Solid Analytic Geometry

CHAPTER 1

COORDINATES AND LINES

1. Vectors. A sequence P = (x1,…, xn) of n numbers xi is called an n-dimensional vector. The elements x1,… , xn are called the coordinates of P and xi is the ith coordinate. We shall limit our attention to real vectors, i.e., to vectors whose coordinates are real numbers.

The vector whose coordinates are all zero is called the zero vector and will be designated by 0. A real vector may be interpreted as a representation, relative to a fixed coordinate system with O as origin, of a point in n-dimensional real Euclidean space.

It may also be interpreted as the line segment directed from O to P. These interpretations have little intuitive significance except for the cases n ≦ 3, and we shall carry out the details in this text for the case n = 3.

The sum P + Q of two vectors P = (x1,…, xn) and Q = (y1,…, yn) is the vector (x1 + y1,…, xn + yn) whose ith coordinate is the sum xi + yi of the ith coordinate of P and the ith coordinate of Q. We leave the verification of the following simple results to the reader:

Lemma 1. Addition of vectors is commutative, that is, P + Q = Q + P for all vectors P and Q.

Lemma 2. Addition of vectors is associative, that is, (P + Q) + R = P + (Q + R) for all vectors P, Q, R.

Lemma 3. The zero vector 0 has the property that P + 0 = P for all vectors P.

Lemma 4. Let P = (x1,…, xn). Then the vector – P = (– x1,…, – xn) has the property that P + (–P) = 0.

Lemma 5. If P and Q are any vectors the equation P + X = Q has the solution X = Q + (– P). We call this vector the difference of Q and P and write X = Q – P. Then the ith coordinate of Q – P is the difference of the ith coordinate of Q and the ith coordinate of P.

EXERCISE

Verify the five lemmas.

2. Scalar multiplication. If a is a number and P = (x1,…, xn) is a vector, we define the scalar product of P by a to be

Evidently 1P = P, (–1)P = –P, 0 P = 0. The reader should verify that

for all scalars a and b and all vectors P and Q.

A sum

of scalar products ajPj of vectors Pj by scalars aj, is called a linear combination of P1,…, Pm. We shall say that P1,…, Pm are linearly independent vectors if it is true that a linear combination a1P1 + · + amPm = 0 if and only if a1,…, am are all zero. If P1,…, Pm are not linearly independent, we shall say that P1,…, Pm are linearly dependent.

Let Ei be the vector whose ith coordinate is 1 and whose other coordinates are all zero. Then

Thus every vector is a linear combination of E1,…, En. If x1E1 + · · · + xnEn = P = 0, then (x1,…, xn) = 0, that is, x1 = x2 = ··· = xn = 0. It follows that E1,…, En are linearly independent.

EXERCISES

1. Show that if P = (x1,…, xn) and Q = (y1,…, yn) are not zero then P and Q are linearly dependent if and only if Q is a scalar multiple of P.

2. Show that if P1,…, Pm are linearly independent and Pm+1 is another vector then P1,…, Pm, Pm+1 are linearly dependent if and only if Pm+1 is a linear combination of P1,…, Pm.

3. Compute the following linear combinations of P1 = (1, –1, 2, 3), P2 = (0, 1, –1, 2), P3 = (–2, l, –1, 2).

(a)2P1 + P2 + P3

(b)P1 + 3P2 – 2P3

(c)3P1 + 2P2 – 4P3

4. Use Exercises 1 and 2 in determining which of the following sets of three vectors are linearly independent sets.

(a)(1, –1, 2), (1, 1, 0), (0, –1, 1)

(b)(2, 1, 1), (1, –1, 1), (5, 4, 2)

(c)(1, 0, –2), (2, –1, 2), (4, –3, 10)

(d)(1, –1, 1), (–1, 2, 1), (–1, 2, 2)

(e)(1, 0, –1, 1), (0, –1, 1, –1), (4, –1, –3, 4)

(f)(5, 1, –2, –6), (1, 1, 0, –2), (2, –1, –1, 0)

(g)(1, 0, 0, 0), (1, 1, 1, 1), (3, 1, 1, 1)

(h)(1, 1, –1, 2), (2, 2, –2, 3), (3, 3, –2, 6)

5. Prove that any three two-dimensional vectors are linearly dependent.

6. Prove that any four three-dimensional vectors are linearly dependent.

3. Inner products. If P = (x1,…, xn) and Q = (y1,…, yn) are any two vectors, we shall call the number

the inner product of P and Q. Evidently, P · Q = Q · P.

The norm of a vector P is defined to be the inner product

If P is any real vector, the number P · P ≧ 0 and has a nonnegative square root

which we shall call the length of P.

A vector P is called a unit vector if P · P = 1. Thus a real unit vector is a vector whose length (and whose norm) is 1.

Lemma 6. Every real nonzero vector is a scalar multiple of exactly two unit vectors. These are the vectors U = t–1P and – U, where t is the length of P. Then if P = tU, where t > 0 and U is a unit vector, the number t is the length of P.

For proof we first let P = tU where U = (u1,…, un) is a unit vector. Then P · P = (tu1)² + · · · + (tun)² = t²(u1² + ··· + ut²) = t², and if t ≧ 0. Conversely, let U = t–1P, where . Then U · U = (t–1x1)² + · · · + (t–1xn)² = t–2(x1² + ··· + xn²) = 1 and U is a unit vector. The vector – U = – t–1P is clearly also a unit vector.

EXERCISES

1. Give the norms and lengths of the following vectors:

(a)(2, 2, –1)

(b)(1, 1, 0)

(c)(1, –4, 8)

(d)(1, –1, 1, –1)

(e)(1, –1, 2, 1)

(f)(3, 2, –1, 1, 1)

2. Give the unit vector for each vector of Exercise 1.

4. The angle between two vectors. If P = (x1,…, xn) and Q = (y1, …, yn) are any two real nonzero vectors, the difference

For (P · P)(Q · Q) = (x1² + · · · + xn²)(y1² + ··· + yn²) is the sum of x1²y1² + x2²y2² + · · + xn²yn² and all expressions of the form (xiyi)² + (xiyi)² for 1 ≦ i < j ≦ n. The square (P · Q)² = (x1y1 + ··· + xnyn)² is the sum of x1²y1² + x2²y2² + · · · + xn²yn² and all products of the form 2xiyixjyj for 1 ≦ i < j ≦ n. The difference then is the sum of all expressions of the form (xiyj)² + (xjyi)² – 2xiyixjyj = (xiyj – xjyi)² for 1 ≦ i < j ≦ n, and must be nonnegative.

The numbers P · P, Q · Q, and (P · Q)² are all positive, and we have shown that

It follows that there exists an angle θ between 0 and 180° such that

We define this angle θ to be the angle between the vectors P and Q.

Two vectors are said to be orthogonal (i.e., perpendicular) if cos θ = 0. Then P and Q are orthogonal if and only if their inner product

Thus, if P and Q are any vectors, we multiply corresponding coordinates and add the products. The sum so obtained is zero if and only if P and Q are orthogonal.

EXERCISES

1. Compute P · Q for each of the following vector pairs P, Q:

(a)(1, 1, –1), (1, 0, 1)

(b)(1, 2, 3), (–1, 1, –1)

(c)(1, 1, 2), (0, –1, 1)

(d)(–1, 0, 1), (2, 1, 1)

(e)(–1, 3, 2), (1, 1, –1)

(f)(1, –1, 1, 1), (1, 1, 1, 0)

(g)(2, 3, –1, 6), (3, –2, 6, 1)

(h)(1, 2, 3, 4), (2, –1, –1, 1)

(i)(4, –6, 1, 2), (1, 2, –1, 2)

(j)(3, 1, –1, 1), (0, 1, 1, 0)

2. Which pairs are orthogonal?

3. Compute cos θ for each nonorthogonal pair.

5. Directed lines. Directed lines are frequently used in the geometry of three-dimensional Euclidean space, i.e., in ordinary solid analytic geometry. Every pair of distinct points P and Q in space determines a line passing through P and Q. We shall use the notation PQ for this line and shall prefix the word ray when we mean the ray PQ, which is the half line from P through Q.

Let us assume that a unit of measurement has been prescribed and that we have measured the length of the line segment joining P to Q in terms of this unit. The result is a real number that is positive if P and Q are different points and is zero only when P and Q coincide. We shall use the notation |PQ| for this measurement of length.

When P and Q are points on a directed line, we shall use the symbol for the signed length of the segment joining P to Q and directed from P toward Q. Then if the direction from P to Q is the positive direction on the line, and if the direction from P to Q is opposite to the positive direction on the line. See Fig. 1 in which and , and note that in all cases .

FIG. 1.

FIG. 2.

If P, Q, R are on a directed line, it should be clear from Figs. 1 and 2 that . This equation may be generalized to the case of any finite number of points on a directed line and the generalized equation is

6. Orthogonal projections. A theorem of solid geometry states that through a given point P there exists precisely one plane perpendicular to a given line L. This plane intersects L in a point P′ such that the line PP′ is perpendicular to L. We shall call P′ the orthogonal projection of P on L.

If P and Q are any two points, we shall designate by the line segment which joins P to Q and which is directed from P to Q. Project P and Q orthogonally on a directed line L, and obtain projections P′ and Q′. Then we define the orthogonal projection of on L to be the signed length . It follows that the orthogonal projection of is the negative of the orthogonal projection of .

A directed broken line joining two points P and Q is the geometric configuration consisting of the directed line segments for any finite number n of points P1,…,Pn. Let us use the notation for such a configuration and define the orthogonal projection of to be the sum

. By formula (7) this sum is equal to P′Q′ and we have proved the following:

Theorem 1. The orthogonal projection of any directed line segment on a directed line L is equal to the orthogonal projection on L of any directed broken line from P to Q.

As we have said, a ray PQ is a half line that begins at the point P, passes through Q, and extends indefinitely. If PQ and PR are two rays from the same point P, there is a unique angle θ between them such that 0 ≦ θ ≦ 180°. We define this angle to be the angle between the ray PQ and the ray PR.

FIG. 3.

Let P and Q be any points and P′ and Q′ be their respective projections on a directed line L, as in Fig. 4. Then Q, P, P′ are three of the vertices of a parallelogram, and we find the fourth vertex R by drawing a line segment P′R such that |P′R| = |PQ|. Define the angle θ between the ray PQ and the line L to be the angle between the ray P′R and the ray from P′ in the positive direction on L. By the standard ratio definition of the cosine of an angle we have

FIG. 4.

We have proved the following:

Theorem 2. Let θ be the angle between and a directed line L. Then the projection of on L is equal to |PQ| cos θ.

7. Rectangular coordinates in ordinary space. A rectangular Cartesian coordinate system in ordinary three-dimensional real Euclidean space is a certain one-to-one correspondence between the points of space and three-dimensional real vectors (x, y, z). The construction of the correspondence begins with the construction of three mutually perpendicular directed lines intersecting at a point O called the origin of the coordinate system (see Fig. 5).

The three lines are called coordinate axes. The first of them is a vertical line directed upward. It is called the z axis. The second line is a horizontal line in the plane of the book and is directed to the right. It is called the y axis. The remaining line is the x axis. It should be thought of as a line perpendicular to the plane of the book and directed toward the reader. The specification of a coordinate system will be completed as soon as a unit of measurement, which will be used for all measurements of lengths of lines, is given. This is usually done by specifying a unit point U on the x axis such that is the unit of length.

FIG. 5.

Let P be any point in space, x be the projection of on the x axis, y be the projection of on the y axis, and z be the projection of on the z axis. Then the vector (x, y, z) is uniquely determined by P and we write P = (x, y, z).

Conversely, if (x, y, z) is given, we can draw a plane perpendicular to the x axis and