7.5 Lines and Planes in 3-Space

INTRODUCTION

In this section we discuss how to find various equations of lines and planes in 3-space.

Lines: Vector Equation

As in the plane, any two distinct points in 3-space determine only one line between them. To find an equation of the line through P1(x1, y1, z1) and P2(x2, y2, z2), let us assume that P(x, y, z) is any point on the line. In FIGURE 7.5.1, if r = , r1 = , and r2 = , we see that vector a = r2 − r1 is parallel to vector r − r2. Thus,

r − r2 = t(r2 − r1). (1)

If we write

a = r2 − r1 = 〈x2 − x1, y2 − y1, z2 − z1〉 = 〈a1, a2, a3〉,(2)

then (1) implies that a vector equation for the line a is

r = r2 + ta.

The scalar t is called a parameter and the nonzero vector a is called a direction vector; the components a1, a2, and a3 of the direction vector a are called direction numbers for the line.

3 points P (x y z), 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) are marked on a line in 3-space. The position vectors that connect to these points are r, r subscript 1, and r subscript 2. Vector a joins P subscript 1 and P subscript 2. Vector r minus r subscript 2 joins P subscript 2 and P. It is in the same line as vector a.

FIGURE 7.5.1 Line through distinct points in 3-space

Since r − r1 is also parallel to a, an alternative vector equation for the line is r = r1 + ta. Indeed, r = r1 + t(−a) and r = r1 + t(ka), k a nonzero scalar, are also equations for a.

Alternative forms of the vector equation.

EXAMPLE 1 Vector Equation of a Line

Find a vector equation for the line through (2, −1, 8) and (5, 6, −3).

SOLUTION

Define a = 〈2 − 5, − 1 − 6, 8 − (−3)〉 = 〈−3, −7, 11〉. The following are three different vector equations for the line:

〈x, y, z〉 = 〈2, −1, 8〉 + t〈−3, −7, 11〉 (3)

〈x, y, z〉 = 〈5, 6, −3〉 + t〈−3, −7, 11〉 (4)

〈x, y, z〉 = 〈5, 6, −3〉 + t〈3, 7, −11〉. (5) ≡

Parametric Equations

By writing (1) as

and equating components, we obtain

x = x2 + a1t, y = y2 + a2t, z = z2 + a3t. (6)

The equations in (6) are called parametric equations for the line through P1 and P2. As the parameter t increases from − to , we can think of the point P(x, y, z) tracing out the entire line. If the parameter t is restricted to a closed interval [t0, t1], then P(x, y, z) traces out a line segment starting at the point corresponding to t0 and ending at the point corresponding to t1. For example, in Figure 7.5.1, if −1 ≤ t ≤ 0, then P(x, y, z) traces out the line segment starting at P1(x1, y1, z1) and ending at P2(x2, y2, z2).

EXAMPLE 2 Parametric Equations of a Line

Find parametric equations for the line in Example 1.

SOLUTION

From (3), it follows that

x = 2 − 3t, y = −1 − 7t, z = 8 + 11t. (7)

An alternative set of parametric equations can be obtained from (5):

x = 5 + 3t, y = 6 + 7t, z = −3 − 11t.(8) ≡

Note that the value t = 0 in (7) gives (2, −1, 8), whereas in (8), t = −1 must be used to obtain the same point.

EXAMPLE 3 Vector Parallel to a Line

Find a vector a that is parallel to the line a whose parametric equations are x = 4 + 9t, y = −14 + 5t, z = 1 − 3t.

SOLUTION

The coefficients (or a nonzero constant multiple of the coefficients) of the parameter in each equation are the components of a vector that is parallel to the line. Thus, a = 9i + 5j − 3k is parallel to a and hence is a direction vector of the line. ≡

Symmetric Equations

From (6), observe that we can clear the parameter by writing

provided that the three direction numbers a1, a2, and a3 are nonzero. The resulting equations

(9)

are said to be symmetric equations for the line through P1 and P2.

EXAMPLE 4 Symmetric Equations of a Line

Find symmetric equations for the line through (4, 10, −6) and (7, 9, 2).

SOLUTION

Define a1 = 7 − 4 = 3, a2 = 9 − 10 = −1, and a3 = 2 − (−6) = 8. It follows from (9) that symmetric equations for the line are

≡

If one of the direction numbers a1, a2, or a3 is zero in (6), we use the remaining two equations to eliminate the parameter t. For example, if a1 = 0, a2 ≠ 0, a3 ≠ 0, then (6) yields

In this case,

are symmetric equations for the line.

EXAMPLE 5 Symmetric Equations of a Line

Find symmetric equations for the line through (5, 3, 1) and (2, 1, 1).

SOLUTION

Define a1 = 5 − 2 = 3, a2 = 3 − 1 = 2, and a3 = 1 − 1 = 0. From the preceding discussion, it follows that symmetric equations for the line are

.

In other words, the symmetric equations describe a line in the plane z = 1. ≡

A line in space is also determined by specifying a point P1(x1, y1, z1) and a nonzero direction vector a. Through the point P1, there passes only one line a parallel to the given vector. If P(x, y, z) is a point on the line a, shown in FIGURE 7.5.2, then, as before,

A position vector a and a point P subscript 1 (x subscript 1, y subscript 1, z subscript 1) are graphed in 3-space. A vector parallel to a from P subscript 1 ends at point P (x y z). A line L subscript a passes through P subscript 1 and P. The position vectors to these points are marked.

FIGURE 7.5.2 Line determined by a point P and vector a

EXAMPLE 6 Line Parallel to a Vector

Write vector, parametric, and symmetric equations for the line through (4, 6, −3) and parallel to a = 5i − 10j + 2k.

SOLUTION

With a1 = 5, a2 = −10, and a3 = 2 we have immediately

≡

Planes: Vector Equation

FIGURE 7.5.3(a) illustrates the fact that through a given point P1(x1, y1, z1) there pass an infinite number of planes. However, as shown in Figure 7.5.3(b), if a point P1 and a vector n are specified, there is only one plane containing P1 with n normal, or perpendicular, to the plane. Moreover, if P(x, y, z) is any point on , and r = , r1 = , then, as shown in Figure 7.5.3(c), r − r1 is in the plane. It follows that a vector equation of the plane is

n · (r − r1) = 0. (10)

Graph a. 3 perpendicular planes pass through a point. Graph b. 2 parallel planes with normal vector n are graphed. Plane P contains a point P subscript 1. Graph c. 2 points are marked on a plane P with normal vector n. The vector r minus r subscript 1 connects P subscript 1 (x subscript 1, y subscript 1, z subscript 1) to P (x y z).

FIGURE 7.5.3 Vector n is perpendicular to a plane

Cartesian Equation

Specifically, if the normal vector is n = ai + bj + ck, then (10) yields a Cartesian equation of the plane containing P1(x1, y1, z1):

a(x − x1) + b(y − y1) + c(z − z1) = 0. (11)

Equation (11) is sometimes called the point-normal form of the equation of a plane.

EXAMPLE 7 Equation of a Plane

Find an equation of the plane with normal vector n = 2i + 8j − 5k containing the point (4, −1, 3).

SOLUTION

It follows immediately from the point-normal form (11) that an equation of the plane is 2(x − 4) + 8(y + 1) − 5(z − 3) = 0 or 2x + 8y − 5z + 15 = 0. ≡

Equation (11) can always be written as ax + by + cz + d = 0 by identifying d = −ax1 − by1 − cz1. Conversely, we shall now prove that any linear equation

ax + by + cz + d = 0, a, b, c not all zero (12)

is a plane.

THEOREM 7.5.1 Plane with Normal Vector

The graph of any equation ax + by + cz + d = 0, a, b, c not all zero, is a plane with the normal vector n = ai + bj + ck.

PROOF:

Suppose x0, y0, and z0 are numbers that satisfy the given equation. Then, ax0 + by0 + cz0 + d = 0 implies that d = −ax0 − by0 − cz0. Replacing this latter value of d in the original equation gives, after simplifying, a(x − x0) + b(y − y0) + c(z − z0) = 0, or, in terms of vectors,

[ai + bj + ck] · [(x − x0)i + (y − y0) j + (z − z0)k] = 0.

This last equation implies that ai + bj + ck is normal to the plane containing the point (x0, y0, z0) and to the vector (x − x0)i + (y − y0)j + (z − z0)k. ≡

EXAMPLE 8 A Vector Normal to a Plane

A vector normal to the plane 3x − 4y + 10z − 8 = 0 is n = 3i − 4j + 10k. ≡

Of course, a nonzero scalar multiple of a normal vector is still perpendicular to the plane.

Three noncollinear points P1, P2, and P3 also determine a plane.* To obtain an equation of the plane, we need only form two vectors between two pairs of points. As shown in FIGURE 7.5.4, their cross product is a vector normal to the plane containing these vectors. If P(x, y, z) represents any point on the plane, and r = , r1 = , r2 = , r3 = , then r − r1 (or, for that matter, r − r2 or r − r3) is in the plane. Hence,

[(r2 − r1) × (r3 − r1)] · (r − r1) = 0(13)

is a vector equation of the plane. Do not memorize the last formula. The procedure is the same as (10) with the exception that the vector n normal to the plane is obtained by means of the cross product.

4 points are marked on a plane in 3-space. The vectors joining P subscript 1 to the other points are as follows. P subscript 1 to P, r minus r subscript 1. P subscript 1 to P subscript 2, r subscript 2 minus r subscript 1. P subscript 1 to P subscript 3, r subscript 3 minus r subscript 1. A vector normal to the plane at P subscript 1 is labeled cross product of r subscript 2 minus r subscript 1 and r subscript 3 minus r subscript 1.

FIGURE 7.5.4 Vectors r2 − r1 and r3 − r1 are in the plane, and their cross product is normal to the plane

EXAMPLE 9 Three Points That Determine a Plane

Find an equation of the plane that contains (1, 0, −1), (3, 1, 4), and (2, −2, 0).

SOLUTION

We need three vectors. Pairing the points on the left as shown yields the vectors on the right. The order in which we subtract is irrelevant.

Now,

is a vector normal to the plane containing the given points. Hence, a vector equation of the plane is (u × v) · w = 0. The latter equation yields

≡

Graphs

The graph of (12) with one or even two variables missing is still a plane. For example, we saw in Section 7.2 that the graphs of

x = x0, y = y0, z = z0,

where x0, y0, z0 are constants, are planes perpendicular to the x-, y-, and z-axes, respectively. In general, to graph a plane, we should try to find

(i) the x-, y-, and z-intercepts and, if necessary,

(ii) the trace of the plane in each coordinate plane.

A trace of a plane in a coordinate plane is the line of intersection of the plane with a coordinate plane.

EXAMPLE 10 Graph of a Plane

Graph the equation 2x + 3y + 6z = 18.

SOLUTION

Setting:y = 0, z = 0 gives x = 9

x = 0, z = 0 gives y = 6

x = 0, y = 0 gives z = 3.

As shown in FIGURE 7.5.5, we use the x-, y-, and z-intercepts (9, 0, 0), (0, 6, 0), and (0, 0, 3) to draw the graph of the plane in the first octant. ≡

A plane labeled 2 x + 3 y + 6 z equals 18 is graphed in 3-space. It intersects the axes at (9, 0, 0), (0, 6, 0), and (0, 0, 3).

FIGURE 7.5.5 Plane in Example 10

EXAMPLE 11 Graph of a Plane

Graph the equation 6x + 4y = 12.

SOLUTION

In two dimensions, the graph of the equation is a line with x-intercept (2, 0) and y-intercept (3, 0). However, in three dimensions, this line is the trace of a plane in the xy-coordinate plane. Since z is not specified, it can be any real number. In other words, (x, y, z) is a point on the plane provided that x and y are related by the given equation. As shown in FIGURE 7.5.6, the graph is a plane parallel to the z-axis. ≡

A plane labeled 6 x + 4 y equals 12 is graphed in 3-space. It is parallel to the z-axis and intersects the x and y axes at (2, 0, 0) and (0, 3, 0).

FIGURE 7.5.6 Plane in Example 11

EXAMPLE 12 Graph of a Plane

Graph the equation x + y − z = 0.

SOLUTION

First observe that the plane passes through the origin (0, 0, 0). Now, the trace of the plane in the xz-plane (y = 0) is z = x, whereas its trace in the yz-plane (x = 0) is z = y. Drawing these two lines leads to the graph given in FIGURE 7.5.7. ≡

A plane labeled x + y minus z equals 0 is graphed in 3-space. It is between z = x, z = y, and the origin.

FIGURE 7.5.7 Plane in Example 12

Two planes and that are not parallel must intersect in a line . See FIGURE 7.5.8. Example 13 will illustrate one way of finding parametric equations for the line of intersection. In Example 14 we shall see how to find a point of intersection (x0, y0, z0) of a plane and a line . See FIGURE 7.5.9.

2 planes P subscript 1 and P subscript 2 intersect in a line L.

FIGURE 7.5.8 Planes intersect in a line

A plane P and a line L intersect at a point (x subscript 0, y subscript 0, z subscript 0).

FIGURE 7.5.9 Point of intersection of a plane and a line

EXAMPLE 13 Line of Intersection of Two Planes

Find parametric equations for the line of intersection of

2x − 3y + 4z = 1

x − y − z = 5.

SOLUTION

In a system of two equations and three unknowns, we choose one variable arbitrarily, say z = t, and solve for x and y from

2x − 3y = 1 − 4t

x − y = 5 + t.

Proceeding, we find x = 14 + 7t, y = 9 + 6t, z = t. These are parametric equations for the line of intersection of the given planes. ≡

EXAMPLE 14 Point of Intersection of a Line and a Plane

Find the point of intersection of the plane 3x − 2y + z = −5 and the line x = 1 + t, y = −2 + 2t, z = 4t.

SOLUTION

If (x0, y0, z0) denotes the point of intersection, then we must have 3x0 − 2y0 + z0 = −5 and x0 = 1 + t0, y0 = −2 + 2t0, z0 = 4t0, for some number t0. Substituting the latter equations into the equation of the plane gives

3(1 + t0) − 2(−2 + 2t0) + 4t0 = −5 or t0 = −4.

From the parametric equations for the line, we then obtain x0 = −3, y0 = −10, and z0 = −16. The point of intersection is (−3, −10, −16). ≡

REMARKS

In everyday speech, the words orthogonal, perpendicular, and normal are often used interchangeably in the sense that two objects touch, intersect, or abut at a 90° angle. But in recent years an unwritten convention has arisen to use these terms in specific mathematical contexts. As a general rule, we say that two vectors are orthogonal, two lines (or two planes) are perpendicular, and that a vector is normal to a plane.

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

In Problems 1–6, find a vector equation for the line through the given points.

  1. (1, 2, 1), (3, 5, −2)
  2. (0, 4, 5), (−2, 6, 3)
  3. (, −, 1), (−, , −)
  4. (10, 2, −10), (5, −3, 5)
  5. (1, 1, −1), (−4, 1, −1)
  6. (3, 2, 1), (, 1, −2)

In Problems 7–12, find parametric equations for the line through the given points.

  1. (2, 3, 5), (6, −1, 8)
  2. (2, 0, 0), (0, 4, 9)
  3. (1, 0, 0), (3, −2, −7)
  4. (0, 0, 5), (−2, 4, 0)
  5. (4, , ), (−6, − , )
  6. (−3, 7, 9), (4, −8, −1)

In Problems 13–18, find symmetric equations for the line through the given points.

  1. (1, 4, −9), (10, 14, −2)
  2. (, 0, −), (1, 3, )
  3. (4, 2, 1), (−7, 2, 5)
  4. (−5, −2, −4), (1, 1, 2)
  5. (5, 10, −2), (5, 1, −14)
  6. (, −, ), (, , −)

In Problems 19–22, find parametric and symmetric equations for the line through the given point parallel to the given vector.

  1. (4, 6, −7), a = 〈3, , −〉
  2. (1, 8, −2), a = −7i − 8j
  3. (0, 0, 0), a = 5i + 9j + 4k
  4. (0, −3, 10), a = 〈12, −5, −6〉
  5. Find parametric equations for the line through (6, 4, −2) that is parallel to the line x/2 = (1 − y)/3 = (z − 5)/6.
  6. Find symmetric equations for the line through (4, −11, −7) that is parallel to the line x = 2 + 5t, y = −1 + t, z = 9 − 2t.
  7. Find parametric equations for the line through (2, −2, 15) that is parallel to the xz-plane and the xy-plane.
  8. Find parametric equations for the line through (1, 2, 8) that is (a) parallel to the y-axis, and (b) perpendicular to the xy-plane.
  9. Show that the lines given by r = t〈1, 1, 1〉 and r = 〈6, 6, 6〉 + t〈−3, −3, −3〉 are the same.
  10. Let a and b be lines with direction vectors a and b, respectively. a and b are orthogonal if a and b are orthogonal and parallel if a and b are parallel. Determine which of the following lines are orthogonal and which are parallel.
    1. r = 〈1, 0, 2〉 + t〈9, −12, 6〉
    2. x = 1 + 9t, y = 12t, z = 2 − 6t
    3. x = 2t, y = −3t, z = 4t
    4. x = 5 + t, y = 4t, z = 3 + t
    5. x = 1 + t, y = t, z = 2 − t

In Problems 29 and 30, determine the points of intersection of the given line and the three coordinate planes.

  1. x = 4 − 2t, y = 1 + 2t, z = 9 + 3t

In Problems 31–34, determine whether the given lines intersect. If so, find the point of intersection.

  1. x = 4 + t, y = 5 + t, z = −1 + 2t
    x = 6 + 2s, y = 11 + 4s, z = −3 + s
  2. x = 1 + t, y = 2 − t, z = 3t
    x = 2 − s, y = 1 + s, z = 6s
  3. x = 2 − t, y = 3 + t, z = 1 + t
    x = 4 + s, y = 1 + s, z = 1 − s
  4. x = 3 − t, y = 2 + t, z = 8 + 2t
    x = 2 + 2s, y = −2 + 3s, z = −2 + 8s

The angle between two lines a and b is the angle between their direction vectors a and b. In Problems 35 and 36, find the angle between the given lines.

  1. x = 4 − t, y = 3 + 2t, z = −2t
    x = 5 + 2s, y = 1 + 3s, z = 5 − 6s

In Problems 37 and 38, the given lines lie in the same plane. Find parametric equations for the line through the indicated point that is perpendicular to this plane.

  1. x = 3 + t, y = −2 + t, z = 9 + t
    x = 1 − 2s, y = 5 + s, z = −2 − 5s; (4, 1, 6)

In Problems 39–44, find an equation of the plane that contains the given point and is perpendicular to the indicated vector.

  1. (5, 1, 3); 2i − 3j + 4k
  2. (1, 2, 5); 4i − 2j
  3. (6, 10, −7); −5i + 3k
  4. (0, 0, 0); 6i − j + 3k
  5. (, , − ); 6i + 8j − 4k
  6. (−1, 1, 0); −i + j − k

In Problems 45–50, find, if possible, an equation of a plane that contains the given points.

  1. (3, 5, 2), (2, 3, 1), (−1, −1, 4)
  2. (0, 1, 0), (0, 1, 1), (1, 3, −1)
  3. (0, 0, 0), (1, 1, 1), (3, 2, −1)
  4. (0, 0, 3), (0, −1, 0), (0, 0, 6)
  5. (1, 2, −1), (4, 3, 1), (7, 4, 3)
  6. (2, 1, 2), (4, 1, 0), (5, 0, −5)

In Problems 51–60, find an equation of the plane that satisfies the given conditions.

  1. Contains (2, 3, −5) and is parallel to x + y − 4z = 1
  2. Contains the origin and is parallel to 5x − y + z = 6
  3. Contains (3, 6, 12) and is parallel to the xy-plane
  4. Contains (−7, −5, 18) and is perpendicular to the y-axis
  5. Contains the lines x = 1 + 3t, y = 1 − t, z = 2 + t; x = 4 + 4s, y = 2s, z = 3 + s
  6. Contains the lines ;
    r = 〈1, −1, 5〉 + t〈1, 1, −3〉
  7. Contains the parallel lines x = 1 + t, y = 1 + 2t, z = 3 + t; x = 3 + s, y = 2s, z = −2 + s
  8. Contains the point (4, 0, −6) and the line x = 3t, y = 2t, z = −2t
  9. Contains (2, 4, 8) and is perpendicular to the line x = 10 − 3t, y = 5 + t, z = 6 − t
  10. Contains (1, 1, 1) and is perpendicular to the line through (2, 6, −3) and (1, 0, −2)
  11. Let and be planes with normal vectors n1 and n2, respectively. and are orthogonal if n1 and n2 are orthogonal and parallel if n1 and n2 are parallel. Determine which of the following planes are orthogonal and which are parallel.
    1. 2x − y + 3z = 1
    2. x + 2y + 2z = 9
    3. x + y − z = 2
    4. −5x + 2y + 4z = 0
    5. −8x − 8y + 12z = 1
    6. −2x + y − 3z = 5
  12. Find parametric equations for the line that contains (−4, 1, 7) and is perpendicular to the plane −7x + 2y + 3z = 1.
  13. Determine which of the following planes are perpendicular to the line x = 4 − 6t, y = 1 + 9t, z = 2 + 3t.
    1. 4x + y + 2z = 1
    2. 2x − 3y + z = 4
    3. 10x − 15y − 5z = 2
    4. −4x + 6y + 2z = 9
  14. Determine which of the following planes are parallel to the line (1 − x)/2 = (y + 2)/4 = z − 5.
    1. x − y + 3z = 1
    2. 6x − 3y = 1
    3. x − 2y + 5z = 0
    4. −2x + y − 2z = 7

In Problems 65–68, find parametric equations for the line of intersection of the given planes.

In Problems 69–72, find the point of intersection of the given plane and line.

  1. 2x − 3y + 2z = −7; x = 1 + 2t, y = 2 − t, z = −3t
  2. x + y + 4z = 12; x = 3 − 2t, y = 1 + 6t, z = 2 − t
  3. x + y − z = 8; x = 1, y = 2, z = 1 + t
  4. x − 3y + 2z = 0; x = 4 + t, y = 2 + t, z = 1 + 5t

In Problems 73 and 74, find parametric equations for the line through the indicated point that is parallel to the given planes.

In Problems 75 and 76, find an equation of the plane that contains the given line and is orthogonal to the indicated plane.

  1. x = 4 + 3t, y = −t, z = 1 + 5t; x + y + z = 7

In Problems 77–82, graph the given equation.

  1. 5x + 2y + z = 10
  2. 3x + 2z = 9
  3. −y − 3z + 6 = 0
  4. 3x + 4y − 2z − 12 = 0
  5. −x + 2y + z = 4
  6. x − y − 1 = 0

 

* If you ever sit at a four-legged table that rocks, you might consider replacing it with a three-legged table.