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8.7 Planes

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DEFINITION 8.7.1 A plane is a set of points P satisfying an equation

of the form

P = P0 + sX + tY; s; t 2 R; (8.19)

where X and Y are non{zero, non{parallel vectors.

For example, the xy{plane consists of the points P = (x; y; 0) and corre-

sponds to the plane equation

P = xi + yj = O + xi + yj:

In terms of coordinates, equation 8.19 takes the form

x = x0 + sa1 + ta2

y = y0 + sb1 + tb2

z = z0 + sc1 + tc2;

where P0 = (x0; y0; z0) and (a1; b1; c1) and (a2; b2; c12 are non{zero and

non{proportional.

THEOREM 8.7.1 Let A; B; C be three non{collinear points. Then there

is one and only one plane through these points, namely the plane given by

the equation

P = A + s

AB +t

AC; (8.20)

or equivalently -

AP= s

AB +t

AC : (8.21)

(See Figure 8.15.)

8.7. PLANES 177

¢®

O»»»»»»»»»»»»:

©©©©*

P - PPPPPq

PPPPPP

©©©©

AB0= s

AB;

AC0= t

AP= s

AB +t

Figure 8.15: Vector equation for the plane ABC.

Proof. First note that equation 8.20 is indeed the equation of a plane

through A; B and C, as

AB and

AC are non{zero and non{parallel and

(s; t) = (0; 0); (1; 0) and (0; 1) give P = A; B and C, respectively. Call

this plane P.

Conversely, suppose P = P0 + sX + tY is the equation of a plane Q

passing through A; B; C. Then A = P0 + s0X + t0Y , so the equation for

Q may be written

P = A + (s ¡ s0)X + (t ¡ t0)Y = A + s0X + t0Y ;

so in e®ect we can take P0 = A in the equation of Q. Then the fact that B

and C lie on Q gives equations

B = A + s1X + t1Y; C = A + s2X + t2Y;

or -

AB= s1X + t1Y;

AC= s2X + t2Y: (8.22)

Then equations 8.22 and equation 8.20 show that

P µ Q:

Conversely, it is straightforward to show that because

AB and

AC are not

parallel, we have ¯¯¯¯

s1 t1

s2 t2

¯¯¯¯

6= 0:

178 CHAPTER 8. THREE{DIMENSIONAL GEOMETRY

¢®

¤º

¡µ

HHHHHHj

¡@

C P

AD=

AB £

AD ¢

AP= 0

Figure 8.16: Normal equation of the plane ABC.

Hence equations 8.22 can be solved for X and Y as linear combinations of -

AB and

AC, allowing us to deduce that

Q µ P:

Hence

Q = P:

THEOREM 8.7.2 (Normal equation for a plane) Let

A = (x1; y1; z1); B = (x2; y2; z2); C = (x3; y3; z3)

be three non{collinear points. Then the plane through A; B; C is given by

AP ¢(

AB £

AC) = 0; (8.23)

or equivalently, ¯¯¯¯¯¯

x ¡ x1 y ¡ y1 z ¡ z1

x2 ¡ x1 y2 ¡ y1 z2 ¡ z1

x3 ¡ x1 y3 ¡ y1 z3 ¡ z1

¯¯¯¯¯¯

= 0; (8.24)

where P = (x; y; z). (See Figure 8.16.)

8.7. PLANES 179

¢®

»»»»»»»»

AK ai + bj + ck

ax + by + cz = d

Figure 8.17: The plane ax + by + cz = d.

REMARK 8.7.1 Equation 8.24 can be written in more symmetrical form

as ¯¯¯¯¯¯¯¯

x y z 1

x1 y1 z1 1

x2 y2 z2 1

x3 y3 z3 1

¯¯¯¯¯¯¯¯

= 0: (8.25)

Proof. Let P be the plane through A; B; C. Then by equation 8.21, we

have P 2 P if and only if

AP is a linear combination of

AB and

AC and so

by lemma 8.6.1(i), using the fact that

AB £

AC6= 0 here, if and only if

is perpendicular to

AB £

AC. This gives equation 8.23.

Equation 8.24 is the scalar triple product version of equation 8.23, taking

into account the equations

AP = (x ¡ x1)i + (y ¡ y1)j + (z ¡ z1)k;

AB = (x2 ¡ x1)i + (y2 ¡ y1)j + (z2 ¡ z1)k;

AC = (x3 ¡ x1)i + (y3 ¡ y1)j + (z3 ¡ z1)k:

REMARK 8.7.2 Equation 8.24 gives rise to a linear equation in x; y and

ax + by + cz = d;

180 CHAPTER 8. THREE{DIMENSIONAL GEOMETRY

where ai + bj + ck 6= 0. For

¯¯¯¯¯¯

x ¡ x1 y ¡ y1 z ¡ z1

x2 ¡ x1 y2 ¡ y1 z2 ¡ z1

x3 ¡ x1 y3 ¡ y1 z3 ¡ z1

¯¯¯¯¯¯

x y z

x2 ¡ x1 y2 ¡ y1 z2 ¡ z1

x3 ¡ x1 y3 ¡ y1 z3 ¡ z1

¯¯¯¯¯¯

x1 y1 z1

x2 ¡ x1 y2 ¡ y1 z2 ¡ z1

x3 ¡ x1 y3 ¡ y1 z3 ¡ z1

¯¯¯¯¯¯

(8.26)

and expanding the ¯rst determinant on the right{hand side of equation 8.26

along row 1 gives an expression

ax + by + cz

where

a =

¯¯¯¯

y2 ¡ y1 z2 ¡ z1

y3 ¡ y1 z3 ¡ z1

¯¯¯¯

; b = ¡

¯¯¯¯

x2 ¡ x1 z2 ¡ z1

x3 ¡ x1 z3 ¡ z1

¯¯¯¯

; c =

¯¯¯¯

x2 ¡ x1 y2 ¡ y1

x3 ¡ x1 y3 ¡ y1

¯¯¯¯

But a; b; c are the components of

AB £

AC, which in turn is non{zero, as

A; B; C are non{collinear here.

Conversely if ai + bj + ck 6= 0, the equation

ax + by + cz = d

does indeed represent a plane. For if say a 6= 0, the equation can be solved

for x in terms of y and z:

5 =

4 ¡d

5 + y

4 ¡b

5 + z

4 ¡c

5 ;

which gives the plane

P = P0 + yX + zY;

where P0 = (¡d

a ; 0; 0) and X = ¡b

a i + j and Y = ¡c

a i + k are evidently

non{parallel vectors.

REMARK 8.7.3 The plane equation ax+by+cz = d is called the normal

form, as it is easy to prove that if P1 and P2 are two points in the plane,

then ai + bj + ck is perpendicular to

P1P2. Any non{zero vector with this

property is called a normal to the plane. (See Figure 8.17.)

8.7. PLANES 181

By lemma 8.6.1(ii), it follows that every vector X normal to a plane

through three non{collinear points A; B; C is parallel to

AB £

AC, since

X is perpendicular to

AB and

AC.

EXAMPLE 8.7.1 Show that the planes

x + y ¡ 2z = 1 and x + 3y ¡ z = 4

intersect in a line and ¯nd the distance from the point C = (1; 0; 1) to this

line.

Solution. Solving the two equations simultaneously gives

x = ¡

z; y =

2 ¡

z; (8.27)

where z is arbitrary. Hence

xi + yj + zk = ¡

i ¡

j + z(

i ¡

j + k);

which is the equation of a line L through A = (¡1

2 ; ¡3

2 ; 0) and having

direction vector 5

2 i ¡ 1

2 j + k.

We can now proceed in one of three ways to ¯nd the closest point on L

to A.

One way is to use equation 8.17 with B de¯ned by

AB=

i ¡

j + k:

Another method minimizes the distance CP, where P ranges over L.

A third way is to ¯nd an equation for the plane through C, having

2 i ¡ 1

2 j + k as a normal. Such a plane has equation

5x ¡ y + 2z = d;

where d is found by substituting the coordinates of C in the last equation.

d = 5 £ 1 ¡ 0 + 2 £ 1 = 7:

We now ¯nd the point P where the plane intersects the line L. Then

will be perpendicular to L and CP will be the required shortest distance

from C to L. We ¯nd using equations 8.27 that

5(¡

z) ¡ (

2 ¡

z) + 2z = 7;

182 CHAPTER 8. THREE{DIMENSIONAL GEOMETRY

¢®

»»»»»»»»

a1i + b1j + c1k

a2i + b2j + c2k

a1x + b1y + c1z = d1 a2x + b2y + c2z = d2

XXXXXXXX

¢ ¢

¢¸

Figure 8.18: Line of intersection of two planes.

so z = 11

15 . Hence P = ( 4

3 ; 17

15 ; 11

15 ).

It is clear that through a given line and a point not on that line, there

passes exactly one plane. If the line is given as the intersection of two planes,

each in normal form, there is a simple way of ¯nding an equation for this

plane. More explicitly we have the following result:

THEOREM 8.7.3 Suppose the planes

a1x + b1y + c1z = d1 (8.28)

a2x + b2y + c2z = d2 (8.29)

have non{parallel normals. Then the planes intersect in a line L.

Moreover the equation

¸(a1x + b1y + c1z ¡ d1) + ¹(a2x + b2y + c2z ¡ d2) = 0; (8.30)

where ¸ and ¹ are not both zero, gives all planes through L.

(See Figure 8.18.)

Proof. Assume that the normals a1i + b1j + c1k and a2i + b2j + c2k are

non{parallel. Then by theorem 8.4.3, not all of

¢1 =

¯¯¯¯

a1 b1

a2 b2

¯¯¯¯

; ¢2 =

¯¯¯¯

b1 c1

b2 c2

¯¯¯¯

; ¢3 =

¯¯¯¯

a1 c1

a2 c2

¯¯¯¯

(8.31)

8.7. PLANES 183

are zero. If say ¢1 6= 0, we can solve equations 8.28 and 8.29 for x and y in

terms of z, as we did in the previous example, to show that the intersection

forms a line L.

We next have to check that if ¸ and ¹ are not both zero, then equa-

tion 8.30 represents a plane. (Whatever set of points equation 8.30 repre-

sents, this set certainly contains L.)

(¸a1 + ¹a2)x + (¸b1 + ¹b2)y + (¸c1 + ¹c2)z ¡ (¸d1 + ¹d2) = 0:

Then we clearly cannot have all the coe±cients

¸a1 + ¹a2; ¸b1 + ¹b2; ¸c1 + ¹c2

zero, as otherwise the vectors a1i + b1j + c1k and a2i + b2j + c2k would be

parallel.

Finally, if P is a plane containing L, let P0 = (x0; y0; z0) be a point not

on L. Then if we de¯ne ¸ and ¹ by

¸ = ¡(a2x0 + b2y0 + c2z0 ¡ d2); ¹ = a1x0 + b1y0 + c1z0 ¡ d1;

then at least one of ¸ and ¹ is non{zero. Then the coordinates of P0 satisfy

equation 8.30, which therefore represents a plane passing through L and P0

and hence identical with P.

EXAMPLE 8.7.2 Find an equation for the plane through P0 = (1; 0; 1)

and passing through the line of intersection of the planes

x + y ¡ 2z = 1 and x + 3y ¡ z = 4:

Solution. The required plane has the form

¸(x + y ¡ 2z ¡ 1) + ¹(x + 3y ¡ z ¡ 4) = 0;

where not both of ¸ and ¹ are zero. Substituting the coordinates of P0 into

this equation gives

¡2¸ + ¹(¡4) = 0; ¸ = ¡2¹:

So the required equation is

¡2¹(x + y ¡ 2z ¡ 1) + ¹(x + 3y ¡ z ¡ 4) = 0;

¡x + y + 3z ¡ 2 = 0:

Our ¯nal result is a formula for the distance from a point to a plane.

184 CHAPTER 8. THREE{DIMENSIONAL GEOMETRY

¢®

»»»»»»»»

ai + bj + ck

ax + by + cz = d

Figure 8.19: Distance from a point P0 to the plane ax + by + cz = d.

THEOREM 8.7.4 (Distance from a point to a plane)

Let P0 = (x0; y0; z0) and P be the plane

ax + by + cz = d: (8.32)

Then there is a unique point P on P such that

P0P is normal to P. Morever

P0P = jax0 + by0 + cz0 ¡ d p j a2 + b2 + c2

(See Figure 8.19.)

Proof. The line through P0 normal to P is given by

P = P0 + t(ai + bj + ck);

or in terms of coordinates

x = x0 + at; y = y0 + bt; z = z0 + ct:

Substituting these formulae in equation 8.32 gives

a(x0 + at) + b(y0 + bt) + c(z0 + ct) = d

t(a2 + b2 + c2) = ¡(ax0 + by0 + cz0 ¡ d);

t = ¡

ax0 + by0 + cz0 ¡ d

a2 + b2 + c2

8.8. PROBLEMS 185

Then

P0P = jj

P0P jj = jjt(ai + bj + ck)jj

= jtj

a2 + b2 + c2

= jax0 + by0 + cz0 ¡ dj

a2 + b2 + c2

= jax0 + by0 + cz0 ¡ d p j a2 + b2 + c2

Other interesting geometrical facts about lines and planes are left to the

problems at the end of this chapter.

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