In this section, we use vectors to derive analytic descriptions of planes inR3. We also show how to solve a variety of distance problems involving “flat objects”
(i.e., points, lines, and planes).
Coordinate Equations of Planes
A planeinR3is determined uniquely by the following geometric information:
a particular point P0(x0,y0,z0) in the plane and a particular vector n= Ai+ Bj+Ckthat is normal (perpendicular) to the plane. In other words, is the set of all points P(x,y,z) in space such that−−→
P0P is perpendicular to n. (See Figure 1.70.) This means thatis defined by the vector equation
z
y x
P
n P0 Π
Figure 1.70 The plane inR3 through the pointP0and perpendicular to the vectorn.
n·−−→
P0P=0. (1)
Since−−→
P0P=(x−x0)i+(y−y0)j+(z−z0)k, equation (1) may be rewritten as
(Ai+Bj+Ck)·((x−x0)i+(y−y0)j+(z−z0)k)=0 or
A(x−x0)+B(y−y0)+C(z−z0)=0. (2)
This is equivalent to
Ax+By+C z= D, whereD= Ax0+By0+C z0.
EXAMPLE 1 The plane through the point (3, 2, 1) with normal vector 2i−j+4k has equation
(2i−j+4k)·((x−3)i+(y−2)j+(z−1)k)=0
⇐⇒2(x−3)−(y−2)+4(z−1)=0
⇐⇒2x−y+4z=8. ◆ Not only does a plane inR3have an equation of the form given by equation (2), but, conversely, any equation of this form must describe a plane. Moreover, it is easy to read off the components of a vector normal to the plane from such an equation: They are just the coefficients ofx,y, andz.
EXAMPLE 2 Given the plane with equation 7x+2y−3z=1, find a normal vector to the plane and identify three points that lie on that plane.
A possible normal vector isn=7i+2j−3k. However, any nonzero scalar multiple ofnwill do just as well. Algebraically, the effect of using a scalar multiple ofnas normal is to multiply equation (2) by such a scalar.
Finding three points in the plane is not difficult. First, lety=z=0 in the defining equation and solve forx:
7x+2·0−3·0=1 ⇐⇒ 7x =1 ⇐⇒ x= 17. Thus1
7,0,0
is a point on the plane. Next, letx =z=0 and solve fory:
7·0+2y−3·0=1 ⇐⇒ y = 12. So
0,12,0
is another point on the plane. Finally, letx = y=0 and solve forz.
You should find that
0,0,−13
lies on the plane. ◆
EXAMPLE 3 Put coordinate axes on R3 so that the z-axis points vertically.
Then a plane inR3 is vertical if its normal vector nis horizontal (i.e., if nis parallel to the x y-plane). This means thatnhas no k-component, so ncan be written in the formAi+Bj. It follows from equation (2) that a vertical plane has an equation of the form
A(x−x0)+B(y−y0)=0. Hence, a nonvertical plane has an equation of the form
A(x−x0)+B(y−y0)+C(z−z0)=0,
whereC =0. ◆
EXAMPLE 4 From high school geometry, you may recall that a plane is determined by three (noncollinear) points. Let’s find an equation of the plane that contains the pointsP0(1,2,0),P1(3,1,2), andP2(0,1,1).
There are two ways to solve this problem. The first approach is algebraic and rather uninspired. From the aforementioned remarks, any plane must have an equation of the formAx+By+C z= Dfor suitable constantsA,B,C, and D. Thus, we need only to substitute the coordinates of P0, P1, and P2 into this equation and solve forA,B,C, andD. We have that
• substitution ofP0gives A+2B= D;
• substitution ofP1gives 3A+B+2C = D; and
• substitution ofP2gives B+C = D.
Hence, we must solve a system of three equations in four unknowns:
⎧⎪
⎨
⎪⎩
A+2B = D
3A+ B+2C = D B+ C = D
. (3)
In general, such a system has either no solution or else infinitely many solutions.
We must be in the latter case, since we know that the three pointsP0, P1, and P2 lie on some plane (i.e., that some set of constants A, B,C, and D must exist).
Furthermore, the existence of infinitely many solutions corresponds to the fact that any particular equation for a plane may be multiplied by a nonzero constant without altering the plane defined. In other words, we can choose a value for one ofA,B,C, orD, and then the other values will be determined. So let’s multiply the first equation given in (3) by 3, and subtract it from the second equation. We
obtain ⎧
⎪⎨
⎪⎩
A+ 2B = D
−5B+2C = −2D
B+ C = D
. (4)
Now, multiply the third equation in (4) by 5 and add it to the second:
⎧⎪
⎨
⎪⎩
A+2B = D
7C =3D B+ C = D
. (5)
Multiply the third equation appearing in (5) by 2 and subtract it from the first:
⎧⎪
⎨
⎪⎩
A −2C = −D 7C = 3D
B+ C = D
. (6)
By adding appropriate multiples of the second equation to both the first and third equations of (6), we find that
⎧⎪
⎨
⎪⎩
A = −17D
7C = 3D
B = 47D
. (7)
Thus, if in (7) we takeD = −7 (for example), then A=1, B= −4,C = −3, and the equation of the desired plane is
x−4y−3z= −7. z
y
x n
P1 P2 P0
Figure 1.71 The plane determined by the pointsP0,P1, andP2in Example 4.
The second method of solution is cleaner and more geometric. The idea is to make use of equation (1). Therefore, we need to know the coordinates of a particular point on the plane (no problem—we are given three such points) and a vectornnormal to the plane. The vectors−−→
P0P1and−−→
P0P2both lie in the plane.
(See Figure 1.71.) In particular, the normal vectorn must be perpendicular to them both. Consequently, the cross product provides just what we need. That is, we may take
n= −−→P0P1× −−→P0P2 =(2i−j+2k)×(−i−j+k)
=
i j k
2 −1 2
−1 −1 1
=i−4j−3k.
If we takeP0(1,2,0) to be the particular point in equation (1), we find that the equation we desire is
(i−4j−3k)·((x−1)i+(y−2)j+zk)=0 or
(x−1)−4(y−2)−3z =0.
This is the same equation as the one given by the first method. ◆
z
x y
x − 2y + z = 4
2x + y + 3z = −7
Figure 1.72 The line of intersection of the planes x−2y+z=4 and
2x+y+3z= −7 in Example 5.
EXAMPLE 5 Consider the two planes having equations x−2y+z=4 and 2x+y+3z= −7. We determine a set of parametric equations for their line of intersection. (See Figure 1.72.) We use Proposition 2.1. Thus, we need to find a point on the line and a vector parallel to the line. To find the point on the line, we note that the coordinates (x,y,z) of any such point must satisfy the system of simultaneous equations given by the two planes
x−2y+z= 4
2x+y+3z= −7 . (8)
From the equations given in (8), it is not too difficult to produce a single solution (x,y,z). For example, if we let z=0 in (8), we obtain the simpler system
x−2y= 4
2x+y= −7 . (9)
The solution to the system of equations (9) is readily calculated to bex= −2, y= −3. Thus, (−2,−3,0) are the coordinates of a point on the line.
To find a vector parallel to the line of intersection, note that such a vector must be perpendicular to the two normal vectors to the planes. The normal vectors to the planes arei−2j+kand 2i+j+3k. Therefore, a vector parallel to the line of intersection is given by
(i−2j+k)×(2i+j+3k)= −7i−j+5k.
Hence, Proposition 2.1 implies that a vector parametric equation for the line is r(t)=(−2i−3j)+t(−7i−j+5k),
and a standard set of parametric equations is
⎧⎪
⎨
⎪⎩
x = −7t−2 y = −t −3 z =5t
.
◆
Parametric Equations of Planes
Another way to describe a plane inR3 is by a set of parametric equations. First, suppose that a=(a1,a2,a3) andb=(b1,b2,b3) are two nonzero, nonparallel vectors in R3. Then a andbdetermine a plane in R3 that passes through the origin. (See Figure 1.73.) To find the coordinates of a point P(x,y,z) in this plane, draw a parallelogram whose sides are parallel toaandband that has two opposite vertices at the origin and atP, as shown in Figure 1.74. Then there must exist scalarssandt so that the position vector ofPis−→
O P=sa+tb. The plane
z
y
x
a b
Figure 1.73 The plane through the origin determined by the vectorsaandb.
z
y x
a b tb
sa P
Figure 1.74 For the pointPin the plane shown,−→
O P =sa+tb for appropriate scalarssandt.
may be described as
x∈R3 |x=sa+tb;s,t ∈R .
Now, suppose that we seek to describe a general plane(i.e., one that does not necessarily pass through the origin). Let
c=(c1,c2,c3)= −−→O P0
denote the position vector of a particular point P0 inand letaandbbe two (nonzero, nonparallel) vectors that determine the plane through the origin parallel to. By parallel translatingaandbso that their tails are at the head ofc(as in Figure 1.75), we adapt the preceding discussion to see that the position vector of any point P(x,y,z) inmay be described as
−→O P=sa+tb+c. z
y x
c b
a P
P0 Π
Figure 1.75 The plane passing throughP0(c1,c2,c3) and parallel toaandb.
To summarize, we have shown the following:
PROPOSITION 5.1 A vector parametric equation for the plane containing the point P0(c1,c2,c3) (whose position vector is−−→
O P0=c) and parallel to the nonzero, nonparallel vectorsaandbis
x(s,t)=sa+tb+c. (10) By taking components in formula (10), we readily obtain a set of parametric equations for:
⎧⎪
⎨
⎪⎩
x =sa1+tb1+c1 y=sa2+tb2+c2 z =sa3+tb3+c3
. (11)
Compare formula (10) with that of equation (1) in Proposition 2.1. We need to use two parameterss andt to describe a plane (instead of a single parameter t that appears in the vector parametric equation for a line) because a plane is a two-dimensional object.
EXAMPLE 6 We find a set of parametric equations for the plane that passes through the point (1,0,−1) and is parallel to the vectors 3i−kand 2i+5j+2k.
From formula (10), any point on the plane is specified by x(s,t)=s(3i−k)+t(2i+5j+2k)+(i−k)
=(3s+2t +1)i+5tj+(2t−s−1)k. The individual parametric equation may be read off as
⎧⎪
⎨
⎪⎩
x =3s+2t +1 y=5t
z =2t −s−1 .
◆
Distance Problems
Cross products and vector projections provide convenient ways to understand a range of distance problems involving lines and planes: Several examples follow.
What is important about these examples are the vector techniques for solving geometric problems that they exhibit, not the general formulas that may be derived from them.
a B
P0
BP0 − proja BP0
Figure 1.76 A general configuration for finding the distance between a point and a line, using vector projections.
EXAMPLE 7 (Distance between a point and a line) We find the distance between the pointP0(2,1,3) and the linel(t)=t(−1,1,−2)+(2,3,−2) in two ways.
METHOD 1. From the vector parametric equations for the given line, we read off a point B on the line—namely, (2,3,−2)—and a vector aparallel to the line—namely, a=(−1,1,−2). Using Figure 1.76, the length of the vector
−−→B P0−proja−−→
B P0 provides the desired distance between P0 and the line. Thus, we calculate that
−−→B P0 =(2,1,3)−(2,3,−2)
=(0,−2,5);
proja−−→
B P0 =
a·−−→
B P0 a·a
a
= (−1,1,−2)·(0,−2,5) (−1,1,−2)·(−1,1,−2)
(−1,1,−2)
=(2,−2,4). The desired distance is
−−→B P0−proja−−→
B P0 = (0,−2,5)−(2,−2,4) = (−2,0,1) =√ 5.
a B
P0
D
θ
Figure 1.77 Another general configuration for finding the distance between a point and a line.
METHOD 2. In this case, we use a little trigonometry. If θ denotes the angle between the vectorsaand−−→
B P0as in Figure 1.77, then sinθ = D
−−→B P0,
whereDdenotes the distance betweenP0and the line. Hence, D = −−→B P0sinθ = a −−→B P0sinθ
a = a× −−→B P0 a .
Therefore, we calculate a× −−→B P0 =
i j k
−1 1 −2
0 −2 5
=i+5j+2k, so that the distance sought is
D= i+5j+2k
−i+j−2k =
√30
√6 =√ 5,
which agrees with the answer obtained by Method 1. ◆
n P1 D
P2 Π2
Π1
Figure 1.78 The general configuration for finding the distanceDbetween two parallel planes.
EXAMPLE 8 (Distance between parallel planes) The planes 1: 2x−2y+z=5 and 2: 2x−2y+z=20 are parallel. (Why?) We see how to compute the distance between them.
Using Figure 1.78 as a guide, we see that the desired distanceDis given by projn−−→
P1P2, whereP1 is a point on1, P2 is a point on2, andnis a vector normal to both planes.
First, the vectornthat is normal to both planes may be read directly from the equation for either1or2asn=2i−2j+k. It is not hard to find a point P1on1: the point P1(0,0,5) will do. Similarly, takeP2(0,0,20) for a point on 2. Then
−−→P1P2 =(0,0,15), and calculate
projn−−→
P1P2 =
n·−−→
P1P2
n·n
n= (2,−2,1)·(0,0,15) (2,−2,1)·(2,−2,1)
(2,−2,1)
= −159(2,−2,1)
= −53(2,−2,1). Hence, the distanceDthat we seek is
D= projn−−→
P1P2 = 53√
9=5. ◆
EXAMPLE 9 (Distance between two skew lines) Find the distance between the two skew lines
l1(t)=t(2,1,3)+(0,5,−1) and l2(t)=t(1,−1,0)+(−1,2,0). (Two lines inR3are said to beskewif they are neither intersecting nor parallel.
It follows that the lines must lie in parallel planes and that the distance between the lines is equal to the distance between the planes.)
B1
B2
a1
a2 l2
l1
Figure 1.79 Configuration for determining the distance between two skew lines in Example 9.
To solve this problem, we need to findprojn−−→
B1B2, the length of the projec- tion of the vector between a point on each line onto a vectornthat is perpendicular to both lines, hence, also perpendicular to the parallel planes that contain the lines.
(See Figure 1.79.)
From the vector parametric equations for the lines, we read that the point B1(0,5,−1) is on the first line andB2(−1,2,0) is on the second. Hence,
−−→B1B2 =(−1,2,0)−(0,5,−1)=(−1,−3,1).
For a vectornthat is perpendicular to both lines, we may usen=a1×a2, where a1=(2,1,3) is a vector parallel to the first line anda2=(1,−1,0) is parallel to the second. (We may read these vectors from the parametric equations.) Thus,
n=a1×a2=
i j k
2 1 3
1 −1 0
=3i+3j−3k, and so,
projn−−→
B1B2 =
n·−−→
B1B2
n·n
n= (−1,−3,1)·(3,3,−3) (3,3,−3)·(3,3,−3)
(3,3,−3)
= −1527(3,3,−3)
= −53(1,1,−1). The desired distance isprojn−−→
B1B2 = 53√
3. ◆