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Coordinated Multivariable Calculus

Section 1.5 Lines and Planes in Space

In single variable calculus, we learn that a differentiable function is locally linear. In other words, if we zoom in on the graph of a differentiable function at a point, the graph looks like the tangent line to the function at that point. Linear functions played important roles in single variable calculus, useful in approximating differentiable functions, in approximating roots of functions (see Newton’s Method), and approximating solutions to first order differential equations (see Euler’s Method). In multivariable calculus, we will soon study curves in space; differentiable curves turn out to be locally linear as well. In addition, as we study functions of two variables, we will see that such a function is locally linear at a point if the surface defined by the function looks like a plane (the tangent plane) as we zoom in on the graph.
Consequently, it is important for us to understand both lines and planes in space, as these define the linear functions in R2 and R3. (Recall that a function is linear if it is a polynomial function whose terms all have degree less than or equal to 1. For example, x defines a single variable linear function and x+y a two variable linear function, but xy is not linear since it has degree two, the sum of the degress of its factors.) We begin our work by considering some familiar ideas in R2 but from a new perspective.

Preview Activity 1.5.1.

We are familiar with equations of lines in the plane in the form y=mx+b, where m is the slope of the line and (0,b) is the y-intercept. In this activity, we explore a more flexible way of representing lines that we can use not only in the plane, but in higher dimensions as well.
To begin, consider the line through the point (2,βˆ’1) with slope 23 as shown in Figure 1.5.1.
Figure 1.5.1. The line through (2,βˆ’1) with slope 23.
  1. Suppose we increase x by 1 from the point (2,βˆ’1). How does the y-value change? What is the point on the line with x-coordinate 3?
  2. Suppose we decrease x by 3.25 from the point (2,βˆ’1). How does the y-value change? What is the point on the line with x-coordinate βˆ’1.25?
  3. Now, suppose we increase x by some arbitrary value 3t from the point (2,βˆ’1). How does the y-value change? What is the point on the line with x-coordinate 2+3t?
  4. Observe that the slope of the line is related to any vector whose y-component divided by the x-component is the slope of the line. For the line in this exercise, we might use the vector ⟨3,2⟩, which describes the direction of the line. Explain why the terminal points of the vectors r(t), where
    r(t)=⟨2,βˆ’1⟩+⟨3,2⟩t,
    trace out the graph of the line through the point (2,βˆ’1) with slope 23.
  5. Now we extend this vector approach to R3 and consider a second example. Let L be the line in R3 through the point (1,0,2) in the direction of the vector ⟨2,βˆ’1,4⟩. Find the coordinates of three distinct points on line L. Explain your thinking.
  6. Find a vector in the form
    r(t)=⟨x0,y0,z0⟩+⟨a,b,c⟩t
    whose terminal points trace out the line L that is described in (e). That is, you should be able to locate any point on the line by determining a corresponding value of t.

Subsection 1.5.1 Lines in Space

In two-dimensional space, a non-vertical line is defined to be the set of points satisfying the equation
y=mx+b,
for some constants m and b. The value of m (the slope) tells us how the dependent variable changes for every one unit increase in the independent variable, while the point (0,b) is the y-intercept and anchors the line to a location on the y-axis. Alternatively, we can think of the slope as being related to the vector ⟨1,m⟩, which tells us the direction of the line, as shown on the left in Figure 1.5.2. Thus, we can identify a line in space by fixing a point P and a direction v, as shown on the right. Since we also have vectors in space to provide direction, this same idea of a point and a direction determining a line works in Rn for any n.
Figure 1.5.2. A vector description of a line

Definition 1.5.3.

A line in space is the set of terminal points of vectors emanating from a given point P that are parallel to a fixed vector v.
The fixed vector v in the definition is called a direction vector for the line. As we saw in Preview Activity 1.5.1, to find an equation for a line through point P in the direction of vector v, observe that any vector parallel to v will have the form tv for some scalar t. So, any vector emanating from the point P in a direction parallel to the vector v will be of the form
(1.5.1)OP→+vt
for some scalar t (where O is the origin).
Figure 1.5.4. A line in 2-space.
Figure 1.5.4 presents three images of a line in two-space in which we can identify the vector OPβ†’ and the vector tv as in Equation (1.5.1). Here, OPβ†’ is the fixed vector shown in blue, while the direction vector v is the vector parallel to the vector shown in green (that is, the green vector represents tv, and the line is traced out by the terminal points of the magenta vector). In other words, the tips (terminal points) of the magenta vectors (the vectors of the form OPβ†’+tv) trace out the line as t changes.
In particular, the terminal points of the vectors of the form in (1.5.1) define a linear function r in space of the following form, which is valid in any dimension.

The vector form of a line.

The vector form of a line through the point P in the direction of the vector v is
(1.5.2)r(t)=r0+tv,
where r0 is the position vector OP→ from the origin to the point P.
Of course, it is common to represent lines in the plane using the slope-intercept equation y=mx+b. The vector form of the line, described above, is an alternative way to represent lines that has the following two advantages. First, in two dimensions, we are able to represent vertical lines, whose slope m is not defined, using a vertical direction vector, such as v=⟨0,1⟩. Second, this description of lines works in any dimension even though there is no concept of the slope of a line in more than two dimensions.
Figure 1.5.5. A line in 3-space.

Activity 1.5.2.

Let P1=(1,2,βˆ’1) and P2=(βˆ’2,1,βˆ’2). Let L be the line in R3 through P1 and P2, and note that three snapshots of this line are shown in Figure 1.5.5.
  1. Find a direction vector for the line L.
  2. Find a vector equation of L in the form r(t)=r0+tv.
  3. Consider the vector equation s(t)=βŸ¨βˆ’5,0,βˆ’3⟩+t⟨6,2,2⟩. What is the direction of the line given by s(t)? Is this new line parallel to line L?
  4. Do r(t) and s(t) represent the same line, L? Explain.
Answer.
  1. Note that there are infinitely many choices for a direction vector for the line L. We will choose the direction vector v that points from P1=(1,2,βˆ’1) to P2=(βˆ’2,1,βˆ’2), so v=βŸ¨βˆ’3,βˆ’1,βˆ’1⟩.
  2. We will let r0 be the position vector for P1, so r0=⟨1,2,βˆ’1⟩. Then a vector equation of L is r(t)=r0+tv=⟨1,2,βˆ’1⟩+tβŸ¨βˆ’3,βˆ’1,βˆ’1⟩.
  3. The direction the line given by s(t) is ⟨6,2,2⟩, which is a multiple of βŸ¨βˆ’3,βˆ’1,βˆ’1⟩ (the direction vector for L) so this new line is parallel to L.
  4. We know that r(t) and s(t) represent parallel lines. Two parallel lines will either have no points in common or all points in common, so we want to either show that there is a point on one line that is not on the other, or show that they have a point in common. We know that P1=(1,2,βˆ’1) is a point on the line represented by r(t), so we can check if it is on s(t). That is, we want to see if there is a t value with ⟨1,2,βˆ’1⟩=s(t), or
    ⟨1,2,βˆ’1⟩=βŸ¨βˆ’5,0,βˆ’3⟩+t⟨6,2,2⟩⟨6,2,2⟩=t⟨6,2,2⟩
    which is satisfied by t=1. Then r(t) and s(t) both represent L.

Subsection 1.5.2 The Parametric Equations of a Line

The vector form of a line, r(t)=r0+tv in Equation (1.5.2), describes a line as the set of terminal points of the vectors r(t). If we write this in terms of components letting
r(t)=⟨x(t),y(t),z(t)⟩,   r0=⟨x0,y0,z0⟩,    and    v=⟨a,b,c⟩,
then we can equate the components on both sides of r(t)=r0+tv to obtain the equations
x(t)=x0+at,     y(t)=y0+bt,      and    z(t)=z0+ct,
which describe the coordinates of the points on the line. The variable t represents an arbitrary scalar and is called a parameter. In particular, we use the following language.

The parametric equations of a line.

The parametric equations for a line through the point P=(x0,y0,z0) in the direction of the vector v=⟨a,b,c⟩ are
x(t)=x0+at,     y(t)=y0+bt,     z(t)=z0+ct.
Notice that there are many different parametric equations for the same line. For example, choosing another point P on the line or another direction vector v produces another set of parametric equations. It is sometimes useful to think of t as a time parameter and the parametric equations as telling us where we are on the line at each time. In this way, the parametric equations describe a particular walk taken along the line; there are, of course, many possible ways to walk along a line.

Activity 1.5.3.

Let P1=(1,2,βˆ’1) and P2=(βˆ’2,1,βˆ’2), and let L be the line in R3 through P1 and P2, which is the same line as in Activity 1.5.2.
  1. Find parametric equations of the line L.
  2. Does the point (1,2,1) lie on L? If so, what value of t results in this point?
  3. Consider another line, K, whose parametric equations are
    x(s)=11+4s,  y(s)=1βˆ’3s,  z(s)=3+2s.
    What is the direction of the line K?
  4. Do the lines L and K intersect? If so, provide the point of intersection and the t and s values, respectively, that result in the point. If not, explain why.
Answer.
  1. From Activity 1.5.2 we know that L is represented by r(t)=⟨1,2,βˆ’1⟩+tβŸ¨βˆ’3,βˆ’1,βˆ’1⟩=⟨1βˆ’3t,2βˆ’t,βˆ’1βˆ’t⟩. Then L is parameterized by the equations
    x(t)=1βˆ’3t,  y(t)=2βˆ’t,  z(t)=βˆ’1βˆ’t.
  2. If there is some value of t so that the point (1,2,1) lies on L, then we would need 1=x(t)=1βˆ’3t, so t=0. However, when t=0 we get that z(t)=βˆ’1, not 1, so (1,2,1) is not on L.
  3. The line K points in the direction ⟨4,βˆ’3,2⟩.
  4. If the lines L and K intersect, then there exist some s and t values so that the x, y and z coordinates of L and K are the same. That is, we get the system of equations
    1βˆ’3t=11+4s,  2βˆ’t=1βˆ’3s,  βˆ’1βˆ’t=3+2s.
    Solving the second and third equations for t, we get t=1+3s and t=βˆ’4βˆ’2s so 1+3s=βˆ’4βˆ’2s or s=βˆ’1. Plugging s=βˆ’1 back into t=1+3s we get t=βˆ’2. We can see that these s and t values also satisfy the first equation in our system of equations since 1βˆ’3t=7=11+4s.
    Then the lines L and K intersect when s=βˆ’1 and t=βˆ’2, at the point (7,4,1).

Subsection 1.5.3 Planes in Space

Now that we have a way of describing lines, we would like to develop a means of describing planes in three dimensions. We studied the coordinate planes and planes parallel to them in Section 1.1. Each of those planes had one of the variables x, y, or z equal to a constant. We can note that any vector in a plane with x constant is orthogonal to the vector ⟨1,0,0⟩, any vector in a plane with y constant is orthogonal to the vector ⟨0,1,0⟩, and any vector in a plane with z constant is orthogonal to the vector ⟨0,0,1⟩. This idea works in general to define a plane.

Definition 1.5.6.

A plane p in space is the set of all terminal points of vectors emanating from a given point P0 perpendicular to a fixed vector n, as shown in Figure 1.5.7.
Figure 1.5.7. A point P0 on a plane p with a normal vector n
The definition allows us to find the equation of a plane. Assume that n=⟨a,b,c⟩, P0=(x0,y0,z0), and that P=(x,y,z) is an arbitrary point on the plane. Since the vector P0Pβ†’ lies in the plane, it must be perpendicular to n. This means that
0=(nβ‹…P0Pβ†’=(nβ‹…(⟨x,y,zβŸ©βˆ’βŸ¨x0,y0,z0⟩)=(nβ‹…βŸ¨xβˆ’x0,yβˆ’y0,zβˆ’z0⟩=(a(xβˆ’x0)+b(yβˆ’y0)+c(zβˆ’z0).
The fixed vector n perpendicular to the plane is frequently called a normal vector to the plane. We may now summarize as follows.

Equations of a plane.

  • The scalar equation of the plane with normal vector n=⟨a,b,c⟩ containing the point P0=(x0,y0,z0) is
    (1.5.3)a(xβˆ’x0)+b(yβˆ’y0)+c(zβˆ’z0)=0.
  • The vector equation of the plane with normal vector n=⟨a,b,c⟩ containing the points P0=(x0,y0,z0) and P=(x,y,z) is
    (1.5.4)n⋅P0P→=0.
We may take the scalar equation of a plane a little further and note that since
a(xβˆ’x0)+b(yβˆ’y0)+c(zβˆ’z0)=0,
it equivalently follows that
ax+by+cz=ax0+by0+cz0.
That is, we may write an equation of a plane as ax+by+cz=d where d=nβ‹…βŸ¨x0,y0,z0⟩.
For instance, if we would like to describe the plane passing through the point P0=(4,βˆ’2,1) and perpendicular to the vector n=⟨1,2,1⟩, we have
⟨1,2,1βŸ©β‹…βŸ¨x,y,z⟩=⟨1,2,1βŸ©β‹…βŸ¨4,βˆ’2,1⟩
x+2y+z=1.
Notice that the coefficients of x, y, and z in this description give a vector perpendicular to the plane. For instance, if we are presented with the plane
βˆ’2x+yβˆ’3z=4,
we know that n=βŸ¨βˆ’2,1,βˆ’3⟩ is a vector perpendicular to the plane.

Activity 1.5.4.

  1. Write a scalar equation of the plane p1 passing through the point (0,2,4) and perpendicular to the vector n=⟨2,βˆ’1,1⟩.
  2. Is the point (2,0,2) on the plane p1?
  3. Write a scalar equation of the plane p2 that is parallel to p1 and passing through the point (3,0,4). (Hint: Compare normal vectors of the planes.)
  4. Write a parametric description of the line l passing through the point (2,0,2) and perpendicular to the plane p3 described by the equation x+2yβˆ’2z=7.
  5. Find the point at which l intersects the plane p3.
Answer.
  1. 2(xβˆ’0)βˆ’1(yβˆ’2)+1(zβˆ’4)=0
  2. Plugging the point (2,0,2) into the equation in part 1, we get
    2(2βˆ’0)βˆ’1(0βˆ’2)+1(2βˆ’4)=4β‰ 0
    so it is not oh the plane p1.
  3. For the plane p2 to be parallel to p1, it must have the same normal vector: n=⟨2,βˆ’1,1⟩. We also know that p2 contains the point (3,0,4), so we have all we need for the scalar equation of p2:
    2(xβˆ’3)βˆ’1(yβˆ’0)+1(zβˆ’4)=0.
  4. We can read the normal vector for p3 from the coefficients of x+2yβˆ’2z=7. The vector ⟨1,2,βˆ’2⟩ is normal to p3. We can use this vector and the point (2,0,2) to get the parametric equations
    x(t)=2+t,  y(t)=2t,  z(t)=2βˆ’2t.
  5. We want to first find a t value so that the point (x(t),y(t),z(t)) satisfies x+2yβˆ’2z=7. That is
    x+2yβˆ’2z=7(2+t)+2(2t)βˆ’2(2βˆ’2t)=7βˆ’2+9t=7t=1
    so l intersects the plane p3 when t=1, at the point (3,2,0).
Just as two distinct points in space determine a line, three non-collinear points in space determine a plane. Consider three points P0, P1, and P2 in space, not all lying on the same line as shown in Figure 1.5.8.
Figure 1.5.8. A plane determined by three points P0, P1, and P2
Observe that the vectors P0P1β†’ and P0P2β†’ both lie in the plane p. If we form their cross-product
n=P0P1β†’Γ—P0P2β†’,
we obtain a normal vector to the plane p. Therefore, if P is any other point on p, it then follows that P0P→ will be perpendicular to n, and we have, as before, the equation
(1.5.5)n⋅P0P→=0.

Activity 1.5.5.

Let P0=(1,2,βˆ’1), P1=(1,0,βˆ’1), and P2=(0,1,3) and let p be the plane containing P0, P1, and P2.
  1. Determine the components of the vectors P0P1β†’ and P0P2β†’.
  2. Find a normal vector n to the plane p.
  3. Find a scalar equation of the plane p.
  4. Consider a second plane, q, with scalar equation βˆ’3(xβˆ’1)+4(y+3)+2(zβˆ’5)=0. Find two different points on plane q, as well as a vector m that is normal to q.
  5. The angle between two planes is the acute angle between their respective normal vectors. What is the angle between planes p and q?
Answer.
  1. We get the vectors P0P1β†’=⟨0,βˆ’2,0⟩ and P0P2β†’=βŸ¨βˆ’1,βˆ’1,4⟩.
  2. The vectors P0P1β†’ and P0P2β†’ are (non-parallel) vectors in p, so their cross product will be a normal vector to p.
    n=P0P1β†’Γ—P0P2β†’=βŸ¨βˆ’8βˆ’0,βˆ’(0βˆ’0),0βˆ’2⟩=βŸ¨βˆ’8,0,βˆ’2⟩.
  3. The plane p contains P0=(1,2,βˆ’1) and is normal to n=βŸ¨βˆ’8,0,βˆ’2⟩ so a scalar equation is
    βˆ’8(xβˆ’1)+0(yβˆ’2)βˆ’2(z+1)=0
    or
    βˆ’8xβˆ’2z+6=0.
  4. From the statement of the scalar equation we know that the point Q0=(1,βˆ’3,5) is on q and the vector m=βŸ¨βˆ’3,4,2⟩ is normal to q. There are infinitely many other points on q, we can pick x=y=0 to get βˆ’3(0βˆ’1)+4(0+3)+2(zβˆ’5)=0 which solves to z=βˆ’2.5, and the point Q1=(0,0,βˆ’2.5) is on q.
  5. The angle between planes p and q is the angle between the vectors n=βŸ¨βˆ’8,0,βˆ’2⟩ and m=βŸ¨βˆ’3,4,2⟩, which can be found using the trig formula for dot product or the magnitude of the cross product. Using the dot product we get
    ΞΈ=cosβˆ’1⁑(nβ‹…m|n||m|)=cosβˆ’1⁑(24+0βˆ’464+49+16+4)β‰ˆ1.1036
    radians.

Subsection 1.5.4 Summary

  • While lines in R3 do not have a slope, like lines in R2 they can be characterized by a point and a direction vector. Indeed, we define a line in space to be the set of terminal points of vectors emanating from a given point that are parallel to a fixed vector.
  • Vectors play a critical role in representing the equation of a line. In particular, the terminal points of the vector r(t)=r0+tv define a linear function r in space through the terminal point of the vector r0 in the direction of the vector v, tracing out a line in space.
  • A plane in space is the set of all terminal points of vectors emanating from a given point perpendicular to a fixed vector.
  • If P1, P2, and P3 are non-collinear points in space, the vectors P1P2β†’ and and P1P3β†’ are vectors in the plane and the vector n=P1P2β†’Γ—P1P3β†’ is a normal vector to the plane. So any point P in the plane satisfies the equation PP1β†’β‹…n=0. If we let P=(x,y,z), n=⟨a,b,c⟩ be the normal vector, and P1=(x0,y0,z0), we can also represent the plane with the equation
    a(xβˆ’x0)+b(yβˆ’y0)+c(zβˆ’z0)=0.

Exercises 1.5.5 Exercises

1.

Rewrite the vector equation r(t)=(3+3t)i+(βˆ’5βˆ’3t)j+(βˆ’1+3t)k as the corresponding parametric equations for the line.
x(t)=
y(t)=
z(t)=

2.

Find the vector and parametric equations for the line through the point P(4, 2, 4) and parallel to the vector 0iβˆ’3jβˆ’3k.
Vector Form: r=⟨, , 4 ⟩+t⟨ , , -3 ⟩
Parametric form (parameter t, and passing through P when t = 0):
x=x(t)=
y=y(t)=
z=z(t)=

3.

Consider the line which passes through the point P(5, -2, -2), and which is parallel to the line x=1+2t,y=2+2t,z=3+6t
Find the point of intersection of this new line with each of the coordinate planes:
xy-plane: (,, )
xz-plane: (,, )
yz-plane: (,, )

4.

Find the point at which the line βŸ¨βˆ’5,5,2⟩+t⟨5,βˆ’3,βˆ’3⟩ intersects the plane βˆ’2x+4y+z=βˆ’18.
(, , )

5.

Find an equation of a plane containing the three points (-3, 2, -4), (-1, -2, -1), (-1, -1, 1) in which the coefficient of x is -11.

6.

Find an equation for the plane containing the line in the xy-plane where y=2, and the line in the xz-plane where z=3.
equation:

7.

Find the angle in radians between the planes 2x+z=1 and βˆ’3y+z=1.

8.

A store sells CDs at one price and DVDs at another price. The figure below shows the revenue (in dollars) of the music store as a function of the number, c, of CDs and the number, d, of DVDs that it sells. The values of the revenue are shown on each line.
(Hint: for this problem there are many possible ways to estimate the requisite values; you should be able to find information from the figure that allows you to give an answer that is essentially exact.)
(a) What is the price of a CD? dollars
(b) What is the price of a DVD? dollars

9.

The table below gives the number of calories burned per minute for someone roller-blading, as a function of the person’s weight in pounds and speed in miles per hour [from the August 28,1994, issue of Parade Magazine].
calories burned per minute
weightβˆ–speed 8 9 10 11
120 4.2 5.8 7.4 8.9
140 5.1 6.7 8.3 9.9
160 6.1 7.7 9.2 10.8
180 7 8.6 10.2 11.7
200 7.9 9.5 11.1 12.6
(a) Suppose that a 140 lb person and a 200 person both go 10 miles, the first at 10 mph and the second at 9 mph.
How many calories does the 140 lb person burn?
How many calories does the 200 lb person burn?
(b) We might also be interested in the number of calories each person burns per pound of their weight.
How many calories per pound does the 140 lb person burn?
How many calories per pound does the 200 lb person burn?

10.

The vector and parametric forms of a line allow us to easily describe line segments in space.
Let P1=(1,2,βˆ’1) and P2=(βˆ’2,1,βˆ’2), and let L be the line in R3 through P1 and P2 as in Activity 1.5.2.
  1. What value of the parameter t makes (x(t),y(t),z(t))=P1? What value of t makes (x(t),y(t),z(t))=P2?
  2. What t values describe the line segment between the points P1 and P2?
  3. What about the line segment (along the same line) from (7,4,1) to (βˆ’8,βˆ’1,βˆ’4)?
  4. Now, consider a segment that lies on a different line: parameterize the segment that connects point R=(4,βˆ’2,7) to Q=(βˆ’11,4,27) in such a way that t=0 corresponds to point Q, while t=2 corresponds to R.

11.

This exercise explores key relationships between a pair of lines. Consider the following two lines: one with parametric equations x(s)=4βˆ’2s, y(s)=βˆ’2+s, z(s)=1+3s, and the other being the line through (βˆ’4,2,17) in the direction v=βŸ¨βˆ’2,1,5⟩.
  1. Find a direction vector for the first line, which is given in parametric form.
  2. Find parametric equations for the second line, written in terms of the parameter t.
  3. Show that the two lines intersect at a single point by finding the values of s and t that result in the same point. Then find the point of intersection.
  4. Find the acute angle formed where the two lines intersect, noting that this angle will be given by the acute angle between their respective direction vectors.
  5. Find an equation for the plane that contains both of the lines described in this problem.

12.

This exercise explores key relationships between a pair of planes. Consider the following two planes: one with scalar equation 4xβˆ’5y+z=βˆ’2, and the other which passes through the points (1,1,1), (0,1,βˆ’1), and (4,2,βˆ’1).
  1. Find a vector normal to the first plane.
  2. Find a scalar equation for the second plane.
  3. Find the angle between the planes, where the angle between them is defined by the angle between their respective normal vectors.
  4. Find a point that lies on both planes.
  5. Since these two planes do not have parallel normal vectors, the planes must intersect, and thus must intersect in a line. Observe that the line of intersection lies in both planes, and thus the direction vector of the line must be perpendicular to each of the respective normal vectors of the two planes. Find a direction vector for the line of intersection for the two planes.
  6. Determine parametric equations for the line of intersection of the two planes.

13.

In this problem, we explore how we can use what we know about vectors and projections to find the distance from a point to a plane.
Let p be the plane with equation z=βˆ’4x+3y+4, and let Q=(4,βˆ’1,8).
  1. Show that Q does not lie in the plane p.
  2. Find a normal vector n to the plane p.
  3. Find the coordinates of a point P in p.
  4. Find the components of PQ→. Draw a picture to illustrate the objects found so far.
  5. Explain why |compnPQ→| gives the distance from the point Q to the plane p. Find this distance.