7.2 Vectors in 3-Space

INTRODUCTION

In the plane, or 2-space, one way of describing the position of a point P is to assign to it coordinates relative to two mutually orthogonal, or perpendicular, axes called the x- and y-axes. If P is the point of intersection of the line x = a (perpendicular to the x-axis) and the line y = b (perpendicular to the y-axis), then the ordered pair (a, b) is said to be the rectangular or Cartesian coordinates of the point. See FIGURE 7.2.1. In this section we extend the notions of Cartesian coordinates and vectors to three dimensions.

A point P is marked on quadrant 1 of a coordinate plane. A vertical line from P intersects the x axis at a distance a from the origin. A horizontal line from P intersects the y axis at a distance b from the origin. The coordinates of P are (a, b).

FIGURE 7.2.1 Rectangular coordinates in 2-space

Rectangular Coordinate System in 3-Space

In three dimensions, or 3-space, a rectangular coordinate system is constructed using three mutually orthogonal axes. The point at which these axes intersect is called the origin O. These axes, shown in FIGURE 7.2.2(a), are labeled in accordance with the so-called right-hand rule: If the fingers of the right hand, pointing in the direction of the positive x-axis, are curled toward the positive y-axis, then the thumb will point in the direction of a new axis perpendicular to the plane of the x- and y-axes. This new axis is labeled the z-axis. The dashed lines in Figure 7.2.2(a) represent the negative axes. Now, if

x = a, y = b, z = c

are planes perpendicular to the x-axis, y-axis, and z-axis, respectively, then the point P at which these planes intersect can be represented by an ordered triple of numbers (a, b, c) said to be the rectangular or Cartesian coordinates of the point. The numbers a, b, and c are, in turn, called the x-, y-, and z-coordinates of P(a, b, c). See Figure 7.2.2(b).

Graph a. 3 perpendicular axes define a coordinate system. The axes intersect at the origin O. The x axis and the y axis are on the horizontal plane. Using the right hand and curling the fingers from the positive x axis to the y axis, the positive z axis lies along the thumb. Graph b. A box is graphed on a 3-dimensional coordinate system. One vertex is at (0, 0, 0) and the opposite vertex is at P (a, b, c). Plane x = a is parallel to the y z plane. Plane x = b is parallel to the x z plane. Plane x = c is parallel to the x y plane.

FIGURE 7.2.2 Rectangular coordinates in 3-space

Octants

Each pair of coordinate axes determines a coordinate plane. As shown in FIGURE 7.2.3, the x- and y-axes determine the xy-plane, the x- and z-axes determine the xz-plane, and so on. The coordinate planes divide 3-space into eight parts known as octants. The octant in which all three coordinates of a point are positive is called the first octant. There is no agreement for naming the other seven octants.

In 3-space, the x y, y z, and x z planes intersect to form 8 octants.

FIGURE 7.2.3 Octants

The following table summarizes the coordinates of a point either on a coordinate axis or in a coordinate plane. As seen in the table, we can also describe, say, the xy-plane by the simple equation z = 0. Similarly, the xz-plane is y = 0 and the yz-plane is x = 0.

EXAMPLE 1 Graphs of Three Points

Graph the points (4, 5, 6), (3, −3, −1), and (−2, −2, 0).

SOLUTION

Of the three points shown in FIGURE 7.2.4, only (4, 5, 6) is in the first octant. The point (−2, −2, 0) is in the xy-plane. ≡

Point (4, 5, 6) is in the first octant. Points (3, negative 3, negative 1) and (negative 2, negative 2, 0) are on the x y plane.

FIGURE 7.2.4 Points in Example 1

Distance Formula

To find the distance between two points P1(x1, y1, z1) and P2(x2, y2, z2) in 3-space, let us first consider their projection onto the xy-plane. As seen in FIGURE 7.2.5, the distance between (x1, y1, 0) and (x2, y2, 0) follows from the usual distance formula in the plane and is . If the coordinates of P3 are (x2, y2, z1), then the Pythagorean theorem applied to the right triangle P1P2P3 yields

or (1)

3 points are graphed in 3-space: P subscript 1 (x subscript 1, y subscript 1, z subscript 1), P subscript 2 (x subscript 2, y subscript 2, z subscript 2) and P subscript 3 (x subscript 2, y subscript 2, z subscript 1). The distance between P subscript 1 and P subscript 2 is d, and the distance between P 3 and P 2 is the absolute value of z subscript 2 minus z subscript 1. The distance between the projections of these points on the x y plane is the square root of the expression, x subscript 2 minus x subscript 1, squared plus y subscript 2 minus y subscript 1, squared.

FIGURE 7.2.5 Distance d between two points in 3-space

EXAMPLE 2 Distance Between Two Points

Find the distance between (2, −3, 6) and (−1, −7, 4).

SOLUTION

Choosing P2 as (2, −3, 6) and P1 as (−1, −7, 4), formula (1) gives

≡

Midpoint Formula

The formula for finding the midpoint of a line segment between two points in 2-space carries over in an analogous fashion to 3-space. If P1(x1, y1, z1) and P2(x2, y2, z2) are two distinct points, then the coordinates of the midpoint of the line segment between them are

(2)

EXAMPLE 3 Coordinates of a Midpoint

Find the coordinates of the midpoint of the line segment between the two points in Example 2.

SOLUTION

From (2) we obtain

≡

Vectors in 3-Space

A vector a in 3-space is any ordered triple of real numbers

a = 〈a1, a2, a3〉,

where a1, a2, and a3 are the components of the vector. The set of all vectors in 3-space will be denoted by the symbol R3. The position vector of a point P(x1, y1, z1) in space is the vector = 〈x1, y1, z1〉 whose initial point is the origin O and whose terminal point is P. See FIGURE 7.2.6.

A position vector O P is graphed in 3-space, where the coordinates of P are (x subscript 1, y subscript 1, z subscript 1).

FIGURE 7.2.6 Position vector

The component definitions of addition, subtraction, scalar multiplication, and so on are natural generalizations of those given for vectors in R2. Moreover, the vectors in R3 possess all the properties listed in Theorem 7.1.1.

DEFINITION 7.2.1 Component Definitions in 3-Space

Let a = 〈a1, a2, a3〉 and b = 〈b1, b2, b3〉 be vectors in R3.

  1. Addition: a + b = 〈a1 + b1, a2 + b2, a3 + b3〉
  2. Scalar multiplication: ka = 〈ka1, ka2, ka3〉
  3. Equality: a = b if and only if a1 = b1, a2 = b2, a3 = b3
  4. Negative: −b = (−1)b = 〈−b1, −b2, −b3〉
  5. Subtraction: a − b = a + (−b) = 〈a1 − b1, a2 − b2, a3 − b3〉
  6. Zero vector: 0 = 〈0, 0, 0〉
  7. Magnitude: =

If and are the position vectors of the points P1(x1, y1, z1) and P2(x2, y2, z2), then the vector is given by

= − = 〈x2 − x1, y2 − y1, z2 − z1〉. (3)

As in 2-space, can be drawn either as a vector whose initial point is P1 and whose terminal point is P2 or as position vector with terminal point

P(x2 − x1, y2 − y1, z2 − z1).

See FIGURE 7.2.7.

2 vectors O P subscript 1 and O P subscript 2 are position vectors in octant 1. The coordinates of the end points are P subscript 1 (x subscript 1, y subscript 1, z subscript 1) and P subscript 2 (x subscript 2, y subscript 2, z subscript 2). The vector joining P subscript 1 to P subscript 2 is labeled P sub 1 P sub 2. When displaced to the origin, the vector is labeled O P.

FIGURE 7.2.7 and are the same vector

EXAMPLE 4 Vector Between Two Points

Find the vector if the points P1 and P2 are given by P1(4, 6, −2) and P2(1, 8, 3), respectively.

SOLUTION

If the position vectors of the points are = 〈4, 6, −2〉 and = 〈1, 8, 3〉, then from (3) we have

= − = 〈1 − 4, 8 − 6, 3 − (−2)〉 = 〈−3, 2, 5〉. ≡

EXAMPLE 5 A Unit Vector

Find a unit vector in the direction of a = 〈−2, 3, 6〉.

SOLUTION

Since a unit vector has length 1, we first find the magnitude of a and then use the fact that is a unit vector in the direction of a. The magnitude of a is

A unit vector in the direction of a is

≡

The i, j, k Vectors

We saw in the preceding section that the set of two unit vectors i = 〈1, 0〉 and j = 〈0, 1〉 constitute a basis for the system of two-dimensional vectors. That is, any vector a in 2-space can be written as a linear combination of i and j: a = a1i + a2j. Likewise any vector a = 〈a1, a2, a3〉 in 3-space can be expressed as a linear combination of the unit vectors

i = 〈1, 0, 0〉, j = 〈0, 1, 0〉, k = 〈0, 0, 1〉.

To see this we use (i) and (ii) of Definition 7.2.1 to write

that is,

The vectors i, j, and k illustrated in FIGURE 7.2.8(a) are called the standard basis for the system of three-dimensional vectors. In Figure 7.2.8(b) we see that a position vector a = a1i + a2j + a3k is the sum of the vectors a1i, a2j, and a3k, which lie along the coordinate axes and have the origin as a common initial point.

Graph a. Unit vectors i, j and k are graphed on the 3 axes of 3-space, where i is along the positive x axis, j along the positive y axis, and k along the positive z axis. Graph b. A position vector measures a units. A dotted line from the end point intersects the x axis at a point. The vector joining the origin to the point is labeled a subscript 1 i. A dotted line from the end point of the position vector intersects the y axis at a point. The vector joining the origin to the point is labeled a subscript 2 j. Another line from the end point of the position vector intersects the z axis at a point. The vector joining the origin to the point is labeled a subscript 2 k.

FIGURE 7.2.8 i, j, and k form a basis for R3

EXAMPLE 6 Using the i, j, k Vectors

The vector a = 〈7, −5, 13〉 is the same as a = 7i − 5j + 13k. ≡

When the third dimension is taken into consideration, any vector in the xy-plane is equivalently described as a three-dimensional vector that lies in the coordinate plane z = 0. Although the vectors 〈a1, a2〉 and 〈a1, a2, 0〉 are technically not equal, we shall ignore the distinction. That is why, for example, we denoted 〈1, 0〉 and 〈1, 0, 0〉 by the same symbol i. But to avoid any possible confusion, hereafter we shall always consider a vector a three-dimensional vector, and the symbols i and j will represent only 〈1, 0, 0〉 and 〈0, 1, 0〉, respectively. Similarly, a vector in either the xy-plane or the xz-plane must have one zero component. In the yz-plane, a vector

b = 〈0, b2, b3〉 is written b = b2j + b3k.

In the xz-plane, a vector

c = 〈c1, 0, c3〉 is the same as c = c1i + c3k.

EXAMPLE 7 Vector in xz-Plane

(a) The vector a = 5i + 3k = 5i + 0j + 3k lies in the xz-plane and can also be written as a = 〈5, 0, 3〉.

(b) 5i + 3k = ≡

EXAMPLE 8 Linear Combination

If a = 3i − 4j + 8k and b = i − 4k, find 5a − 2b.

SOLUTION

We treat b as a three-dimensional vector and write, for emphasis, b = i + 0j − 4k.

From

we get ≡

7.2 Exercises Answers to selected odd-numbered problems begin on page ANS-16.

In Problems 1–6, graph the given point. Use the same coordinate axes.

  1. (1, 1, 5)
  2. (0, 0, 4)
  3. (3, 4, 0)
  4. (6, 0, 0)
  5. (6, −2, 0)
  6. (5, −4, 3)

In Problems 7–10, describe geometrically all points P(x, y, z) that satisfy the given condition.

  1. z = 5
  2. x = 1
  3. x = 2, y = 3
  4. x = 4, y = −1, z = 7
  5. Give the coordinates of the vertices of the rectangular parallelepiped whose sides are the coordinate planes and the planes x = 2, y = 5, z = 8.
  6. In FIGURE 7.2.9, two vertices are shown of a rectangular parallelepiped having sides parallel to the coordinate planes. Find the coordinates of the remaining six vertices.
    A parallelepiped is graphed in a coordinate system. The coordinates of 2 opposite vertices are (3, 3, 4) and (negative 1, 6, 7).

    FIGURE 7.2.9 Rectangular parallelepiped in Problem 12

  7. Consider the point P(−2, 5, 4).
    1. If lines are drawn from P perpendicular to the coordinate planes, what are the coordinates of the point at the base of each perpendicular?
    2. If a line is drawn from P to the plane z = −2, what are the coordinates of the point at the base of the perpendicular?
    3. Find the point in the plane x = 3 that is closest to P.
  8. Determine an equation of a plane parallel to a coordinate plane that contains the given pair of points.
    1. (3, 4, −5), (−2, 8, −5)
    2. (1, −1, 1), (1, −1, −1)
    3. (−2, 1, 2), (2, 4, 2)

In Problems 15–20, describe the locus of points P(x, y, z) that satisfy the given equation(s).

  1. xyz = 0
  2. x2 + y2 + z2 = 0
  3. (x + 1)2 + (y − 2)2 + (z + 3)2 = 0
  4. (x − 2)(z − 8) = 0
  5. z2 − 25 = 0
  6. x = y = z

In Problems 21 and 22, find the distance between the given points.

  1. (3, −1, 2), (6, 4, 8)
  2. (−1, −3, 5), (0, 4, 3)
  3. Find the distance from the point (7, −3, −4) to (a) the yz-plane and (b) the x-axis.
  4. Find the distance from the point (−6, 2, −3) to (a) the xz-plane and (b) the origin.

In Problems 25–28, the given three points form a triangle. Determine which triangles are isosceles and which are right triangles.

  1. (0, 0, 0), (3, 6, −6), (2, 1, 2)
  2. (0, 0, 0), (1, 2, 4), (3, 2, 2)
  3. (1, 2, 3), (4, 1, 3), (4, 6, 4)
  4. (1, 1, −1), (1, 1, 1), (0, −1, 1)

In Problems 29 and 30, use the distance formula to prove that the given points are collinear.

  1. P1(1, 2, 0), P2(−2, −2, −3), P3(7, 10, 6)
  2. P1(2, 3, 2), P2(1, 4, 4), P3(5, 0, −4)

In Problems 31 and 32, solve for the unknown.

  1. P1(x, 2, 3), P2(2, 1, 1); d(P1, P2) =
  2. P1(x, x, 1), P2(0, 3, 5); d(P1, P2) = 5

In Problems 33 and 34, find the coordinates of the midpoint of the line segment between the given points.

  1. (1, 3, ), (7, −2, )
  2. (0, 5, −8), (4, 1, −6)
  3. The coordinates of the midpoint of the line segment between P1(x1, y1, z1) and P2(2, 3, 6) are (−1, −4, 8). Find the coordinates of P1.
  4. Let P3 be the midpoint of the line segment between P1(−3, 4, 1) and P2(−5, 8, 3). Find the coordinates of the midpoint of the line segment (a) between P1 and P3 and (b) between P3 and P2.

In Problems 37–40, find the vector .

  1. P1(3, 4, 5), P2(0, −2, 6)
  2. P1(−2, 4, 0), P2(6, , 8)
  3. P1(0, −1, 0), P2(2, 0, 1)
  4. P1(, , 5), P2(−, −, 12)

In Problems 41–48, a = 〈1, −3, 2〉, b = 〈−1, 1, 1〉, and c = 〈2, 6, 9〉. Find the indicated vector or scalar.

  1. a + (b + c)
  2. 2a − (b − c)
  3. b + 2(a − 3c)
  4. 4(a + 2c) − 6b
  8. a + b
  9. Find a unit vector in the opposite direction of a = 〈10, −5, 10〉.
  10. Find a unit vector in the same direction as a = i − 3j + 2k.
  11. Find a vector b that is four times as long as a = i − j + k in the same direction as a.
  12. Find a vector b for which = that is parallel to a = 〈−6, 3, −2〉 but has the opposite direction.
  13. Using the vectors a and b shown in FIGURE 7.2.10, sketch the “average vector” (a + b).
    2 position vectors a and b are graphed in a coordinate system.

    FIGURE 7.2.10 Vectors for Problem 53