Vectors in 3-D Space: for College-bound Students (Part of Quick Review / Preview Series in Math)

In this short course, we present 'Vectors in 3-D Space' in a systematic way so that students can develop and apply the concept of vectors through concrete objects in 3-D space before they learn more abstract vectors treated in so-called Linear Algebra in colleges and universities. In the U.S., students usually take Calculus AB or BC before graduation from high-schools, but Linear Algebra is rarely offered in many schools in general because of lack of time and they usually encounter geometrical vectors (representing a force, velocity, or acceleration) only through the course of physics. We hope that students will easily grasp the concept of vectors and can immediately apply to 3-D objects such as straight lines, planes, spheres at will. Those who have familiarized themselves with vectors through this textbook will have advantages when they encounter forces, velocities, electric field strength, Lorentz force, magnetic flux density, vector potential, and so on, to name really a few concepts in physics, where vectors are constantly used. Moreover, after 'basis' is introduced, they will know the fact that the familiar real numbers in the 1-D straight line are nothing but vectors in 1-D Space. They will also realize that calculating the distance between the planes in 3-D space can be carried out simply by projection of points on the normal axis as exactly as in calculating the distance between 2 numbers along the x-axis.

A typical example is in order. Take for instance a line on the x-y plane: 3x+4y = 5. In the inner product form, it is expressed
<3, 4 | x, y> = 5 –- (1).
(Here, since neither arrow nor bold face is available, vectors are expressed in terms of the bra-ket notation.) Dividing with the magnitude | n | = sqrt(9+16) = 5 of the normal vector | n > = (3, 4), it becomes the distance form:
<3/5, 4/5 | x, y> = 1, [<=> (3/5)x+(4/5)y = 1],
of which right-hand side denotes the distance from the origin O.
The distance function may be put as h(x, y) = <3/5, 4/5 | x, y> –- (2).
If we substitute (1) into (2), we obtain the distance from O. It means that any point on the line (1) keeps the same distance from O: 1. Now, substitute (-2, -1) into (2). We have h(-2, -1) = -2, which indicates the distance from the origin is 2, but on the negative side of the normal axis. What is the distance from the point (-2, -1) to the line (1)? We have only to subtract and take the absolute value if the result is negative:
|1-(-2)| = 3,
where the calculation can be carried out as exactly as in the x-axis. The axis is in this case the normal axis of the line (1) instead of the x-axis.
If you have learned the distance formula in the class, for instance, and have been accustomed to plugging in a value into the formula without clear understanding, stop such practice. One of the values of learning math consists in acquiring the methods or way of mathematical thinking. We show you what is the essence of mathematical thinking about through the illustrated problems for which answers are described in detail and in many ways for some cases.

As in another textbook in the series, we describe the summary as concisely as possible together with typical sample problems at first. Afterwards, you will make practice by engaging in solving problems in the following exercise sections. Those exercise sections will be divided according to the level of difficulty or whether they contain combined concepts. After finishing the course, we recommend that you engage in solving illustrated problems periodically so that you refresh your understandings on the subject.