{"loaderData":{"0-0-56":{"bookData":{"_id":"6880fd9cb10f6b73b00d7f58","category_id":"630da3e25d2b6424645ccb5a","author_id":["6888dbf10cdd58e3928c9c5d"],"book_name":"Introduction to Vectors and Analytic Geometry","book_image":"/books/CoverImagepng-1753283996943.png","book_description":"

In this exposition, Emil Artin—one of the foremost mathematicians of the 20th century and a gifted expositor—provides a systematic exploration of fundamental concepts in analytic geometry, guiding the reader from the basic notions of points and their numerical representation in space to the more sophisticated definitions of vectors, distances, and angles. Beginning with a reflective analysis of how the abstract ideal of a point relates to measurable physical phenomena, the discussion naturally progresses toward precise mathematical formulations. Through careful derivations—illustrated with intuitive diagrams—the reader revisits essential geometric results such as the theorem of Pythagoras, and further extends these principles into higher-dimensional spaces. The text also introduces vectors rigorously, clarifying their operations, properties, and interpretations in terms of length and direction. It culminates with applications to geometric entities such as lines, planes, and their intersections, providing methods to calculate distances and angles using vectors, scalar products, and vector products. This approach not only emphasizes mathematical clarity but also equips the reader with versatile analytical tools applicable across dimensions.

","have_express_version":true,"is_published":false,"status":"ACTIVE","slug":"introduction-to-vectors-and-analytic-geometry","buy_regular_link":"","buy_express_link":"","chapter_name_type":1,"created_at":"2025-07-21T12:26:24.872Z","__v":0,"sequence":3,"chapter_data":[{"_id":"6880fe0fb10f6b73b00d7f97","book_id":"6880fd9cb10f6b73b00d7f58","chapter_name":"Points","sequence":1,"chapter_content":"

In high school, one learns to think of a point as something which has no size but to which one can assign a definite location. But this concept is so idealized that we cannot realize it in nature. For certainly any dot we draw on the blackboard has dimensions, and the principles of modern physics tell us that the determination of a precise location for an object is impossible. Therefore, the things which we shall call points must come from an ideal world, and that will be the world of numbers.

Let us analyze for a moment the procedure of making a graph. For example, we might be measuring the temperature at various times, say hourly, on a certain day. Then we would describe our data graphically as shown:

Each of the dots we have indicated, or each point of our graph, corresponds to a pair of numbers, e.g., (9 A.M., $30^{\\circ}$ ). The first number indicates the (signed) distance from a fixed vertical line of reference; this point is -3 hours from the noon-line. Similarly the second number indicates the distance above the zero-degree line, in this case +30 degrees.

If we could extend indefinitely the range of available times and temperatures, it is evident that we could represent every point in the plane of the paper by a pair of numbers, and by only one such pair. For instance a point well off to the right and well below the zero-degree line might be represented by (noon two days later, $-379^{\\circ}$ ), or simply (48, - 379). It is seen, then, that once a pair of reference lines (or axes) has been chosen, we can consider a point in the plane simply as a pair of numbers, representing its distances from the reference lines.

Likewise in space, we take three lines which intersect in a point. They then determine three planes, and we can identify every point in space with the triple of numbers consisting of its signed distances from these planes. Proceeding from this motivation, we define a point in n-dimensional space to be an n-tuple
$
\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right) \\quad \\text { of numbers. }
$
Then n-dimensional space is defined to be the totality, or set, of all n-tuples $\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)$ of numbers.

","chapter_express_content":"","public_visibility":true,"allow_part_see":false,"book_slug":"introduction-to-vectors-and-analytic-geometry","slug":"points","regular_authors":"","express_authors":"","created_at":"2025-07-21T12:26:24.875Z","__v":0,"header_data":[]},{"_id":"6880febbb10f6b73b00d7fdd","book_id":"6880fd9cb10f6b73b00d7f58","chapter_name":"Distance","sequence":2,"chapter_content":"

Now we return to two-dimensional space, where the notion of distance is a familiar one. In the figure below, $\\left(x_{1}, x_{2}\\right)$ and $\\left(y_{1}, y_{2}\\right)$ are two points in the plane, and the distance between them is the hypotenuse of the triangle with vertices
$
\\left(x_{1}, x_{2}\\right),\\left(y_{1}, x_{2}\\right) \\text { and }\\left(y_{1}, y_{2}\\right)
$
or
$
\\sqrt{\\left(y_{1}-x_{1}\\right)^{2}+\\left(y_{2}-x_{2}\\right)^{2}}
$
by the theorem of Pythagoras.

It is of some interest to give a very simple and often unfamiliar proof of the theorem of Pythagoras, as follows: We wish to prove that if $a$ and $b$ are the legs of a right triangle and $c$ is its hypotenuse, then $a^{2}+b^{2}=c^{2}$. Let us form a square of side $a+b$, and decompose it in two ways:

From (1), the area of the square is
$
c^{2}+4 \\Delta
$
where $\\Delta$ is the area of a right triangle with legs a and b. From (2), its area is
$
a^{2}+b^{2}+4 \\Delta
$
Thus
$
a^{2}+b^{2}+4 \\Delta=c^{2}+4 \\Delta
$
or$
a^{2}+b^{2}=c^{2}
$
which was to be proved.

By applying the theorem of Pythagoras twice, we see that the distance between two points $\\left(x_{1}, x_{2}, x_{3}\\right)$ and $\\left(y_{1}, y_{2}, y_{3}\\right)$ in 3-space is
$
\\sqrt{\\left(y_{1}-x_{1}\\right)^{2}+\\left(y_{2}-x_{2}\\right)^{2}+\\left(y_{3}-x_{3}\\right)^{2}}
$

Thus we are led to define the distance between two points $\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)$ and $\\left(y_{1}, y_{2}, \\ldots, y_{n}\\right)$ in n-space to be$
\\sqrt{\\left(y_{1}-x_{1}\\right)^{2}+\\left(y_{2}-x_{2}\\right)^{2}+\\ldots+\\left(y_{n}-x_{n}\\right)^{2}}
$

We observe that this is equal to
$
\\sqrt{\\left(x_{1}-y_{1}\\right)^{2}+\\left(x_{2}-y_{2}\\right)^{2}+\\ldots+\\left(x_{n}-y_{n}\\right)^{2}}
$

We also note that the distance between two points is 0 only if the points are not actually distinct, i.e., only if $x_{1}=y_{1}, x_{2}=y_{2}, \\ldots, x_{n}=y_{n}$. (Why?)

NOTE. When we write $b=\\sqrt{a}$, where $a$ is positive or zero, we refer to that number $b$ which is positive or zero and such that $b^{2}=a$.

","chapter_express_content":"","public_visibility":true,"allow_part_see":false,"book_slug":"introduction-to-vectors-and-analytic-geometry","slug":"distance","regular_authors":"","express_authors":"","created_at":"2025-07-21T12:26:24.875Z","__v":0,"header_data":[]},{"_id":"688100b6b10f6b73b00d802f","book_id":"6880fd9cb10f6b73b00d7f58","chapter_name":"Vectors","sequence":3,"chapter_content":"

In physics, one often refers to a vector as a quantity with length and direction, and represents it by an arrow ↗. This can hardly be called a definition, since almost all the terms used are undefined. To make it precise, observe that the “arrow” is determined if we specify: a) its initial point (the tail of the arrow); b) its end point (the head of the arrow).

Thus we are led to define a vector in n-space to be an ordered pair of points in n-space, therefore an ordered pair of n-tuples of numbers. The first point will be called the initial point, the second the end point.

It is likewise customary to regard two vectors as equal if they have the same length and the same direction, thus if their “arrows” satisfy these conditions. Let us look at this in 2-space:

If these two vectors are to be equal in this sense, it means that the two right triangles shown have their hypotenuse and one acute angle equal, therefore are congruent. In particular, the other sides are equal, i.e., \\[ y_{1}-x_{1}=v_{1}-u_{1}, \\quad y_{2}-x_{2}=v_{2}-u_{2} \\]

Therefore let \\[ \\left(\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right), \\quad\\left(y_{1}, y_{2}, \\ldots, y_{n}\\right)\\right) \\] and \\[ \\left(\\left(u_{1}, u_{2}, \\ldots, u_{n}\\right), \\quad\\left(v_{1}, v_{2}, \\ldots, v_{n}\\right)\\right) \\] be two vectors. We define them to be equal if \$y_{1}-x_{1}=v_{1}-u_{1}\$, \$y_{2}-x_{2}=v_{2}-u_{2}, \\ldots, y_{n}-x_{n}=v_{n}-u_{n}\$

It is seen, then, that a vector is determined to within equality by the numbers \$a_{1}=y_{1}-x_{1}, a_{2}=y_{2}-x_{2}, \\ldots, a_{n}=y_{n}-x_{n}\$. Thus one is led to consider a quantity which one writes as \$\\left[a_{1}, a_{2}, \\ldots, a_{n}\\right]\$. The numbers \$a_{1}, a_{2}, \\ldots, a_{n}\$ are called components of the vector, and the expression \$A=\\left[a_{1}, a_{2}, \\ldots, a_{n}\\right]\$ is called the free vector determined by the vector \\[ \\left(\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right), \\quad\\left(y_{1}, y_{2}, \\ldots, y_{n}\\right)\\right] \\] (or by any vector equal to this one).

Now we see how to find the free vector determined by any (ordered pair of points, and also how to find, for a given free vector and a given point, a second point such that the vector with the first point as initial point and the second as end point determines the given free vector. Given \$A=\\left[a_{1}, a_{2}, \\ldots, a_{n}\\right]\$ and initial point \$\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)\$, the desired end point is of course \$\\left(x_{1}+a_{1}, x_{2}+a_{2}, \\ldots, x_{n}+a_{n}\\right)\$. We shall often use the term located vector to refer to a pair of points, and the term “vector” will rapidly be usurped by the free vector.

The principle of addition of vectors is familiar from the parallelograms of forces used in physics. For the moment, let \$A=\\left[a_{1}, a_{2}\\right]\$ and \$B=\\left[b_{1}, b_{2}\\right]\$ be two vectors in 2-space. Let \$A\$ be located with initial point \$\\left(x_{1}, x_{2}\\right)\$, end point \$\\left(y_{1}, y_{2}\\right)\$. Then we have \$y_{1}=x_{1}+a_{1}\$, \$y_{2}=x_{2}+a_{2}\$. We form \$A+B\$ by locating \$B\$ with initial point \$\\left(y_{1}, y_{2}\\right)\$. Let \$\\left(z_{1}, z_{2}\\right)\$ be the end point of \$B\$. Then \$A+B\$ is the vector determined by the pair of points \$\\left(\\left(x_{1}, x_{2}\\right),\\left(z_{1}, z_{2}\\right)\\right)\$.

But \$z_{1}=y_{1}+b_{1}=x_{1}+a_{1}+b_{1}\$, \$z_{2}=y_{2}+b_{2}=x_{2}+a_{2}+b_{2}\$. Thus we have, for the free vectors, \\[ A+B=\\left[a_{1}+b_{1}, a_{2}+b_{2}\\right] \\]

This leads us to define, for arbitrary \$n\$ \\[ \\left[a_{1}, a_{2}, \\ldots, a_{n}\\right]+\\left[b_{1}, b_{2}, \\ldots, b_{n}\\right]=\\left[a_{1}+b_{1}, a_{2}+b_{2}, \\ldots, a_{n}+b_{n}\\right] \\]

Please note that this is a definition, so that it requires no proof. The things that do require proof are the elementary properties of this operation, and they are given as exercises:

\$A+B=B+A\$
\$(A+B)+C=A+(B+C)\$
There is a vector 0 such that for any vector \$A\$, \$A+0=A\$. (How many are there?)
Given any vector \$A\$, there is a vector \$(-A)\$ such that \$A+(-A)=0\$, where 0 is as in iii). (How many are there?)

NOTE. It seems advantageous at this stage to introduce the student to the words associative and commutative, and to the concept of a group. An opportunity is provided to present simple examples of operations which are non-commutative or which fail to satisfy some of the group axioms.

Let \$A=\\left[a_{1}, a_{2}, \\ldots, a_{n}\\right]\$ be a vector in n-space. We define the length of \$A\$ to be the distance between its initial point and its end point when it is located at some point. Let \$A\$ be located with initial point \$\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)\$, end point \$\\left(y_{1}, y_{2}, \\ldots, y_{n}\\right)\$. Then \\[ y_{1}=x_{1}+a_{1}, y_{2}=x_{2}+a_{2}, \\ldots, y_{n}=x_{n}+a_{n} \\] The length of \$A\$ is thus \\[ \\sqrt{\\left(y_{1}-x_{1}\\right)^{2}+\\left(y_{2}-x_{2}\\right)^{2}+\\ldots+\\left(y_{n}-x_{n}\\right)^{2}}=\\sqrt{a_{1}^{2}+a_{2}^{2}+\\ldots+a_{n}^{2}} \\]

From this we see that the point at which we located the vector \$A\$ does not affect its length. We shall denote this number (not a vector) by \$|A|\$. It is sometimes called the absolute value of A. We thus have the rule \\[ |A|=\\sqrt{a_{1}^{2}+a_{2}^{2}+\\ldots+a_{n}^{2}} \\]

We now have means of calculating witا two categories of objects; vectors and numbers (or scalars). We shall introduce more operations, which will mix these two categories. It should, however, be remembered that the two are distinct, and that an equation \$A=2\$, for instance, is as meaningless as the equation \" \$x\$ sheep \$=2\$ oranges\". Due to convention and the lack of suitable symbols, we use 0 to represent both the number zero and the zero-vector of iii). No confusion should arise from equations involving this symbol, however, because the rest of the equation will enable us to interpret the category correctly.

Next we define multiplication of vectors by numbers; if \$A=\\left[a_{1}, a_{2}, \\ldots, a_{n}\\right]\$ and \$b\$ is a number, we define \\[ b A=\\left[b a_{1}, b a_{2}, \\ldots, b a_{n}\\right] \\]

Note, for example, that for \$b=3\$, we have \$3 A=A+A+A\$, as in the case of numbers. Again some properties are given as exercises:

v. \$(a+b) C=a C+b C\$

vi. \$a(B+C)=a B+a C\$

vii. \$0 A=0\$

viii. \$1 A=A,\\ (-1) A=-A\$

ix. \$a(b C)=(a b) C\$.

Consider two (located) vectors \$\\left(\\left(x_{1}, x_{2}\\right),\\left(y_{1}, y_{2}\\right)\\right)\$ and \$\\left(\\left(u_{1}, u_{2}\\right),\\left(v_{1}, v_{2}\\right)\\right)\$ in 2-space. When do we say that they have the same direction? In the following diagram this amounts to saying that the two angles indicated by \$\\angle\$ are equal; that is, they are equal in magnitude and also in sign, where they are measured counterclockwise from the positive \$x\$-axis (i.e., the horizontal axis). Thus the tangents of these angles are equal, or \\[ \\frac{y_{2}-x_{2}}{y_{1}-x_{1}}=\\frac{v_{2}-u_{2}}{v_{1}-u_{1}} \\]

Multiplying by \\[ \\frac{y_{1}-x_{1}}{v_{2}-u_{2}} \\] gives \\[ \\frac{y_{2}-x_{2}}{v_{2}-u_{2}}=\\frac{y_{1}-x_{1}}{v_{1}-u_{1}} \\]

Let us call this number \$b\$. Then \$\\left[y_{1}-x_{1}, y_{2}-x_{2}\\right]=b\\left[v_{1}-u_{1}, v_{2}-u_{2}\\right]\$. Moreover, \$b>0\$ (i.e., \$b\$ is positive), as is evident from the diagram. (We have assumed in performing divisions that the vectors are neither vertical nor horizontal; it is left as an exercise to show that the same type of result holds in these cases.) Thus we are led to make the following definitions:

Let \$A, B\$ be non-zero vectors in n-space. We say that \$A\$ and \$B\$ have the same direction if \$B=b A\$ for some number \$b>0\$. We say that \$A\$ and \$B\$ have opposite directions if \$B=b A\$ for some number \$b<0\$. We say \$B\$ is parallel to \$A\$ if \$A\$ and \$B\$ either have the same direction or have opposite directions, therefore if \$B=b A\$ for some number \$b \\neq 0\$. (Note that then \$A\$ is also parallel to \$B\$.)

","chapter_express_content":"","public_visibility":true,"allow_part_see":false,"book_slug":"introduction-to-vectors-and-analytic-geometry","slug":"vectors","regular_authors":"","express_authors":"","created_at":"2025-07-21T12:26:24.875Z","__v":0,"header_data":[]},{"_id":"688867170cdd58e3928c97cb","book_id":"6880fd9cb10f6b73b00d7f58","chapter_name":"Straight Lines","sequence":4,"chapter_content":"

We have written two types of n-tuples of numbers: the n-tuple \$\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)\$ is a point in n-dimensional space, while the n-tuple \$A=\\left[a_{1}, a_{2}, \\ldots, a_{n}\\right]\$ is a free vector in n-space. We now drop the notation of the square brackets and write \$A=\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)\$. The apparent confusion resulting between points and free vectors is eliminated as follows. Given the point \$\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)\$, consider the free vector determined by the located vector with initial point \$(0,0, \\ldots, 0)\$ and end point \$\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)\$. In our new notation, this vector is also \$\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)\$. Thus one may consider the n-tuple \$\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)\$ as representing a point in n-space, or as representing the vector with \$(0,0, \\ldots, 0)\$ as initial point and this point as end point. By this identification, we may use vector addition to add objects which are ostensibly points, and then identify the resulting vector with a point.

Now we shall define a line in n-space. This is an object of a different category, neither point nor vector nor number. It is of the same category as the whole of n-dimensional space, in that it is a set, or collection of points. Thus to determine a line, we must give a rule which will enable us to decide whether a given point is on that line, or as we say, is a member of that set of points.

It is a familiar property of plane geometry that a line is determined by giving a point on it and a direction (say another line, or a vector) parallel to it. It is of course also determined by giving two distinct points on it, but this property is less convenient for our purposes. Let \$P=\\left(p_{1}, p_{2}\\right)\$ be a point in 2-space, and let \$A=\\left(a_{1}, a_{2}\\right)\$ be a vector, \$A \\neq 0\$. The question we must then answer is: Given a point \$X=\\left(x_{1}, x_{2}\\right)\$, when is it on the line through \$P\$ and parallel to \$A\$ ? It is clear that the line segment \$\\overline{P X}\$, or, rather, the vector with initial point \$P\$ and end point \$X\$, must be parallel to \$A\$.

It is evident from the diagram that this vector, when added to the vector \$P\$ (using our identification of points and vectors), gives the vector \$X\$. Hence the vector with initial point \$P\$ and end point \$X\$ is the vector \$X-P\$ (meaning \$X+(-P)\$; see the exercises on vector addition). Thus we must have \$X-P=t A\$, where \$t\$ is some non-zero number. Of course, if we set \$t=0\$, then \$X=P\$ is on the line. Thus \$X=P+t A\$, where \$t\$ is some number. Moreover, if \$t\$ is any number, the point \$X=P+t A\$ has the property that either \$X=P\$ or \$X-P\$ is parallel to \$A\$, i.e., \$X\$ lies on the desired line.

Now let \$P\$ be a point in n-space, \$A\$ a (non-zero) vector in n-space. We define the line through \$P\$ parallel to \$A\$ to be the set of all points \$X\$ such that \\[ X=P+t A \\] for some number \$t\$, i.e., where \$t\$ ranges through all numbers.

In 2-space, this takes the form \$\\left(x_{1}, x_{2}\\right)=\\left(p_{1}+t a_{1}, p_{2}+t a_{2}\\right)\$. Since \$A \\neq 0\$, either \$a_{1} \\neq 0\$ or \$a_{2} \\neq 0\$. Let us assume \$a_{1} \\neq 0\$. Then \\[ x_{1}=p_{1}+t a_{1}, \\quad \\text { or } \\quad t=\\frac{x_{1}-p_{1}}{a_{1}} \\] and \\[ x_{2}=p_{2}+t a_{2}=p_{2}+\\left(\\frac{x_{1}-p_{1}}{a_{1}}\\right) a_{2}=\\frac{a_{2}}{a_{1}} x_{1}+\\left(p_{2}-p_{1} \\cdot \\frac{a_{2}}{a_{1}}\\right) \\]

If we change our notation and write \$x\$ for \$x_{1}, y\$ for \$x_{2}, m\$ for \$\\frac{a_{2}}{a_{1}}\$ and \$b\$ for \$p_{2}-p_{1} \\frac{a_{2}}{a_{1}}\$, we obtain \\[ y=m x+b \\] which is the ordinary “slope-intercept” equation of a line. It means that the point ( \$x, y\$ ) is on the line if and only if \$y=m x+b\$, where \$m\$ and \$b\$ are the numbers defined above. The form \$P+t A\$ for the equation of a line is customarily called the parametric form. It is used in this course because of its complete generality for all dimensions and for all vectors \$A \\neq 0\$.

Now let \$P=\\left(p_{1}, p_{2}, \\ldots, p_{n}\\right)\$ and \$Q=\\left(q_{1}, q_{2}, \\ldots, q_{n}\\right)\$ be two points. What is the line through \$P\$ and \$Q\$ ? We have a point, say \$P\$, on the line, and a vector parallel to the line, namely the vector determined by the initial point \$P\$ and end point \$Q\$. But we have seen that this vector is just \$Q-P\$. Thus the desired line is the set of points \\[ X=P+t(Q-P) \\] where \$t\$ ranges through all numbers, or the set of points \\[ X=(1-t) P+t Q \\] where \$t\$ ranges through all numbers.

","chapter_express_content":"","public_visibility":true,"allow_part_see":false,"book_slug":"introduction-to-vectors-and-analytic-geometry","slug":"straight-lines","regular_authors":"","express_authors":"","created_at":"2025-07-29T05:29:08.616Z","__v":0,"header_data":[]},{"_id":"68886cc40cdd58e3928c991b","book_id":"6880fd9cb10f6b73b00d7f58","chapter_name":"Scalar Products","sequence":5,"chapter_content":"

The next definition we shall make is that of scalar product, whereby two vectors are combined to give a number.

Let \$A=\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)\$, \$B=\\left(b_{1}, b_{2}, \\ldots, b_{n}\\right)\$. We define the scalar product \$A\\boldsymbol{\\cdot} B\$ by \\[ A \\boldsymbol{\\cdot} B=a_{1} b_{1}+a_{2} b_{2}+\\ldots+a_{n} b_{n} \\]

It is left as an exercise to prove the following:

\$A\\boldsymbol{\\cdot}(B+C)=A\\boldsymbol{\\cdot} B+A\\boldsymbol{\\cdot} C\$
\$A\\boldsymbol{\\cdot} B=B\\boldsymbol{\\cdot} A\$
\$a(B\\boldsymbol{\\cdot} C)=(a B)\\boldsymbol{\\cdot} C=B\\boldsymbol{\\cdot}(a C)\$

In particular, we define \$A^{2}=A\\boldsymbol{\\cdot} A\$. Thus \$A^{2}=a_{1}^{2}+a_{2}^{2}+\\ldots+a_{n}^{2}\$. But if we recall that \\[ |A|=\\sqrt{a_{1}^{2}+a_{2}^{2}+\\ldots+a_{n}^{2}} \\] we have \\[ A^{2}=|A|^{2}. \\]

(Note that this equation makes sense, since both sides are numbers. Note also that if \$A \\neq(0,0, \\ldots, 0), A^{2}>0\$.)

Now let us try to express, in terms of the language we have developed, what is to be meant by saying that two vectors are perpendicular. Suppose that \$A\$ and \$B\$ of the diagram are perpendicular. In particular, this means that if we lay off a segment of length \$|B|\$ to the left of their common initial point, the two line segments marked || are of equal length. But if we consider the broken line segment as a vector directed to the left, we observe that this vector is \$-B\$, and that the two equal line segments, directed as indicated by the arrows, are \$A-B\$ and \$A-(-B)=A+B\$. Thus a necessary condition for perpendicularity is \$|A-B|=|A+B|\$.

Conversely, if this is the case, then \$A\$ is perpendicular to \$B\$ in the sense of plane geometry. Squaring both sides of this equation and using the fact that \$A^{2}=|A|^{2}\$ for any vector \$A\$, we have \\[ (A-B)^{2}=|A-B|^{2}=|A+B|^{2}=(A+B)^{2} \\] Using the rules developed in the exercises, \\[ (A-B)^{2}=A^{2}-2 A\\boldsymbol{\\cdot} B+B^{2},(A+B)^{2}=A^{2}+2 A\\boldsymbol{\\cdot} B+B^{2} \\]

Thus \$A^{2}-2 A\\boldsymbol{\\cdot} B+B^{2}=A^{2}+2 A\\boldsymbol{\\cdot} B+B^{2}\$, or

\\[ -2 A \\boldsymbol{\\cdot} B=2 A \\boldsymbol{\\cdot} B, \\quad \\text { i.e., } \\quad A\\boldsymbol{\\cdot} B=0 \\]

Conversely, if \$A\\boldsymbol{\\cdot} B=0\$, we see that \$|A-B|^{2}=|A+B|^{2}\$. But since both \$|A-B|\$ and \$|A+B|\$ are non-negative, this means that \$|A-B|=|A+B|\$, or that \$A\$ and \$B\$ are perpendicular.

Thus we may regard this result as our definition of perpendicularity in \$n\$ dimensions:

\$A\$ is perpendicular to \$B\$ will mean simply that \$A\\boldsymbol{\\cdot} B=0\$.

Now suppose \$A\$ and \$B\$ are non-zero vectors (recalling that the zero vector is \$(0,0, \\ldots, 0)\$ ). Then the quantity \\[ |B| A-|A| B \\] is a vector. Therefore the number \$(|B| A-|A| B)^{2}\$ is either positive or zero, and is zero only if \$|B| A-|A| B=0\$ (vector), i.e., only if \\[ |B| A=|A| B \\] Now \\[ \\begin{aligned} (|B| A-|A| B)^{2} &=|B|^{2} A^{2}-2|A||B| A \\boldsymbol{\\cdot} B+|A|^{2} B^{2} \\\\ &=|B|^{2}|A|^{2}-2|A||B| A \\boldsymbol{\\cdot} B+|A|^{2}|B|^{2} \\\\ &=2|A||B|(|A||B|-A\\boldsymbol{\\cdot} B) \\end{aligned} \\] and this is \$\\geq 0\$. Since \$2|A||B|\$ is positive, we can cancel it and get \$|A||B|-A B \\geq 0\$, or \\[ A \\boldsymbol{\\cdot} B \\leq|A||B| \\] The two are equal only if \$|B| A=|A| B\$, i.e., only if \$A\$ and \$B\$ have the same direction.

Likewise, \$(|B| A+|A| B)^{2} \\geq 0\$, and is zero only if \$|B| A=-|A| B\$, i.e., only if \$A\$ and \$B\$ have opposite directions. But \\[ (|B| A+|A| B)^{2}=2|A||B|(|A||B|+A \\boldsymbol{\\cdot} B) \\geq 0 \\] giving \\[ |A||B|+A \\boldsymbol{\\cdot} B \\geq 0, \\quad \\text { or } \\quad-|A||B| \\leq A \\boldsymbol{\\cdot} B \\]

Equality holds only if \$A\$ and \$B\$ have opposite directions. Combining the two results, we have in general \\[ -|A||B| \\leq A \\boldsymbol{\\cdot} B \\leq|A||B| \\] and we know exactly when one or the other of the inequalities becomes an equality. This result is known as Schwarz’s inequality.

Dividing the inequality through by \$|A||B|\$ (which is positive), we have

\\[ -1 \\leq \\frac{A \\boldsymbol{\\cdot} B}{|A||B|} \\leq 1 \\]

Therefore the number \$\\dfrac{A \\boldsymbol{\\cdot} B}{|A||B|}\$ can be the cosine of some angle. If we require that this angle \$\\theta\$ be between \$0^{\\circ}\$ and \$180^{\\circ}\$ (or in radians, between 0 and \$\\pi\$ ), then \$\\theta\$ is uniquely determined by giving \$\\cos \\theta\$. Thus we define the angle between \$A\$ and \$B\$ to be that angle \$\\theta\$ between \$0^{\\circ}\$ and \$180^{\\circ}\$ such that \\[ \\cos \\theta=\\frac{A \\boldsymbol{\\cdot} B}{|A||B|}. \\]

Observe that \$\\theta=0^{\\circ}\$ if and only if \$A\$ and \$B\$ have the same direction, \$\\theta=180^{\\circ}\$ if and only if they have opposite directions. So our terminology is consistent in this respect. Plane trigonometry will provide a more complete justification of our choice of definitions. Consider the triangle shown. The law of cosines tells us that \\[ |A-B|^{2}=|A|^{2}+|B|^{2}-2|A||B| \\cos \\theta \\]

Using what we know about vectors, this means that \\[ (A-B)^{2}=A^{2}+B^{2}-2|A||B| \\cos \\theta \\] or \\[ \\begin{gathered} A^{2}-2 A\\boldsymbol{\\cdot} B+B^{2} \\\\ =A^{2}+B^{2}-2|A||B| \\cos \\theta \\end{gathered} \\] i.e., \\[ -2 A \\boldsymbol{\\cdot} B=-2|A||B| \\cos \\theta \\] and finally \\[ \\bbox[5px,border:1px solid black;background-color:#f2f2f2]{A \\boldsymbol{\\cdot} B=|A||B| \\cos \\theta} \\] which agrees with our definition of \$\\theta\$.

","chapter_express_content":"","public_visibility":true,"allow_part_see":false,"book_slug":"introduction-to-vectors-and-analytic-geometry","slug":"scalar-products","regular_authors":"","express_authors":"","created_at":"2025-07-29T05:29:08.616Z","__v":0,"header_data":[]},{"_id":"688869bd0cdd58e3928c9830","book_id":"6880fd9cb10f6b73b00d7f58","chapter_name":"Planes","sequence":6,"chapter_content":"

We shall next define another geometrical configuration in n-space; its category is again that of a set of points. It will be called a plane, and once again it is determined by giving a point and a vector. Our experience in geometry tells us that in ordinary 3-space a plane is completely determined if we know a point on the plane and a direction (i.e., a vector or a line segment) perpendicular to the plane. We shall find an equation which will provide a rule for deciding whether a given point is on our plane. For a given point \$P\$ and a given vector \$A\$, the set of points whose coordinates satisfy this equation will then be called the plane through \$P\$ and perpendicular to \$A\$.

Suppose \$X=\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)\$ is any point on such a plane. Then the line segment joining \$X\$ and \$P\$ must lie in this plane and therefore must itself be perpendicular to A. Considering this segment as a vector \$X-P\$, as indicated in the diagram, we see that \$(X-P)\\boldsymbol{\\cdot} A=0\$, or that \\[ \\boxed{A\\boldsymbol{\\cdot} X=A \\boldsymbol{\\cdot} P} \\] This is then the desired equation, and the plane is the set of all points \$X=\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)\$ satisfying this relation.

For example, the equation of a plane in 3-space would look like \\[ 4 x-7 y+3 z=13 \\] where we use \$X=(x, y, z)\$ instead of \$\\left(x_{1}, x_{2}, x_{3}\\right)\$. This is of the form \$A \\boldsymbol{\\cdot} X=A \\boldsymbol{\\cdot} P\$, where \$A=(4,-7,3)\$. The point \$(1,0,3)\$ is a point of the plane, which we could take as our P. We see from this how to recover a vector perpendicular to a plane from the equation: we simply form a vector whose components are the coefficients of \$x, y\$ and \$z\$, in this notation.

It is also evident how to determine from a pair of equations of this form whether the planes they define are parallel. For this question is equivalent to the question of parallelism for two vectors perpendicular to the planes. Since such vectors are given by reading off the coefficients of \$x, y\$ and \$z\$, we see that the two planes are parallel if and only if these coefficients are proportional.

","chapter_express_content":"","public_visibility":true,"allow_part_see":false,"book_slug":"introduction-to-vectors-and-analytic-geometry","slug":"planes","regular_authors":"","express_authors":"","created_at":"2025-07-29T05:29:08.616Z","__v":0,"header_data":[]},{"_id":"68886a810cdd58e3928c9875","book_id":"6880fd9cb10f6b73b00d7f58","chapter_name":"Vector Product","sequence":7,"chapter_content":"

Because of its applications to physics, we shall define still another operation with vectors, which is again called a product. This product is only defined in case the vectors to be multiplied are vectors in 3-space, and its category is that of a vector. For the latter reason, it is called the vector product. It is written \$A \\times B\$ and is read \" \$A\$ cross \$B\$ \".

Let \$A=\\left(a_{1}, a_{2}, a_{3}\\right), B=\\left(b_{1}, b_{2}, b_{3}\\right)\$. Then we define \\[ A \\times B=\\left(a_{2} b_{3}-a_{3} b_{2}, a_{3} b_{1}-a_{1} b_{3}, a_{1} b_{2}-a_{2} b_{1}\\right) \\]

Exercises

\$A \\times B=-(B \\times A)\$.
\$A \\times(B+C)=(A \\times B)+(A \\times C)\$.
\$(a A) \\times B=a(A \\times B)=A \\times(a B)\$
\$(A \\times B) \\times C=(A\\boldsymbol{\\cdot} C) B-(B \\boldsymbol{\\cdot} C) A\$
\$((A \\times B) \\times C)+((B \\times C) \\times A)+((C \\times A) \\times B)=0\$
\$A \\times B\$ is perpendicular to both \$A\$ and \$B\$.
\$(A \\times B)^{2}=A^{2} B^{2}-(A \\boldsymbol{\\cdot} B)^{2}\$. Use this to give a geometric meaning to \$|A \\times B|\$.

","chapter_express_content":"","public_visibility":true,"allow_part_see":false,"book_slug":"introduction-to-vectors-and-analytic-geometry","slug":"vector-product","regular_authors":"","express_authors":"","created_at":"2025-07-29T05:29:08.616Z","__v":0,"header_data":[]},{"_id":"68886be20cdd58e3928c988b","book_id":"6880fd9cb10f6b73b00d7f58","chapter_name":"Applications and Problems","sequence":8,"chapter_content":"

We shall present a few applications of the technique of vectors to geometry, followed by some problems of more than routine interest. This list should be supplemented by a number of the problems to be found in any of the standard texts on analytic geometry of two and three dimensions.

Intersection of Two Planes in 3-Space

Consider the problem of finding the intersection of two planes in 3-space. In general, this will be a line; we can derive its equation from those of the planes in several ways:

We could solve simultaneously the equations of the planes.
We could find two points which are common to both planes, for instance by choosing two different fixed values for their third coordinates. We then obtain in each case a system of two linear equations in two unknowns (the first and second coordinates), which we solve. Then we find the connecting line.
We could find one point \$P\$ common to both planes, and then find a vector parallel to the line of intersection. The latter is accomplished by observing that any such vector \$A\$ must be perpendicular to each of the known vectors \$B, C\$ perpendicular to the individual planes. From the exercises on vector products, we see that we can take as our vector \$A\$ the vector product \$B \\times C\$. The formula \$P+t A\$ gives the line.

Distance from a Point to a Line in n-Space

Let \$P+t A\$ be a line in n-space, \$Q\$ a point not on the line; let us find the foot of the perpendicular from \$Q\$ to this line.

If \$X\$ is this point, then \$X=P+t_{0} A\$ for some \$t_{0}\$. Moreover, the segment connecting \$X\$ with \$Q\$, or the vector \$X-Q\$, is to be perpendicular to the line \$P+t A\$, therefore to the vector \$A\$. Now \$X-Q=P+t_{0} A-Q\$, giving \\[ \\left(P+t_{0} A-Q\\right)\\boldsymbol{\\cdot} A=0 \\]

This is a linear equation, which we solve for \$t_{0}\$. The desired point is then found by substituting this value for \$t_{0}\$ in \$X=P+t_{0} A\$. The distance between \$Q\$ and the line is simply \$|X-Q|\$.

Distance Between Two Lines

Now let \$P+t A\$ and \$Q+s B\$ be two non-intersecting lines. Let \$X=P+t_{0} A\$ and \$Y=Q+s_{0} B\$ be the endpoints of their common perpendicular. Then \$(X-Y)\\boldsymbol{\\cdot} A=0\$ and \$(X-Y)\\boldsymbol{\\cdot} B=0\$. These give us two linear equations which we solve for \$s_{0}\$ and \$t_{0}\$. The resulting points \$X\$ and \$Y\$ determine the common perpendicular. The distance between the two lines is \$|X-Y|\$.

Angle Between Two Planes

The angle between two planes is found by using the angle between two vectors, i.e., by reading off two vectors perpendicular to the planes from their equations and finding the angle between these vectors. Since two vectors having opposite directions determine the same plane, there are two possible answers, consisting of supplementary angles. We agree to take the angle between \$0^{\\circ}\$ and \$90^{\\circ}\$.

PROBLEMS

1. If a is a number, define its absolute value |a| by \$|a|=\\sqrt{a^{2}}\$. Show that \$|a A|=|a||A|\$, where \$A\$ is a vector.

2. If \$A\$ and \$B\$ are vectors, show that \$(|A|+|B|)^{2}-(A+B)^{2} \\geq 0\$. Use this to prove the triangle inequality: \\[ |A+B| \\leqq|A|+|B|. \\] When can this be an equality?

3. In n-space, let \$A_{1}=(1,0, \\ldots, 0), A_{2}=(0,1,0, \\ldots, 0), \\ldots, A_{n}=(0, \\ldots, 0,1)\$. Show the following: a) If \$a_{1}, a_{2}, \\ldots, a_{n}\$ are numbers such that \$a_{1} A_{1}+a_{2} A_{2}+\\ldots+a_{n} A_{n}=0\$, then \$a_{1}=a_{2}=\\ldots=a_{n}=0\$. b) If \$B\$ is any vector in n-space, there are numbers \$c_{1}, c_{2}, \\ldots, c_{n}\$ such that \$B=c_{1} A_{1}+c_{2} A_{2}+\\ldots+c_{n} A_{n}\$. How many such sets of numbers are there for a given \$B\$ ? c) \$A_{1}^{2}=1\$ and \$A_{1} A_{j}=0\$ if \$1 \\neq j\$, for all 1 and \$j\$. d) If \$B\$ is any vector in n-space and \$B A_{1}=B A_{2}=\\ldots=B A_{n}=0\$, then \$B=0\$. e) If \$A\$ is any vector in n-space, and if \$A=a_{1} A_{1}+a_{2} A_{2}+\\ldots+a_{n} A_{n}\$, then \$|A|^{2}=a_{1}^{2}+a_{2}^{2}+\\ldots+a_{n}^{2}\$ and each \$a_{1}=A A_{1}\$.

4. Show by vectors that the diagonals of a rhombus are perpendicular.

5. Show that the medians of the triangle with vertices \$P, Q\$, \$R\$ meet in \$\\frac{1}{2}(P+Q+R)\$. (Observe that the midpoint of the side \$\\overline{P Q}\$ is \$\\frac{1}{2}(P+Q)\$.

6. What is the plane through \$(-1,2,-3)\$ such that any vector perpendicular to the plane forms equal angles with the vectors \$(1,0,0),(0,1,0)\$ and \$(0,0,1)\$ ?

7. Show that the planes \$x-y+2 z=4\$ and \$2 y-2 x-4 z+7=0\$ are parallel. Find the distance between them.

8. Let \$A\$ and \$B\$ be non-zero vectors in 3-space. Prove that \$A\$ and \$B\$ are parallel if and only if \$A \\times B=0\$.

9. Prove the law of sines using vectors in 3-space.

10. What is the set of points \$X\$ in 3-space such that the vector from \$X\$ to \$(1,3,5)\$ is perpendicular to the vector from \$(-5,-1,3)\$ to \$X\$ ? Show that this is the equation of a sphere with center \$(-2,1,4)\$ and radius \$\\sqrt{ } 14\$.

11. Let \$B\$ and \$W\$ be vectors, different from 0 . Show that there exist vectors \$B_{1}\$ and \$B_{2}\$ such that

\$B_{1}\$ is parallel to \$W\$,
\$B_{2}\$ is perpendicular to \$W\$,
\$B_{1} \\cdot B_{2}=0\$, and
\$B=B_{1}+B_{2}\$ Show that \$B_{1}\$ and \$B_{2}\$ are unique.

12. Let \$R\$ and \$S\$ be the points \$(1,2,3,4)\$ and \$(4,3,2,1)\$; let \$A\$ be the vector \$(1,1,1,1)\$; let \$M\$ be the line through \$R\$ parallel to \$A\$; and notice that \$S\$ does not lie on \$M\$.

Given a point \$Q\$ on \$M\$, compute the distance between \$S\$ and \$Q\$ and show that it is \$\\geq 2 \\sqrt{5}\$.
Show that the distance between \$S\$ and \$Q\$ is precisely \$2 \\sqrt{5}\$ for just one \$Q_{0}\$, being actually \$>2 \\sqrt{5}\$ for all other points \$Q\$. Compute the coordinates of this \$Q_{0}\$.
Show that the vector \$S-Q_{0}\$ is perpendicular to the vector A.

13. Keeping the results of (2) in mind, consider (in n-space) a point \$R\$ and a vector \$A\$ whose length \$|A|\$ is \$>0\$; let \$M\$ be the line through \$R\$ parallel to \$A\$; and let \$S\$ be a point not on \$M\$.

Show that there is just one point \$Q_{0}\$ on \$M\$ such that the vector \$S-Q_{0}\$ is perpendicular to \$A\$ and compute its coordinates.
Given a point \$Q\$ on \$M\$, compute the distance from \$S\$ to \$Q\$ and show that it is \$\\geq\\left|S-Q_{0}\\right|\$. Show that the distance from \$S\$ to \$Q\$ is precisely \$\\left|S-Q_{0}\\right|\$ if and only if \$Q=Q_{0}\$.

14. Compute (in 3-space) the equation of the plane which passes through the points \$(1,0,0),(0,1,0)\$, and \$(0,0,1)\$.

Compute the distance \$|S-Q|\$ between the point \$S=(2,2,2)\$ and a point \$Q\$ in this plane.
Compute the minimum distance and call it \$\\Delta\$.
Show that there is just one point \$Q_{0}\$ in the plane such that the distance \$\\left|S-Q_{0}\\right|=\\Delta\$. Compute the coordinates of \$Q_{0}\$.
Show that the vector \$S-Q_{0}\$ is perpendicular to the plane.

Given three vectors \$A, B\$, and \$C\$ (in n-space) show that the relation \$(A \\cdot B) C=A(B \\cdot C)\$ is not, in general, true.

A cube in n-dimensional space is a set of points ( \$x_{1}, \\ldots\$, \$x_{n}\$ ) satisfying \\[ a_{1} \\leqq x_{1} \\leqq a_{1}+d, a_{2} \\leqq x_{2} \\leqq a_{2}+d, \\ldots, a_{n} \\leqq x_{n} \\leqq a_{n}+d, \\] for some set of \$n\$ numbers \$a_{1}, \\ldots, a_{n}\$ and some \$d>0\$. Then the vertices of the cube are the points of the form \$\\left(b_{1}, \\ldots, b_{n}\\right)\$, where each \$b_{i}\$ is either \$a_{i}\$ or \$a_{i}+d\$. An edge of the cube is a line segment joining two vertices \$\\left(b_{1}, \\ldots, b_{n}\\right)\$ and \$\\left(c_{1}, \\ldots, c_{n}\\right)\$, where \$b_{i}=c_{i}\$ for all except one of the \$i\$’s.

How many vertices and edges has a cube in 2-space? in 3-space? in 4-space? in n-space? How would you define a 2-dimensional face of an n-dimensional cube ( \$n \\geq 2\$ )? a k-dimensional face ( \$n \\geq k\$ )? How many of these are there?

","chapter_express_content":"","public_visibility":true,"allow_part_see":false,"book_slug":"introduction-to-vectors-and-analytic-geometry","slug":"applications-and-problems","regular_authors":"","express_authors":"","created_at":"2025-07-29T05:29:08.616Z","__v":0,"header_data":[]},{"_id":"69476e67ca806ba1072809a2","book_id":"6880fd9cb10f6b73b00d7f58","chapter_name":"Advanced Topics","sequence":9,"chapter_content":"

Intersection of Two Planes in 3-Space

Distance from a Point to a Line in n-Space

Distance Between Two Lines

Angle Between Two Planes

PROBLEMS

More Advanced Topics

Introduction to Vectors and Analytic Geometry

By Unknown Author

Table of Contents

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