Skip to Content
OpenStax Logo
Calculus Volume 3

2.5 Equations of Lines and Planes in Space

Calculus Volume 32.5 Equations of Lines and Planes in Space
Buy book
  1. Preface
  2. 1 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 1.1 Parametric Equations
    3. 1.2 Calculus of Parametric Curves
    4. 1.3 Polar Coordinates
    5. 1.4 Area and Arc Length in Polar Coordinates
    6. 1.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Vectors in Space
    1. Introduction
    2. 2.1 Vectors in the Plane
    3. 2.2 Vectors in Three Dimensions
    4. 2.3 The Dot Product
    5. 2.4 The Cross Product
    6. 2.5 Equations of Lines and Planes in Space
    7. 2.6 Quadric Surfaces
    8. 2.7 Cylindrical and Spherical Coordinates
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  4. 3 Vector-Valued Functions
    1. Introduction
    2. 3.1 Vector-Valued Functions and Space Curves
    3. 3.2 Calculus of Vector-Valued Functions
    4. 3.3 Arc Length and Curvature
    5. 3.4 Motion in Space
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  5. 4 Differentiation of Functions of Several Variables
    1. Introduction
    2. 4.1 Functions of Several Variables
    3. 4.2 Limits and Continuity
    4. 4.3 Partial Derivatives
    5. 4.4 Tangent Planes and Linear Approximations
    6. 4.5 The Chain Rule
    7. 4.6 Directional Derivatives and the Gradient
    8. 4.7 Maxima/Minima Problems
    9. 4.8 Lagrange Multipliers
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  6. 5 Multiple Integration
    1. Introduction
    2. 5.1 Double Integrals over Rectangular Regions
    3. 5.2 Double Integrals over General Regions
    4. 5.3 Double Integrals in Polar Coordinates
    5. 5.4 Triple Integrals
    6. 5.5 Triple Integrals in Cylindrical and Spherical Coordinates
    7. 5.6 Calculating Centers of Mass and Moments of Inertia
    8. 5.7 Change of Variables in Multiple Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Vector Calculus
    1. Introduction
    2. 6.1 Vector Fields
    3. 6.2 Line Integrals
    4. 6.3 Conservative Vector Fields
    5. 6.4 Green’s Theorem
    6. 6.5 Divergence and Curl
    7. 6.6 Surface Integrals
    8. 6.7 Stokes’ Theorem
    9. 6.8 The Divergence Theorem
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  8. 7 Second-Order Differential Equations
    1. Introduction
    2. 7.1 Second-Order Linear Equations
    3. 7.2 Nonhomogeneous Linear Equations
    4. 7.3 Applications
    5. 7.4 Series Solutions of Differential Equations
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 2.5.1. Write the vector, parametric, and symmetric of a line through a given point in a given direction, and a line through two given points.
  • 2.5.2. Find the distance from a point to a given line.
  • 2.5.3. Write the vector and scalar equations of a plane through a given point with a given normal.
  • 2.5.4. Find the distance from a point to a given plane.
  • 2.5.5. Find the angle between two planes.

By now, we are familiar with writing equations that describe a line in two dimensions. To write an equation for a line, we must know two points on the line, or we must know the direction of the line and at least one point through which the line passes. In two dimensions, we use the concept of slope to describe the orientation, or direction, of a line. In three dimensions, we describe the direction of a line using a vector parallel to the line. In this section, we examine how to use equations to describe lines and planes in space.

Equations for a Line in Space

Let’s first explore what it means for two vectors to be parallel. Recall that parallel vectors must have the same or opposite directions. If two nonzero vectors, uu and v,v, are parallel, we claim there must be a scalar, k,k, such that u=kv.u=kv. If uu and vv have the same direction, simply choose k=uv.k=uv. If uu and vv have opposite directions, choose k=uv.k=uv. Note that the converse holds as well. If u=kvu=kv for some scalar k,k, then either uu and vv have the same direction (k>0)(k>0) or opposite directions (k<0),(k<0), so uu and vv are parallel. Therefore, two nonzero vectors uu and vv are parallel if and only if u=kvu=kv for some scalar k.k. By convention, the zero vector 00 is considered to be parallel to all vectors.

As in two dimensions, we can describe a line in space using a point on the line and the direction of the line, or a parallel vector, which we call the direction vector (Figure 2.63). Let LL be a line in space passing through point P(x0,y0,z0).P(x0,y0,z0). Let v=a,b,cv=a,b,c be a vector parallel to L.L. Then, for any point on line Q(x,y,z),Q(x,y,z), we know that PQPQ is parallel to v.v. Thus, as we just discussed, there is a scalar, t,t, such that PQ=tv,PQ=tv, which gives

PQ=tvxx0,yy0,zz0=ta,b,cxx0,yy0,zz0=ta,tb,tc.PQ=tvxx0,yy0,zz0=ta,b,cxx0,yy0,zz0=ta,tb,tc.
2.11
This figure is the first octant of the 3-dimensional coordinate system. There is a line segment passing through two points. The points are labeled “P = (x sub 0, y sub 0, z sub 0)” and “Q = (x, y, z).” There is also a vector in standard position drawn. The vector is labeled “v = <a, b, c>.”
Figure 2.63 Vector v is the direction vector for PQ.PQ.

Using vector operations, we can rewrite Equation 2.11 as

xx0,yy0,zz0=ta,tb,tcx,y,zx0,y0,z0=ta,b,cx,y,z=x0,y0,z0+ta,b,c.xx0,yy0,zz0=ta,tb,tcx,y,zx0,y0,z0=ta,b,cx,y,z=x0,y0,z0+ta,b,c.

Setting r=x,y,zr=x,y,z and r0=x0,y0,z0,r0=x0,y0,z0, we now have the vector equation of a line:

r=r0+tv.r=r0+tv.
2.12

Equating components, Equation 2.11 shows that the following equations are simultaneously true: xx0=ta,xx0=ta, yy0=tb,yy0=tb, and zz0=tc.zz0=tc. If we solve each of these equations for the component variables x,y,andz,x,y,andz, we get a set of equations in which each variable is defined in terms of the parameter t and that, together, describe the line. This set of three equations forms a set of parametric equations of a line:

x=x0+tay=y0+tbz=z0+tc.x=x0+tay=y0+tbz=z0+tc.

If we solve each of the equations for tt assuming a,b,andca,b,andc are nonzero, we get a different description of the same line:

xx0a=tyy0b=tzz0c=t.xx0a=tyy0b=tzz0c=t.

Because each expression equals t, they all have the same value. We can set them equal to each other to create symmetric equations of a line:

xx0a=yy0b=zz0c.xx0a=yy0b=zz0c.

We summarize the results in the following theorem.

Theorem 2.11

Parametric and Symmetric Equations of a Line

A line LL parallel to vector v=a,b,cv=a,b,c and passing through point P(x0,y0,z0)P(x0,y0,z0) can be described by the following parametric equations:

x=x0+ta,y=y0+tb,andz=z0+tc.x=x0+ta,y=y0+tb,andz=z0+tc.
2.13

If the constants a,b,andca,b,andc are all nonzero, then LL can be described by the symmetric equation of the line:

xx0a=yy0b=zz0c.xx0a=yy0b=zz0c.
2.14

The parametric equations of a line are not unique. Using a different parallel vector or a different point on the line leads to a different, equivalent representation. Each set of parametric equations leads to a related set of symmetric equations, so it follows that a symmetric equation of a line is not unique either.

Example 2.45

Equations of a Line in Space

Find parametric and symmetric equations of the line passing through points (1,4,−2)(1,4,−2) and (−3,5,0).(−3,5,0).

Solution

First, identify a vector parallel to the line:

v=−31,54,0(−2)=−4,1,2.v=−31,54,0(−2)=−4,1,2.

Use either of the given points on the line to complete the parametric equations:

x=14t,y=4+t,andz=−2+2t.x=14t,y=4+t,andz=−2+2t.

Solve each equation for tt to create the symmetric equation of the line:

x1−4=y4=z+22.x1−4=y4=z+22.
Checkpoint 2.43

Find parametric and symmetric equations of the line passing through points (1,−3,2)(1,−3,2) and (5,−2,8).(5,−2,8).

Sometimes we don’t want the equation of a whole line, just a line segment. In this case, we limit the values of our parameter t.t. For example, let P(x0,y0,z0)P(x0,y0,z0) and Q(x1,y1,z1)Q(x1,y1,z1) be points on a line, and let p=x0,y0,z0p=x0,y0,z0 and q=x1,y1,z1q=x1,y1,z1 be the associated position vectors. In addition, let r=x,y,z.r=x,y,z. We want to find a vector equation for the line segment between PP and Q.Q. Using PP as our known point on the line, and PQ=x1x0,y1y0,z1z0PQ=x1x0,y1y0,z1z0 as the direction vector equation, Equation 2.12 gives

r=p+t(PQ).r=p+t(PQ).

Using properties of vectors, then

r=p+t(PQ)=x0,y0,z0+tx1x0,y1y0,z1z0=x0,y0,z0+t(x1,y1,z1x0,y0,z0)=x0,y0,z0+tx1,y1,z1tx0,y0,z0=(1t)x0,y0,z0+tx1,y1,z1=(1t)p+tq.r=p+t(PQ)=x0,y0,z0+tx1x0,y1y0,z1z0=x0,y0,z0+t(x1,y1,z1x0,y0,z0)=x0,y0,z0+tx1,y1,z1tx0,y0,z0=(1t)x0,y0,z0+tx1,y1,z1=(1t)p+tq.

Thus, the vector equation of the line passing through PP and QQ is

r=(1t)p+tq.r=(1t)p+tq.

Remember that we didn’t want the equation of the whole line, just the line segment between PP and Q.Q. Notice that when t=0,t=0, we have r=p,r=p, and when t=1,t=1, we have r=q.r=q. Therefore, the vector equation of the line segment between PP and QQ is

r=(1t)p+tq,0t1.r=(1t)p+tq,0t1.
2.15

Going back to Equation 2.12, we can also find parametric equations for this line segment. We have

r=p+t(PQ)x,y,z=x0,y0,z0+tx1x0,y1y0,z1z0=x0+t(x1x0),y0+t(y1y0),z0+t(z1z0).r=p+t(PQ)x,y,z=x0,y0,z0+tx1x0,y1y0,z1z0=x0+t(x1x0),y0+t(y1y0),z0+t(z1z0).

Then, the parametric equations are

x=x0+t(x1x0),y=y0+t(y1y0),z=z0+t(z1z0),0t1.x=x0+t(x1x0),y=y0+t(y1y0),z=z0+t(z1z0),0t1.
2.16

Example 2.46

Parametric Equations of a Line Segment

Find parametric equations of the line segment between the points P(2,1,4)P(2,1,4) and Q(3,−1,3).Q(3,−1,3).

Solution

By Equation 2.16, we have

x=x0+t(x1x0),y=y0+t(y1y0),z=z0+t(z1z0),0t1.x=x0+t(x1x0),y=y0+t(y1y0),z=z0+t(z1z0),0t1.

Working with each component separately, we get

x=x0+t(x1x0)=2+t(32)=2+t,x=x0+t(x1x0)=2+t(32)=2+t,
y=y0+t(y1y0)=1+t(−11)=12t,y=y0+t(y1y0)=1+t(−11)=12t,

and

z=z0+t(z1z0)=4+t(34)=4t.z=z0+t(z1z0)=4+t(34)=4t.

Therefore, the parametric equations for the line segment are

x=2+t,y=12t,z=4t,0t1.x=2+t,y=12t,z=4t,0t1.
Checkpoint 2.44

Find parametric equations of the line segment between points P(−1,3,6)P(−1,3,6) and Q(−8,2,4).Q(−8,2,4).

Distance between a Point and a Line

We already know how to calculate the distance between two points in space. We now expand this definition to describe the distance between a point and a line in space. Several real-world contexts exist when it is important to be able to calculate these distances. When building a home, for example, builders must consider “setback” requirements, when structures or fixtures have to be a certain distance from the property line. Air travel offers another example. Airlines are concerned about the distances between populated areas and proposed flight paths.

Let LL be a line in the plane and let MM be any point not on the line. Then, we define distance dd from MM to LL as the length of line segment MP,MP, where PP is a point on LL such that MPMP is perpendicular to LL (Figure 2.64).

This figure has two line segments. The first line is labeled “L” and has point P on the segment. The second line segment is drawn from point P to point M and is perpendicular to line L. The second line segment is labeled “d.”
Figure 2.64 The distance from point MM to line LL is the length of MP.MP.

When we’re looking for the distance between a line and a point in space, Figure 2.64 still applies. We still define the distance as the length of the perpendicular line segment connecting the point to the line. In space, however, there is no clear way to know which point on the line creates such a perpendicular line segment, so we select an arbitrary point on the line and use properties of vectors to calculate the distance. Therefore, let PP be an arbitrary point on line LL and let vv be a direction vector for LL (Figure 2.65).

This figure has a line segment labeled “L.” On the line segment L there is point P. There is a vector drawn from point P to another point M. Also, from M there is a line segment drawn to line L. This segment is perpendicular to line L. There is also a vector labeled “v” on line segment L. A parallelogram has been formed with vector v, line segment P M, and two other segments back to line L.
Figure 2.65 Vectors PMPM and v form two sides of a parallelogram with base vv and height d,d, which is the distance between a line and a point in space.

By Area of a Parallelogram, vectors PMPM and vv form two sides of a parallelogram with area PM×v.PM×v. Using a formula from geometry, the area of this parallelogram can also be calculated as the product of its base and height:

PM×v=vd.PM×v=vd.

We can use this formula to find a general formula for the distance between a line in space and any point not on the line.

Theorem 2.12

Distance from a Point to a Line

Let LL be a line in space passing through point PP with direction vector v.v. If MM is any point not on L,L, then the distance from MM to LL is

d=PM×vv.d=PM×vv.

Example 2.47

Calculating the Distance from a Point to a Line

Find the distance between t point M=(1,1,3)M=(1,1,3) and line x34=y+12=z3.x34=y+12=z3.

Solution

From the symmetric equations of the line, we know that vector v=4,2,1v=4,2,1 is a direction vector for the line. Setting the symmetric equations of the line equal to zero, we see that point P(3,−1,3)P(3,−1,3) lies on the line. Then,

PM=13,1(−1),33=−2,2,0.PM=13,1(−1),33=−2,2,0.

To calculate the distance, we need to find PM×v:PM×v:

PM×v=|ijk−220421|=(20)i(−20)j+(−48)k=2i+2j12k.PM×v=|ijk−220421|=(20)i(−20)j+(−48)k=2i+2j12k.

Therefore, the distance between the point and the line is (Figure 2.66)

d=PM×vv=22+22+12242+22+12=23821.d=PM×vv=22+22+12242+22+12=23821.
This figure is the first octant of the 3-dimensional coordinate system. There is a 3-dimensional box drawn in the octant. There is a point labeled at (1, 1, 3). There is a line segment labeled “L” inside of the box. Also, there is a perpendicular line segment from the point to line L.
Figure 2.66 Point (1,1,3)(1,1,3) is approximately 2.72.7 units from the line with symmetric equations x34=y+12=z3.x34=y+12=z3.
Checkpoint 2.45

Find the distance between point (0,3,6)(0,3,6) and the line with parametric equations x=1t,y=1+2t,z=5+3t.x=1t,y=1+2t,z=5+3t.

Relationships between Lines

Given two lines in the two-dimensional plane, the lines are equal, they are parallel but not equal, or they intersect in a single point. In three dimensions, a fourth case is possible. If two lines in space are not parallel, but do not intersect, then the lines are said to be skew lines (Figure 2.67).

This figure has two line segments. They are 3-dimensional, are not parallel, and do not intersect. The directions are different and one is above the other.
Figure 2.67 In three dimensions, it is possible that two lines do not cross, even when they have different directions.

To classify lines as parallel but not equal, equal, intersecting, or skew, we need to know two things: whether the direction vectors are parallel and whether the lines share a point (Figure 2.68).

This figure is a table with two rows and two columns. Above the columns is the question “Lines share a common point?” The first column is labeled “yes,” and the second column is labeled “no.” To the left of the rows is the question “Direction vectors are parallel?” The first row is labeled “yes,” and the second row is labeled “no.” The entries of the first row are “equal” and “parallel but not equal.” The entries in the second row are “intersecting” and “skew.”
Figure 2.68 Determine the relationship between two lines based on whether their direction vectors are parallel and whether they share a point.

Example 2.48

Classifying Lines in Space

For each pair of lines, determine whether the lines are equal, parallel but not equal, skew, or intersecting.

  1. L1:x=2s1,y=s1,z=s4L1:x=2s1,y=s1,z=s4
    L2:x=t3,y=3t+8,z=52tL2:x=t3,y=3t+8,z=52t
  2. L1:L1: x=y=zx=y=z
    L2:x32=y=z2L2:x32=y=z2
  3. L1:x=6s1,y=−2s,z=3s+1L1:x=6s1,y=−2s,z=3s+1
    L2:x46=y+3−2=z13L2:x46=y+3−2=z13

Solution

  1. Line L1L1 has direction vector v1=2,1,1;v1=2,1,1; line L2L2 has direction vector v2=1,3,−2.v2=1,3,−2. Because the direction vectors are not parallel vectors, the lines are either intersecting or skew. To determine whether the lines intersect, we see if there is a point, (x,y,z),(x,y,z), that lies on both lines. To find this point, we use the parametric equations to create a system of equalities:
    2s1=t3;s1=3t+8;s4=52t.2s1=t3;s1=3t+8;s4=52t.

    By the first equation, t=2s+2.t=2s+2. Substituting into the second equation yields
    s1=3(2s+2)+8s1=6s+6+85s=−15s=−3.s1=3(2s+2)+8s1=6s+6+85s=−15s=−3.

    Substitution into the third equation, however, yields a contradiction:
    s4=52(2s+2)s4=54s45s=5s=1.s4=52(2s+2)s4=54s45s=5s=1.

    There is no single point that satisfies the parametric equations for L1andL2L1andL2 simultaneously. These lines do not intersect, so they are skew (see the following figure).
    This figure is the 3-dimensional coordinate system. There are two skew lines drawn. They do not intersect and are not parallel.
  2. Line L1 has direction vector v1=1,−1,1v1=1,−1,1 and passes through the origin, (0,0,0).(0,0,0). Line L2L2 has a different direction vector, v2=2,1,1,v2=2,1,1, so these lines are not parallel or equal. Let rr represent the parameter for line L1L1 and let ss represent the parameter for L2:L2:
    x=r y=r z=r x=2s+3 y=s z=s+2. x=r y=r z=r x=2s+3 y=s z=s+2.

    Solve the system of equations to find r=1r=1 and s=1.s=1. If we need to find the point of intersection, we can substitute these parameters into the original equations to get (1,−1,1)(1,−1,1) (see the following figure).
    This figure is the 3-dimensional coordinate system. There are two skew lines drawn. They do not intersect and are not parallel.
  3. Lines L1L1 and L2L2 have equivalent direction vectors: v=6,−2,3.v=6,−2,3. These two lines are parallel (see the following figure).
    This figure is the 3-dimensional coordinate system. There are two skew lines drawn. They do not intersect and are not parallel.
Checkpoint 2.46

Describe the relationship between the lines with the following parametric equations:

x=14t,y=3+t,z=86tx=14t,y=3+t,z=86t
x=2+3s,y=2s,z=−13s.x=2+3s,y=2s,z=−13s.

Equations for a Plane

We know that a line is determined by two points. In other words, for any two distinct points, there is exactly one line that passes through those points, whether in two dimensions or three. Similarly, given any three points that do not all lie on the same line, there is a unique plane that passes through these points. Just as a line is determined by two points, a plane is determined by three.

This may be the simplest way to characterize a plane, but we can use other descriptions as well. For example, given two distinct, intersecting lines, there is exactly one plane containing both lines. A plane is also determined by a line and any point that does not lie on the line. These characterizations arise naturally from the idea that a plane is determined by three points. Perhaps the most surprising characterization of a plane is actually the most useful.

Imagine a pair of orthogonal vectors that share an initial point. Visualize grabbing one of the vectors and twisting it. As you twist, the other vector spins around and sweeps out a plane. Here, we describe that concept mathematically. Let n=a,b,cn=a,b,c be a vector and P=(x0,y0,z0)P=(x0,y0,z0) be a point. Then the set of all points Q=(x,y,z)Q=(x,y,z) such that PQPQ is orthogonal to nn forms a plane (Figure 2.69). We say that nn is a normal vector, or perpendicular to the plane. Remember, the dot product of orthogonal vectors is zero. This fact generates the vector equation of a plane: n·PQ=0.n·PQ=0. Rewriting this equation provides additional ways to describe the plane:

n·PQ=0a,b,c·xx0,yy0,zz0=0a(xx0)+b(yy0)+c(zz0)=0.n·PQ=0a,b,c·xx0,yy0,zz0=0a(xx0)+b(yy0)+c(zz0)=0.
This figure is a parallelogram representing a plane. In the plane is a vector from point P to point Q. Perpendicular to the vector P Q is the vector n.
Figure 2.69 Given a point PP and vector n,n, the set of all points QQ with PQPQ orthogonal to nn forms a plane.

Definition

Given a point PP and vector n,n, the set of all points QQ satisfying the equation n·PQ=0n·PQ=0 forms a plane. The equation

n·PQ=0n·PQ=0
2.17

is known as the vector equation of a plane.

The scalar equation of a plane containing point P=(x0,y0,z0)P=(x0,y0,z0) with normal vector n=a,b,cn=a,b,c is

a(xx0)+b(yy0)+c(zz0)=0.a(xx0)+b(yy0)+c(zz0)=0.
2.18

This equation can be expressed as ax+by+cz+d=0,ax+by+cz+d=0, where d=ax0by0cz0.d=ax0by0cz0. This form of the equation is sometimes called the general form of the equation of a plane.

As described earlier in this section, any three points that do not all lie on the same line determine a plane. Given three such points, we can find an equation for the plane containing these points.

Example 2.49

Writing an Equation of a Plane Given Three Points in the Plane

Write an equation for the plane containing points P=(1,1,−2),P=(1,1,−2), Q=(0,2,1),Q=(0,2,1), and R=(−1,−1,0)R=(−1,−1,0) in both standard and general forms.

Solution

To write an equation for a plane, we must find a normal vector for the plane. We start by identifying two vectors in the plane:

PQ=01,21,1(−2)=−1,1,3QR=−10,−12,01=−1,−3,−1.PQ=01,21,1(−2)=−1,1,3QR=−10,−12,01=−1,−3,−1.

The cross product PQ×QRPQ×QR is orthogonal to both PQPQ and QR,QR, so it is normal to the plane that contains these two vectors:

n=PQ×QR=|ijk−113−1−3−1|=(−1+9)i(1+3)j+(3+1)k=8i4j+4k.n=PQ×QR=|ijk−113−1−3−1|=(−1+9)i(1+3)j+(3+1)k=8i4j+4k.

Thus, n=8,−4,4,n=8,−4,4, and we can choose any of the three given points to write an equation of the plane:

8(x1)4(y1)+4(z+2)=08x4y+4z+4=0.8(x1)4(y1)+4(z+2)=08x4y+4z+4=0.

The scalar equations of a plane vary depending on the normal vector and point chosen.

Example 2.50

Writing an Equation for a Plane Given a Point and a Line

Find an equation of the plane that passes through point (1,4,3)(1,4,3) and contains the line given by x=y12=z+1.x=y12=z+1.

Solution

Symmetric equations describe the line that passes through point (0,1,1)(0,1,1) parallel to vector v1=1,2,1v1=1,2,1 (see the following figure). Use this point and the given point, (1,4,3),(1,4,3), to identify a second vector parallel to the plane:

v2=10,41,3(−1)=1,3,4.v2=10,41,3(−1)=1,3,4.

Use the cross product of these vectors to identify a normal vector for the plane:

n=v1×v2=|ijk121134|=(83)i(41)j+(32)k=5i3j+k.n=v1×v2=|ijk121134|=(83)i(41)j+(32)k=5i3j+k.

The scalar equations for the plane are 5x3(y1)+(z+1)=05x3(y1)+(z+1)=0 and 5x3y+z+4=0.5x3y+z+4=0.

This figure is the 3-dimensional coordinate system. There is a plane sketched. It is vertical, but skew to the z-axis.
Checkpoint 2.47

Find an equation of the plane containing the lines L1L1 and L2:L2:

L1:x=y=zL2:x32=y=z2.L1:x=y=zL2:x32=y=z2.

Now that we can write an equation for a plane, we can use the equation to find the distance dd between a point PP and the plane. It is defined as the shortest possible distance from PP to a point on the plane.

This figure is the sketch of a parallelogram representing a plane. In the plane are points Q and R. there is a broken line from Q to R on the plane. There is a vector n out of the plane at point Q. Also, there is a vector labeled “R P” from point R to point P which is above the plane. This vector is perpendicular to the plane.
Figure 2.70 We want to find the shortest distance from point P to the plane. Let point RR be the point in the plane such that, for any other point in the plane Q,Q, RP<QP.RP<QP.

Just as we find the two-dimensional distance between a point and a line by calculating the length of a line segment perpendicular to the line, we find the three-dimensional distance between a point and a plane by calculating the length of a line segment perpendicular to the plane. Let RR bet the point in the plane such that RPRP is orthogonal to the plane, and let QQ be an arbitrary point in the plane. Then the projection of vector QPQP onto the normal vector describes vector RP,RP, as shown in Figure 2.70.

Theorem 2.13

The Distance between a Plane and a Point

Suppose a plane with normal vector nn passes through point Q.Q. The distance dd from the plane to a point PP not in the plane is given by

d=projnQP=|compnQP|=|QP·n|n.d=projnQP=|compnQP|=|QP·n|n.
2.19

Example 2.51

Distance between a Point and a Plane

Find the distance between point P=(3,1,2)P=(3,1,2) and the plane given by x2y+z=5x2y+z=5 (see the following figure).

This figure is the 3-dimensional coordinate system. There is a point drawn at (3, 1, 2). The point is labeled “P(3, 1, 2).” There is a plane drawn. There is a perpendicular line from the plane to point P(3, 1, 2).

Solution

The coefficients of the plane’s equation provide a normal vector for the plane: n=1,−2,1.n=1,−2,1. To find vector QP,QP, we need a point in the plane. Any point will work, so set y=z=0y=z=0 to see that point Q=(5,0,0)Q=(5,0,0) lies in the plane. Find the component form of the vector from QtoP:QtoP:

QP=35,10,20=−2,1,2.QP=35,10,20=−2,1,2.

Apply the distance formula from Equation 2.19:

d=|QP·n|n=|−2,1,2·1,−2,1|12+(−2)2+12=|−22+2|6=26.d=|QP·n|n=|−2,1,2·1,−2,1|12+(−2)2+12=|−22+2|6=26.
Checkpoint 2.48

Find the distance between point P=(5,−1,0)P=(5,−1,0) and the plane given by 4x+2yz=3.4x+2yz=3.

Parallel and Intersecting Planes

We have discussed the various possible relationships between two lines in two dimensions and three dimensions. When we describe the relationship between two planes in space, we have only two possibilities: the two distinct planes are parallel or they intersect. When two planes are parallel, their normal vectors are parallel. When two planes intersect, the intersection is a line (Figure 2.71).

This figure is two planes that are intersecting. The intersection forms a line segment.
Figure 2.71 The intersection of two nonparallel planes is always a line.

We can use the equations of the two planes to find parametric equations for the line of intersection.

Example 2.52

Finding the Line of Intersection for Two Planes

Find parametric and symmetric equations for the line formed by the intersection of the planes given by x+y+z=0x+y+z=0 and 2xy+z=02xy+z=0 (see the following figure).

This figure is two planes intersecting in the 3-dimensional coordinate system.

Solution

Note that the two planes have nonparallel normals, so the planes intersect. Further, the origin satisfies each equation, so we know the line of intersection passes through the origin. Add the plane equations so we can eliminate the one of the variables, in this case, y:y:

x+y+z=02xy+z=0____________________3x+2z=0.x+y+z=02xy+z=0____________________3x+2z=0.

This gives us x=23z.x=23z. We substitute this value into the first equation to express yy in terms of z:z:

x+y+z=023z+y+z=0y+13z=0y=13z.x+y+z=023z+y+z=0y+13z=0y=13z.

We now have the first two variables, xx and y,y, in terms of the third variable, z.z. Now we define zz in terms of t.t. To eliminate the need for fractions, we choose to define the parameter tt as t=13z.t=13z. Then, z=−3t.z=−3t. Substituting the parametric representation of zz back into the other two equations, we see that the parametric equations for the line of intersection are x=2t,y=t,z=−3t.x=2t,y=t,z=−3t. The symmetric equations for the line are x2=y=z−3.x2=y=z−3.

Checkpoint 2.49

Find parametric equations for the line formed by the intersection of planes x+yz=3x+yz=3 and 3xy+3z=5.3xy+3z=5.

In addition to finding the equation of the line of intersection between two planes, we may need to find the angle formed by the intersection of two planes. For example, builders constructing a house need to know the angle where different sections of the roof meet to know whether the roof will look good and drain properly. We can use normal vectors to calculate the angle between the two planes. We can do this because the angle between the normal vectors is the same as the angle between the planes. Figure 2.72 shows why this is true.

This figure is two parallelograms representing planes. The planes intersect forming angle theta between them. The first plane as vector “n sub 1” normal to the plane. The second vector has vector “n sub 2” normal to the plane. The normal vectors intersect and form the angle theta.
Figure 2.72 The angle between two planes has the same measure as the angle between the normal vectors for the planes.

We can find the measure of the angle θ between two intersecting planes by first finding the cosine of the angle, using the following equation:

cosθ=|n1·n2|n1n2.cosθ=|n1·n2|n1n2.

We can then use the angle to determine whether two planes are parallel or orthogonal or if they intersect at some other angle.

Example 2.53

Finding the Angle between Two Planes

Determine whether each pair of planes is parallel, orthogonal, or neither. If the planes are intersecting, but not orthogonal, find the measure of the angle between them. Give the answer in radians and round to two decimal places.

  1. x+2yz=8and2x+4y2z=10x+2yz=8and2x+4y2z=10
  2. 2x3y+2z=3and6x+2y3z=12x3y+2z=3and6x+2y3z=1
  3. x+y+z=4andx3y+5z=1x+y+z=4andx3y+5z=1

Solution

  1. The normal vectors for these planes are n1=1,2,−1n1=1,2,−1 and n2=2,4,−2.n2=2,4,−2. These two vectors are scalar multiples of each other. The normal vectors are parallel, so the planes are parallel.
  2. The normal vectors for these planes are n1=2,−3,2n1=2,−3,2 and n2=6,2,−3.n2=6,2,−3. Taking the dot product of these vectors, we have
    n1·n2=2,−3,2·6,2,−3=2(6)3(2)+2(−3)=0.n1·n2=2,−3,2·6,2,−3=2(6)3(2)+2(−3)=0.

    The normal vectors are orthogonal, so the corresponding planes are orthogonal as well.
  3. The normal vectors for these planes are n1=1,1,1n1=1,1,1 and n2=1,−3,5:n2=1,−3,5:
    cosθ=|n1·n2|n1n2=|1,1,1·1,−3,5|12+12+1212+(−3)2+52=3105.cosθ=|n1·n2|n1n2=|1,1,1·1,−3,5|12+12+1212+(−3)2+52=3105.

    The angle between the two planes is 1.271.27 rad, or approximately 73°.73°.
Checkpoint 2.50

Find the measure of the angle between planes x+yz=3x+yz=3 and 3xy+3z=5.3xy+3z=5. Give the answer in radians and round to two decimal places.

When we find that two planes are parallel, we may need to find the distance between them. To find this distance, we simply select a point in one of the planes. The distance from this point to the other plane is the distance between the planes.

Previously, we introduced the formula for calculating this distance in Equation 2.19:

d=QP·nn,d=QP·nn,

where QQ is a point on the plane, PP is a point not on the plane, and nn is the normal vector that passes through point Q.Q. Consider the distance from point (x0,y0,z0)(x0,y0,z0) to plane ax+by+cz+k=0.ax+by+cz+k=0. Let (x1,y1,z1)(x1,y1,z1) be any point in the plane. Substituting into the formula yields

d=|a(x0x1)+b(y0y1)+c(z0z1)|a2+b2+c2=|ax0+by0+cz0+k|a2+b2+c2.d=|a(x0x1)+b(y0y1)+c(z0z1)|a2+b2+c2=|ax0+by0+cz0+k|a2+b2+c2.

We state this result formally in the following theorem.

Theorem 2.14

Distance from a Point to a Plane

Let P(x0,y0,z0)P(x0,y0,z0) be a point. The distance from PP to plane ax+by+cz+k=0ax+by+cz+k=0 is given by

d=|ax0+by0+cz0+k|a2+b2+c2.d=|ax0+by0+cz0+k|a2+b2+c2.

Example 2.54

Finding the Distance between Parallel Planes

Find the distance between the two parallel planes given by 2x+yz=22x+yz=2 and 2x+yz=8.2x+yz=8.

Solution

Point (1,0,0)(1,0,0) lies in the first plane. The desired distance, then, is

d=|ax0+by0+cz0+k|a2+b2+c2=|2(1)+1(0)+(−1)(0)+(−8)|22+12+(−1)2=66=6.d=|ax0+by0+cz0+k|a2+b2+c2=|2(1)+1(0)+(−1)(0)+(−8)|22+12+(−1)2=66=6.
Checkpoint 2.51

Find the distance between parallel planes 5x2y+z=65x2y+z=6 and 5x2y+z=−3.5x2y+z=−3.

Student Project

Distance between Two Skew Lines

This figure shows a system of pipes running in different directions in an industrial plant. Two skew pipes are highlighted in red.
Figure 2.73 Industrial pipe installations often feature pipes running in different directions. How can we find the distance between two skew pipes?

Finding the distance from a point to a line or from a line to a plane seems like a pretty abstract procedure. But, if the lines represent pipes in a chemical plant or tubes in an oil refinery or roads at an intersection of highways, confirming that the distance between them meets specifications can be both important and awkward to measure. One way is to model the two pipes as lines, using the techniques in this chapter, and then calculate the distance between them. The calculation involves forming vectors along the directions of the lines and using both the cross product and the dot product.

The symmetric forms of two lines, L1L1 and L2,L2, are

L1:xx1a1=yy1b1=zz1c1L2:xx2a2=yy2b2=zz2c2.L1:xx1a1=yy1b1=zz1c1L2:xx2a2=yy2b2=zz2c2.

You are to develop a formula for the distance dd between these two lines, in terms of the values a1,b1,c1;a2,b2,c2;x1,y1,z1;andx2,y2,z2.a1,b1,c1;a2,b2,c2;x1,y1,z1;andx2,y2,z2. The distance between two lines is usually taken to mean the minimum distance, so this is the length of a line segment or the length of a vector that is perpendicular to both lines and intersects both lines.

  1. First, write down two vectors, v1v1 and v2,v2, that lie along L1L1 and L2,L2, respectively.
  2. Find the cross product of these two vectors and call it N.N. This vector is perpendicular to v1andv2,v1andv2, and hence is perpendicular to both lines.
  3. From vector N,N, form a unit vector nn in the same direction.
  4. Use symmetric equations to find a convenient vector v12v12 that lies between any two points, one on each line. Again, this can be done directly from the symmetric equations.
  5. The dot product of two vectors is the magnitude of the projection of one vector onto the other—that is, A·B=ABcosθ,A·B=ABcosθ, where θθ is the angle between the vectors. Using the dot product, find the projection of vector v12v12 found in step 44 onto unit vector nn found in step 3. This projection is perpendicular to both lines, and hence its length must be the perpendicular distance dd between them. Note that the value of dd may be negative, depending on your choice of vector v12v12 or the order of the cross product, so use absolute value signs around the numerator.
  6. Check that your formula gives the correct distance of |−25|/1981.78|−25|/1981.78 between the following two lines:
    L1:x52=y34=z13L2:x63=y15=z7.L1:x52=y34=z13L2:x63=y15=z7.
  7. Is your general expression valid when the lines are parallel? If not, why not? (Hint: What do you know about the value of the cross product of two parallel vectors? Where would that result show up in your expression for d?)d?)
  8. Demonstrate that your expression for the distance is zero when the lines intersect. Recall that two lines intersect if they are not parallel and they are in the same plane. Hence, consider the direction of nn and v12.v12. What is the result of their dot product?
  9. Consider the following application. Engineers at a refinery have determined they need to install support struts between many of the gas pipes to reduce damaging vibrations. To minimize cost, they plan to install these struts at the closest points between adjacent skewed pipes. Because they have detailed schematics of the structure, they are able to determine the correct lengths of the struts needed, and hence manufacture and distribute them to the installation crews without spending valuable time making measurements.
    The rectangular frame structure has the dimensions 4.0×15.0×10.0m4.0×15.0×10.0m (height, width, and depth). One sector has a pipe entering the lower corner of the standard frame unit and exiting at the diametrically opposed corner (the one farthest away at the top); call this L1.L1. A second pipe enters and exits at the two different opposite lower corners; call this L2L2 (Figure 2.74).
    This figure is a three-dimensional box in an x y z coordinate system. The box has dimensions x = 10 m, y = 15 m, and z = 4 m. Line L1 passes through a main diagonal of the box from the origin to the far corner. Line L2 passes through a diagonal in the base of the box with x-intercept 10 and y-intercept 15.
    Figure 2.74 Two pipes cross through a standard frame unit.

    Write down the vectors along the lines representing those pipes, find the cross product between them from which to create the unit vector n,n, define a vector that spans two points on each line, and finally determine the minimum distance between the lines. (Take the origin to be at the lower corner of the first pipe.) Similarly, you may also develop the symmetric equations for each line and substitute directly into your formula.

Section 2.5 Exercises

In the following exercises, points PP and QQ are given. Let LL be the line passing through points PP and Q.Q.

  1. Find the vector equation of line L.L.
  2. Find parametric equations of line L.L.
  3. Find symmetric equations of line L.L.
  4. Find parametric equations of the line segment determined by PP and Q.Q.
243.

P(−3,5,9),P(−3,5,9), Q(4,−7,2)Q(4,−7,2)

244.

P(4,0,5),Q(2,3,1)P(4,0,5),Q(2,3,1)

245.

P(−1,0,5),P(−1,0,5), Q(4,0,3)Q(4,0,3)

246.

P(7,−2,6),P(7,−2,6), Q(−3,0,6)Q(−3,0,6)

For the following exercises, point PP and vector vv are given. Let LL be the line passing through point PP with direction v.v.

  1. Find parametric equations of line L.L.
  2. Find symmetric equations of line L.L.
  3. Find the intersection of the line with the xy-plane.
247.

P(1,−2,3),P(1,−2,3), v=1,2,3v=1,2,3

248.

P(3,1,5),P(3,1,5), v=1,1,1v=1,1,1

249.

P(3,1,5),P(3,1,5), v=QR,v=QR, where Q(2,2,3)Q(2,2,3) and R(3,2,3)R(3,2,3)

250.

P(2,3,0),P(2,3,0), v=QR,v=QR, where Q(0,4,5)Q(0,4,5) and R(0,4,6)R(0,4,6)

For the following exercises, line LL is given.

  1. Find point PP that belongs to the line and direction vector vv of the line. Express vv in component form.
  2. Find the distance from the origin to line L.L.
251.

x=1+t,y=3+t,z=5+4t,x=1+t,y=3+t,z=5+4t, tt

252.

x=y+1,z=2x=y+1,z=2

253.

Find the distance between point A(−3,1,1)A(−3,1,1) and the line of symmetric equations

x=y=z.x=y=z.

254.

Find the distance between point A(4,2,5)A(4,2,5) and the line of parametric equations

x=−1t,y=t,z=2,x=−1t,y=t,z=2, t.t.

For the following exercises, lines L1L1 and L2L2 are given.

  1. Verify whether lines L1L1 and L2L2 are parallel.
  2. If the lines L1L1 and L2L2 are parallel, then find the distance between them.
255.

L1:x=1+t,y=t,z=2+t,L1:x=1+t,y=t,z=2+t, t,t, L2:x3=y1=z3L2:x3=y1=z3

256.

L1:x=2,y=1,z=t,L1:x=2,y=1,z=t, L2:x=1,y=1,z=23t,L2:x=1,y=1,z=23t, tt

257.

Show that the line passing through points P(3,1,0)P(3,1,0) and Q(1,4,−3)Q(1,4,−3) is perpendicular to the line with equation x=3t,y=3+8t,z=−7+6t,x=3t,y=3+8t,z=−7+6t, t.t.

258.

Are the lines of equations x=−2+2t,y=−6,z=2+6tx=−2+2t,y=−6,z=2+6t and x=−1+t,y=1+t,z=t,x=−1+t,y=1+t,z=t, t,t, perpendicular to each other?

259.

Find the point of intersection of the lines of equations x=−2y=3zx=−2y=3z and x=−5t,y=−1+t,z=t11,x=−5t,y=−1+t,z=t11, t.t.

260.

Find the intersection point of the x-axis with the line of parametric equations

x=10+t,y=22t,z=−3+3t,x=10+t,y=22t,z=−3+3t, t.t.

For the following exercises, lines L1L1 and L2L2 are given. Determine whether the lines are equal, parallel but not equal, skew, or intersecting.

261.

L1:x=y1=zL1:x=y1=z and L2:x2=y=z2L2:x2=y=z2

262.

L1:x=2t,y=0,z=3,L1:x=2t,y=0,z=3, tt and L2:x=0,y=8+s,z=7+s,L2:x=0,y=8+s,z=7+s, ss

263.

L1:x=−1+2t,y=1+3t,z=7t,L1:x=−1+2t,y=1+3t,z=7t, tt and L2:x1=23(y4)=27z2L2:x1=23(y4)=27z2

264.

L1:3x=y+1=2zL1:3x=y+1=2z and L2:x=6+2t,y=17+6t,z=9+3t,L2:x=6+2t,y=17+6t,z=9+3t, tt

265.

Consider line LL of symmetric equations x2=y=z2x2=y=z2 and point A(1,1,1).A(1,1,1).

  1. Find parametric equations for a line parallel to LL that passes through point A.A.
  2. Find symmetric equations of a line skew to LL and that passes through point A.A.
  3. Find symmetric equations of a line that intersects LL and passes through point A.A.
266.

Consider line LL of parametric equations x=t,y=2t,z=3,x=t,y=2t,z=3, t.t.

  1. Find parametric equations for a line parallel to LL that passes through the origin.
  2. Find parametric equations of a line skew to LL that passes through the origin.
  3. Find symmetric equations of a line that intersects LL and passes through the origin.

For the following exercises, point PP and vector nn are given.

  1. Find the scalar equation of the plane that passes through PP and has normal vector n.n.
  2. Find the general form of the equation of the plane that passes through PP and has normal vector n.n.
267.

P(0,0,0),P(0,0,0), n=3i2j+4kn=3i2j+4k

268.

P(3,2,2),P(3,2,2), n=2i+3jkn=2i+3jk

269.

P(1,2,3),P(1,2,3), n=1,2,3n=1,2,3

270.

P(0,0,0),P(0,0,0), n=−3,2,−1n=−3,2,−1

For the following exercises, the equation of a plane is given.

  1. Find normal vector nn to the plane. Express nn using standard unit vectors.
  2. Find the intersections of the plane with the axes of coordinates.
  3. Sketch the plane.
271.

[T] 4x+5y+10z20=04x+5y+10z20=0

272.

3x+4y12=03x+4y12=0

273.

3x2y+4z=03x2y+4z=0

274.

x+z=0x+z=0

275.

Given point P(1,2,3)P(1,2,3) and vector n=i+j,n=i+j, find point QQ on the x-axis such that PQPQ and nn are orthogonal.

276.

Show there is no plane perpendicular to n=i+jn=i+j that passes through points P(1,2,3)P(1,2,3) and Q(2,3,4).Q(2,3,4).

277.

Find parametric equations of the line passing through point P(−2,1,3)P(−2,1,3) that is perpendicular to the plane of equation 2x3y+z=7.2x3y+z=7.

278.

Find symmetric equations of the line passing through point P(2,5,4)P(2,5,4) that is perpendicular to the plane of equation 2x+3y5z=0.2x+3y5z=0.

279.

Show that line x12=y+13=z24x12=y+13=z24 is parallel to plane x2y+z=6.x2y+z=6.

280.

Find the real number αα such that the line of parametric equations x=t,y=2t,z=3+t,x=t,y=2t,z=3+t, tt is parallel to the plane of equation αx+5y+z10=0.αx+5y+z10=0.

For the following exercises, points P,Q,andRP,Q,andR are given.

  1. Find the general equation of the plane passing through P,Q,andR.P,Q,andR.
  2. Write the vector equation n·PS=0n·PS=0 of the plane at a., where S(x,y,z)S(x,y,z) is an arbitrary point of the plane.
  3. Find parametric equations of the line passing through the origin that is perpendicular to the plane passing through P,Q,andR.P,Q,andR.
281.

P(1,1,1),Q(2,4,3),P(1,1,1),Q(2,4,3), and R(−1,−2,−1)R(−1,−2,−1)

282.

P(−2,1,4),Q(3,1,3),P(−2,1,4),Q(3,1,3), and R(−2,1,0)R(−2,1,0)

283.

Consider the planes of equations x+y+z=1x+y+z=1 and x+z=0.x+z=0.

  1. Show that the planes intersect.
  2. Find symmetric equations of the line passing through point P(1,4,6)P(1,4,6) that is parallel to the line of intersection of the planes.
284.

Consider the planes of equations y+z2=0y+z2=0 and xy=0.xy=0.

  1. Show that the planes intersect.
  2. Find parametric equations of the line passing through point P(−8,0,2)P(−8,0,2) that is parallel to the line of intersection of the planes.
285.

Find the scalar equation of the plane that passes through point P(−1,2,1)P(−1,2,1) and is perpendicular to the line of intersection of planes x+yz2=0x+yz2=0 and 2xy+3z1=0.2xy+3z1=0.

286.

Find the general equation of the plane that passes through the origin and is perpendicular to the line of intersection of planes x+y+2=0x+y+2=0 and z3=0.z3=0.

287.

Determine whether the line of parametric equations x=1+2t,y=−2t,z=2+t,x=1+2t,y=−2t,z=2+t, tt intersects the plane with equation 3x+4y+6z7=0.3x+4y+6z7=0. If it does intersect, find the point of intersection.

288.

Determine whether the line of parametric equations x=5,y=4t,z=2t,x=5,y=4t,z=2t, tt intersects the plane with equation 2xy+z=5.2xy+z=5. If it does intersect, find the point of intersection.

289.

Find the distance from point P(1,5,−4)P(1,5,−4) to the plane of equation 3xy+2z6=0.3xy+2z6=0.

290.

Find the distance from point P(1,−2,3)P(1,−2,3) to the plane of equation (x3)+2(y+1)4z=0.(x3)+2(y+1)4z=0.

For the following exercises, the equations of two planes are given.

  1. Determine whether the planes are parallel, orthogonal, or neither.
  2. If the planes are neither parallel nor orthogonal, then find the measure of the angle between the planes. Express the answer in degrees rounded to the nearest integer.
291.

[T] x+y+z=0,x+y+z=0, 2xy+z7=02xy+z7=0

292.

5x3y+z=4,5x3y+z=4, x+4y+7z=1x+4y+7z=1

293.

x5yz=1,x5yz=1, 5x25y5z=−35x25y5z=−3

294.

[T] x3y+6z=4,x3y+6z=4, 5x+yz=45x+yz=4

295.

Show that the lines of equations x=t,y=1+t,z=2+t,x=t,y=1+t,z=2+t, t,t, and x2=y13=z3x2=y13=z3 are skew, and find the distance between them.

296.

Show that the lines of equations x=−1+t,y=−2+t,z=3t,x=−1+t,y=−2+t,z=3t, t,t, and x=5+s,y=−8+2s,z=7s,x=5+s,y=−8+2s,z=7s, ss are skew, and find the distance between them.

297.

Consider point C(−3,2,4)C(−3,2,4) and the plane of equation 2x+4y3z=8.2x+4y3z=8.

  1. Find the radius of the sphere with center CC tangent to the given plane.
  2. Find point P of tangency.
298.

Consider the plane of equation xyz8=0.xyz8=0.

  1. Find the equation of the sphere with center CC at the origin that is tangent to the given plane.
  2. Find parametric equations of the line passing through the origin and the point of tangency.
299.

Two children are playing with a ball. The girl throws the ball to the boy. The ball travels in

the air, curves 33 ft to the right, and falls 55 ft away from the girl (see the following figure). If the plane that contains the trajectory of the ball is perpendicular to the ground, find its equation.

This figure is the image of two children throwing a ball. The path of the ball is represented with an arc. The distance from the child throwing the ball to the point where the ball hits is 5 feet. The distance from the second child to where the ball hits is 3 feet.
300.

[T] John allocates dd dollars to consume monthly three goods of prices a,b,andc.a,b,andc. In this context, the budget equation is defined as ax+by+cz=d,ax+by+cz=d, where x0,y0,x0,y0, and z0z0 represent the number of items bought from each of the goods. The budget set is given by {(x,y,z)|ax+by+czd,x0,y0,z0},{(x,y,z)|ax+by+czd,x0,y0,z0}, and the budget plane is the part of the plane of equation ax+by+cz=dax+by+cz=d for which x0,y0,x0,y0, and z0.z0. Consider a=$8,a=$8, b=$5,b=$5, c=$10,c=$10, and d=$500.d=$500.

  1. Use a CAS to graph the budget set and budget plane.
  2. For z=25,z=25, find the new budget equation and graph the budget set in the same system of coordinates.
301.

[T] Consider r(t)=sint,cost,2tr(t)=sint,cost,2t the position vector of a particle at time t[0,3],t[0,3], where the components of r are expressed in centimeters and time is measured in seconds. Let OPOP be the position vector of the particle after 11 sec.

  1. Determine the velocity vector v(1)v(1) of the particle after 11 sec.
  2. Find the scalar equation of the plane that is perpendicular to v(1)v(1) and passes through point P.P. This plane is called the normal plane to the path of the particle at point P.P.
  3. Use a CAS to visualize the path of the particle along with the velocity vector and normal plane at point P.P.
302.

[T] A solar panel is mounted on the roof of a house. The panel may be regarded as positioned at the points of coordinates (in meters) A(8,0,0),A(8,0,0), B(8,18,0),B(8,18,0), C(0,18,8),C(0,18,8), and D(0,0,8)D(0,0,8) (see the following figure).

This figure is a picture of a rectangular solar grid on a roof. The corners of the rectangle are labeled A, B, C, D. There are two vectors, the first is from A to D. The second is from A to B.
  1. Find the general form of the equation of the plane that contains the solar panel by using points A,B,andC,A,B,andC, and show that its normal vector is equivalent to AB×AD.AB×AD.
  2. Find parametric equations of line L1L1 that passes through the center of the solar panel and has direction vector s=13i+13j+13k,s=13i+13j+13k, which points toward the position of the Sun at a particular time of day.
  3. Find symmetric equations of line L2L2 that passes through the center of the solar panel and is perpendicular to it.
  4. Determine the angle of elevation of the Sun above the solar panel by using the angle between lines L1L1 and L2.L2.
Citation/Attribution

Want to cite, share, or modify this book? This book is Creative Commons Attribution-NonCommercial-ShareAlike License 4.0 and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/calculus-volume-3/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/calculus-volume-3/pages/1-introduction
Citation information

© Mar 30, 2016 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License 4.0 license. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.