec b=n\vec a+m\vec b$). For an external division point, similarly, if we start from $\vec{AP}=\dfrac{m}{m-n}\vec{AB}$ ($m\ne n$) and rearrange, the position vector

is obtained.

7.3.3 Vector operations in components

When vectors are given by their components, addition and scalar multiplication can be replaced by componentwise calculations. If $\vec a=(a_1,a_2)$ and $\vec b=(b_1,b_2)$, the following formulas hold.

These follow directly by substituting $\vec a=a_1\vec e_x+a_2\vec e_y$ and $\vec b=b_1\vec e_x+b_2\vec e_y$ and simplifying with the commutative and associative laws of Theorem 7.1 and the distributive laws of Theorem 7.3 (for example, $\vec a+\vec b=(a_1\vec e_x+a_2\vec e_y)+(b_1\vec e_x+b_2\vec e_y)=(a_1+b_1)\vec e_x+(a_2+b_2)\vec e_y$). Also, the vector joining the $2$ points $A(x_1,y_1),B(x_2,y_2)$ is $\vec{AB}=\vec{OB}-\vec{OA}=(x_2-x_1,\ y_2-y_1)$, and by the Pythagorean theorem its magnitude is $|\vec{AB}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ (this is exactly the distance formula for two points that we learned in Chapter 4).

7.4 The Inner Product of Vectors

7.4.1 Definition of the inner product

Addition and scalar multiplication of vectors were operations whose results were again vectors. Here we introduce a new operation that produces a single real number (a scalar) from two vectors. When we compute the “work done” in physics, if a force $\vec F$ and a displacement $\vec s$ point in the same direction, the work is $|\vec F||\vec s|$; but a force acting in a misaligned direction does work only through the part projected onto the direction of the displacement. The inner product is this idea of “discounting the misalignment of direction by $\cos$,” formulated as an operation on vectors in general.

7.4.2 Perpendicularity of vectors and the inner product

When $\theta=90^\circ$, we have $\cos90^\circ=0$, so the inner product is $0$. Conversely, if neither of $\vec a,\vec b$ is $\vec 0$, then $|\vec a||\vec b|\ne0$, so the inner product is $0$ only when $\cos\theta=0$, that is, only when $\theta=90^\circ$.

7.4.3 Component representation of the inner product

The definition of the inner product involves the angle $\theta$, so it is hard to compute directly from the components alone. So let us rewrite the inner product as an expression in the components only, using the law of cosines that we learned in Chapter 3. Let $\vec a=(a_1,a_2)$ and $\vec b=(b_1,b_2)$. Take $O(0,0)$ as the initial point, and choose the points $A(a_1,a_2)$ and $B(b_1,b_2)$ so that $\vec{OA}=\vec a$ and $\vec{OB}=\vec b$. Applying the law of cosines to triangle $OAB$ gives

(the last equality is just a direct substitution of $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$ from Definition 7.7). On the other hand, $|AB|^2$ can also be computed from the components, using the formula for the distance between two points (Section 7.3.3).

(we regrouped the terms using $|\vec a|^2=a_1^2+a_2^2$ and $|\vec b|^2=b_1^2+b_2^2$). Comparing the two computations of $|AB|^2$, we get $|\vec a|^2+|\vec b|^2-2\,\vec a\cdot\vec b=|\vec a|^2+|\vec b|^2-2(a_1b_1+a_2b_2)$; subtracting $|\vec a|^2+|\vec b|^2$ from both sides and dividing by $-2$ gives the following formula.

7.4.4 Properties of the inner product

7.5 Vector Equations of Lines and Normal Vectors (Plane Geometry)

7.5.1 Vector equation of a line

Until now, we have written the equation of a line in the form $y=ax+b$. Here we learn how to represent a line using vectors.

Suppose there is a vector $\vec a=(\alpha,\beta)$ (other than $\vec 0$). Let $t$ range over all real numbers, and let the vector representing the moving point $(x,y)$ be $t\vec a=(t\alpha,t\beta)=(x,y)$.

Comparing components, we obtain

If $\alpha\ne0$, the first equation gives $t=\dfrac{x}{\alpha}$, and substituting this into the second equation, we get

This is nothing other than the equation of the line through the origin with slope $\dfrac{\beta}{\alpha}$ (the case $b=0$ of $y=ax+b$). Thus $(x,y)=t\vec a$ is a vector equation of a line.

Next, we consider a general line, which does not necessarily pass through the origin. Let $A(\alpha,\beta)$ and $P(p,q)$ be points, and consider the line through the point $P$ parallel to $\vec a=\vec{PA}=(\alpha,\beta)$.

Let the moving point be $(x,y)$. By the triangle rule (Section 7.1.1),

that is,

From the first equation, $t=\dfrac{x-p}{\alpha}$ (with $\alpha\ne0$), and substituting this into the second equation $y=\beta t+q$, we obtain

so this is indeed a line with slope $\dfrac{\beta}{\alpha}$ passing through the point $P(p,q)$ (this agrees with the formula $y-q=m(x-p)$ for “the line through the point $(p,q)$ with slope $m$,” which has been familiar since junior high school).

7.5.2 Lines and normal vectors

We can also construct the equation of a line by using a vector perpendicular to the line.

Let $A(x_1,y_1)$ be a fixed point and $P(x,y)$ a moving point. Let $\vec n=(a,b)$ be a vector perpendicular to $\vec{AP}$. Suppose the point $P$ moves along some line and the point $A$ lies on that line.

Then, by Theorem 7.6 (perpendicularity and the inner product), $\vec n\cdot\vec{AP}=0$, that is, $\vec n\cdot(\vec{OP}-\vec{OA})=0$, and in terms of components this is written

The equation we have obtained is that of the line on which the points $A$ and $P$ lie.

7.6 Equations of Planes, Lines, and Spheres in Space

From here on we deal with vectors in space (three dimensions). Components are written in parentheses, as in $(X,Y,Z)$ (the original notes also use square brackets, $[X,Y,Z]$, but in this book we standardize on parentheses so that the notation is consistent with the plane case).

7.6.1 Planes and normal vectors

In space, we consider a vector perpendicular not to a line but to a plane. Let $A(x_1,y_1,z_1)$ be a fixed point on a certain plane and $P(x,y,z)$ a moving point. Let $\vec n=(a,b,c)$ be a vector perpendicular to $\vec{AP}$. The point $P$ moves on that plane.

Then, by exactly the same argument as in the plane, $\vec n\cdot\vec{AP}=0$, that is, $\vec n\cdot(\vec{OP}-\vec{OA})=0$, so that the relation

holds.

7.6.2 The equation of a line in space

In coordinate space, the vector equation of the line that passes through the point $A(x_0,y_0,z_0)$ and has direction vector $\vec d=(X,Y,Z)$ can be written, using the vector $\vec P=(x,y,z)$ that represents the moving point, as

(Here $\vec d=(X,Y,Z)$ is a vector parallel to the line.) Written in components, this is

That is, we have the following system of equations:

When none of $X,Y,Z$ is $0$, solving each equation in the form $t=\cdots$ gives

(Since the three expressions all represent the same $t$, we link the right-hand sides to one another with equal signs.)

7.6.3 The equation of a sphere

In coordinate space, let us find the equation of the sphere with center $A(x_0,y_0,z_0)$ and radius $r\ (\gt0)$. Using the vector $\vec P$ that represents the moving point, and noting that the sphere is “the set of points whose distance from the center $A$ is always $r$,” we can write its equation as

Now let $\vec P=(x,y,z)$. Then $\vec P-\vec{OA}=(x-x_0,\ y-y_0,\ z-z_0)$. Squaring both sides (in an equation containing an absolute value, squaring removes the square root and makes the equation easier to handle) gives

and, rewriting the left side using the component representation of the magnitude of a vector (Theorem 7.7 (IV) and Formula 7.3 extended to space), we find that

holds.

7.7 Conditions for Collinearity, Concurrency, and Coplanarity, and the Area of a Triangle

7.7.1 Vector equation of a line (the form through two points)

We now restate the vector equations of lines learned so far, as the form given by a fixed point and a direction vector and as the form through two distinct points.

Statement ② is obtained from ① by adopting $\vec d=\vec{AB}=\vec{OB}-\vec{OA}$ as the direction vector, which gives $\vec{OP}=\vec{OA}+t(\vec{OB}-\vec{OA})=(1-t)\vec{OA}+t\vec{OB}$ (if we let $s=1-t$, this takes the form $s+t=1$). Let us use a figure to see why the condition $s+t=1$ is imposed.

7.7.2 Conditions for collinearity, concurrency, and coplanarity

In ①, the first equivalence is exactly Theorem 7.4 (the condition for parallelism) together with the definition itself, and the second equivalence is obtained by setting $k=t$ in Theorem 7.8②. Statement ③ extends the idea of Theorem 7.8②—“the line through two points has $s+t=1$”—by one dimension to the plane spanned by the three points $A,B,C$. (For the same reason that a point in a plane could be expressed in exactly one way as a linear combination of two linearly independent vectors—Theorem 7.5—a point on a plane in space is also characterized as a linear combination of the linearly independent vectors $\vec{OA},\vec{OB},\vec{OC}$ in which the sum of the coefficients is 1.)

7.7.3 Area of a triangle

Let us find the area of triangle $OAB$ by using the inner product. If $\theta$ denotes the angle between the two sides, the area formula we already know from high school mathematics is

Here, since $0^\circ\le\theta\le180^\circ$ implies $\sin\theta\ge0$, we can write $\sin\theta=\sqrt{1-\cos^2\theta}$ (solving $\sin^2\theta+\cos^2\theta=1$ for $\sin\theta$ and choosing the positive root, since the sign is fixed by $\sin\theta\ge0$). Therefore,

(To bring the factor $|\vec{OA}||\vec{OB}|$ in front of the radical inside it, we squared it and multiplied.) Finally, using Definition 7.7, $\vec{OA}\cdot\vec{OB}=|\vec{OA}||\vec{OB}|\cos\theta$, to replace $\big(|\vec{OA}||\vec{OB}|\cos\theta\big)^2$ inside the radical by an expression in the inner product, we find that

holds.

7.8 Summary and Exercises

7.8.1 Summary

• A vector is a quantity that has a “magnitude” and a “direction.” Addition follows the triangle rule, and scalar multiplication is the operation of multiplying the length by $|k|$ while keeping the same direction ($k\gt0$) or reversing the direction ($k\lt0$).
• If there are two vectors that are not $\vec 0$ and are not parallel to each other (that is, linearly independent), then any vector in the plane can be expressed in exactly one way as a linear combination of them (Theorem 7.5). In space, three linearly independent vectors are needed.
• The component representation consists of the coefficients in the decomposition with respect to the standard unit vectors $\vec e_x,\vec e_y(,\vec e_z)$. The position vectors of internal and external division points are given by Formula 7.1.
• The inner product $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$ is given in components by $a_1b_1+a_2b_2(+a_3b_3)$ (Formula 7.3). That the inner product is $0$ is equivalent to the vectors being perpendicular (Theorem 7.6).
• Lines and planes can each be expressed in two ways: by a vector equation (the form using a parameter $t$) and by a form using a normal vector. For a sphere in space, start from $|\vec P-\vec{OA}|=r$ and square both sides to obtain an equation in the coordinates.
• That a point $P$ lies on line $AB$ is equivalent to $\vec{OP}=s\vec{OA}+t\vec{OB}\ (s+t=1)$. The condition for lying on plane $ABC$ is the three-term version of this, with $s+t+u=1$.
• The area of a triangle is $S=\dfrac12\sqrt{|\vec{OA}|^2|\vec{OB}|^2-(\vec{OA}\cdot\vec{OB})^2}$, and in components $S=\dfrac12|x_1y_2-x_2y_1|$.

7.8.2 Exercises

7.8.3 References