Section 2.4: Equations of Lines and PlanesAn equation of three variable F (x, y, z) = 0 is called an equation of a

surface S if

(x1, y1, z1) 2 S if and only if F (x1, y1, z1) = 0.

For instance,x2 + y2 + z2 = 1

is the equation of the unit sphere centered at the origin. The graph of asystem of two equations

F (x, y, z) = 0, G (x, y, z) = 0

represents the intersection of two surfaces represented by F (x, y, z) = 0 andby G (x, y, z) = 0, respectively, and is usually a curve.

A) Lines in R3.A line l is determined by two elements: one point P0 on the line l and

a direction ~v of l,i.e., any vector that is parallel to l. The goal here is todescribe the line using algebra so that one is able to digitize it. Suppose thatthe coordinate of the point P0 on the line and a direction ~v are given as:

P0 (x0, y0, z0) is a given point on l

~v = ha, b, ci is parallel to l.

Consider any point P (x, y, z) in the space. Let¡¡!P0P be the vector connecting

P0 and P. If P is located exactly on the line, then¡¡!P0P is parallel to the line

l,and thus it is parallel to ~v. On the other hand, if P is o¤ the line, then,since P0 is on the line,

¡¡!P0P cannot possibly be parallel to the line. Therefore,¡¡!

P0P cannot possibly be parallel to ~v. We just concluded that

P is on l if and only if¡¡!P0P is parallel to ~v .

Now

¡¡!P0P = hx, y, zi ¡ hx0, y0, z0i = hx ¡ x0, y ¡ y0, z ¡ z0i

~v = ha, b, ci .

1

P(x,y,z)

l

V

z

O

x

y

Po

So ¡¡!P0P // ~v () ¡¡!

P0P = t~v (for a constant t)

which is equivalent to

x ¡ x0a

=y ¡ y0

b=

z ¡ z0c

.

We called these three equation symmetric form of the system of equationsfor line l.

If we setx ¡ x0

a=

y ¡ y0b

=z ¡ z0

c= t,

which is equivalent to

x ¡ x0a

= t

y ¡ y0b

= t

z ¡ z0c

= t,

2

Or

x = x0 + at

y = y0 + bt

z = z0 + ct,

We call it the parametric form of the system of equations for line l. Thissystem can be written in the form of vector equation:

~r = ¡!r0 + t~v, ~r = hx, y, zi , ¡!r0 = hx0, y0, z0i .

Example 4.1. (a) Find the equation of the line passing through (5, ¡1, 3),having direction ~v = h1, 0, ¡2i . Express answer in (i) symmetric form, (ii)vector form, and (iii) parametric form.

(b) Find two other points on the line.Solution: (a) (i)

x ¡ 51

=y + 1

0=

z ¡ 3¡2

(ii)~r = h5, ¡1, 3i + t h1, 0, ¡2i

(iii)

x = 5 + t

y = ¡1z = 3¡ 2t

(b) Take t = 1, (x, y, z) = (6, ¡1, 1) . Take t = ¡1, (x, y, z) = (4, ¡1, 5) .Example 4.2. (a) Find the equation, in symmetric form, of the line l

passing through A (2, 4, ¡3) and B (3,¡1, 1) . (b) Determine where the linel intersects xy ¡ plane.

Solution. (a) The line is parallel to vector¡!AB. So we choose this vector

as the direction of l,

~v =¡!AB = h3, ¡1, 1i ¡ h2, 4, ¡3i = h1, ¡5, 4i ,

and A (2, 4, ¡3) as the point on l. Thus, the equation is

x ¡ 21

=y ¡ 4¡5 =

z + 3

4.

3

(b) If this line crosses xy ¡ plane somewhere at (x, y, z) , then z = 0. Sothis point (x, y, 0) satis…es the line equation, i.e.,

x ¡ 21

=y ¡ 4¡5 =

0 + 3

4.

We solve this system to obtain

x = 2 +3

4=11

4

y = 4¡ 5µ3

4

¶=1

4

z = 0.

Example 4.3. Given two lines

l1 : x = 1 + t, y = ¡2 + 3t, z = 4¡ t

l2 : x = 2t, y = 3 + t, z = ¡3 + 4t.

Determine whether they intersect each other, or they are parallel, or neither(skew lines).

Solution: First of all, in each line equation, "t" is a parameter (or freevariable) that can be chosen arbitrarily. Therefore, the parameter "t" in theequations for line l1 is DIFFERENT from the parameter "t" in the equationsfor line l2. To clarify this issue, we rewrite as

l1 : x = 1 + t, y = ¡2 + 3t, z = 4¡ t

l2 : x = 2s, y = 3 + s, z = ¡3 + 4s,

and intersection of these two lines consists of solutions of the following systemof six equations,

x = 1 + t, y = ¡2 + 3t, z = 4¡ t

x = 2s, y = 3 + s, z = ¡3 + 4s,

for …ve variables: x, y, z, t, s. Two lines intersect each other if and only If thissystem has a solution. If, for instance, (x0, y0, z0, t0, s0) is a solution, thenthe …rst three components, (x0, y0, z0) is a point of intersection.

4

We now proceed to solution the system by eliminating x,y,z:

1 + t = 2s (1)

¡2 + 3t = 3 + s (2)

4¡ t = ¡3 + 4s. (3)

There are three equations with two unknowns. We start with two equations,for instance, the …rst and the second equation:

1 + t = 2s

¡2 + 3t = 3 + s.

This can be easily solved as, by subtracting 2 times the second equation fromthe …rst equation, i.e.,

7¡ 5t = ¡6 =) t =11

5,

s =1 + t

2=8

5.

We need to verify that the solution, t =11

5, s =

8

5, from the …rst two

equations (1) & (2), satis…es the third equation (3). So

LHS of (3) = 4¡ t = 4¡ 11

5=9

5

RHS of (3) = ¡3 + 4s = ¡3 + 4µ8

5

¶=17

5.

Apparently,

t =11

5, s =

8

5

is not a solution of the entire system (1)-(3). We thus conclude that thesetwo line cannot possibly intersect. Answer: skew lines

5

Po Plane π

P(x,y,z)

l

N

z

O

x

y

B) Equations of Plane.De…nition. Any vector that is perpendicular to a plane is called a normal

vector to the plane.There are two normal directions (opposite to each other) to a given plane.

For any given vector ~n, there are in…nite many parallel planes that are allhaving ~n as their normal vector. If we also know a point on the plane,then, this plane is uniquely determined. In other words, a plane π can bedetermined by a point P0 (x0, y0, z0) on the plane and a vector as its normalvector ~n = hA, B, Ci.

For any point P (x, y, z) , if this point P is on the plane π , then the linesegment P0P entirely lies on the plane. Consequently, vector

¡¡!P0P = ~r¡ ¡!r0 = hx ¡ x0, y ¡ y0, z ¡ z0i , where ~r = hx, y, zi , ¡!r0 = hx0, y0, z0i ,

is perpendicular to the normal vector ~n. On the other hand, if P is o¤ the

6

plane π, then ~r =¡¡!P0P is not perpendicular to ~n. We conclude that

P 2 π (P belongs to π) () ¡¡!P0P ¢ ~n = 0,

or(~r ¡ ¡!r0 ) ¢ ~n = 0. (Vector Equation)

We call it vector equation of the plane π.In terms of components,

hx ¡ x0, y ¡ y0, z ¡ z0i ¢ hA, B, Ci = 0.

We obtain scalar form of equation of plane π :

A (x ¡ x0) +B (y ¡ y0) + C (z ¡ z0) = 0, (Scalar Equation)

orAx+By + Cz +D = 0. (Linear Equation)

In 3D spaces, any linear equation as above represents a plane with a normalvector ~n = hA, B, Ci . (In 2D, any linear equation is a straight line.)

Example 4.4. Find the equation of the plane passing through P0 (2, 4,¡1)having a normal vector ~n = h2, 3, 4i .

Solution: A = 2, B = 3, C = 4.The equation is

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

or2x+ 3y + 4z ¡ 12 = 0.

Example 4.5. Find the equation of the plane passing through P (1, 3, 2) ,Q (3, ¡1, 6) , R (5, 2, 0) .

Solution: Let

~u =¡!PR = h5, 2, 0i ¡ h1, 3, 2i = h4, ¡1, ¡2i

~v =¡!PQ = h3,¡1, 6i ¡ h1, 3, 2i = h2, ¡4, 4i

7

Q

R P

V

U

The vector

~u £ ~v =

¯̄̄̄¯̄ i j k4 ¡1 ¡22 ¡4 4

¯̄̄̄¯̄

=

¯̄̄̄ ¡1 ¡2¡4 4

¯̄̄̄~i ¡

¯̄̄̄4 ¡22 4

¯̄̄̄~j +

¯̄̄̄4 ¡12 ¡4

¯̄̄̄~k

= ¡12~i ¡ 20~j ¡ 14~k= ¡2

³6~i+ 10~j + 7~k

´is perpendicular to both ~u and ~v.Thus,

~n =³6~i+ 10~j + 7~k

´is perpendicular to π . Now, we take this normal vector and one point P (1, 3, 2)(you may choose Q or R,instead), and the equation is

6 (x ¡ 1) + 10 (y ¡ 3) + 7 (z ¡ 2) = 0

or6x+ 10y + 7z ¡ 50 = 0.

Note that if we chose Q (3, ¡1, 6) as the known point, then the equationwould be

6 (x ¡ 3) + 10 (y + 1) + 7 (z ¡ 6) = 0or

6x+ 10y + 7z ¡ 50 = 0.

8

Example 4.6. Find the intersection, if any, of the line

x = 2 + 3t, y = ¡4t, z = 5 + t

and the plane4x+ 5y ¡ 2z = 18.

Solution: We need to solve the system of all four equations

x = 2 + 3t

y = ¡4tz = 5 + t

4x+ 5y ¡ 2z = 18

for x, y, z, t. To this end, we substitute the …rst three equations into the lastone. This leads to

4 (2 + 3t) + 5 (¡4t)¡ 2 (5 + t) = 18,

which can be simpli…ed to

¡10t ¡ 2 = 18.

Sot = ¡2,

and

x = 2 + 3t = ¡4y = ¡4t = 8z = 5 + t = 3.

Answer: the intersection is (¡4, 8, 3) .Example 4.7. Given two planes

π1 : x+ y + z = 1

π2 : x ¡ 2y + 3z = 1.

Find (a) the line of intersection, and (b) the angle between two planes.

9

n1 Plane π1

Line l

n2

Plane π2

Solution: Plane π1 and plane π2 have normal vectors ~n1 and ~n2,respectively,as

~n1 = h1, 1, 1i~n2 = h1, ¡2, 3i .

The line is on both planes and thus is perpendicular to both normal vectors.The direction of the line is

~v = ~n1 £ ~n2 =

¯̄̄̄¯̄ i j k1 1 11 ¡2 3

¯̄̄̄¯̄

=

¯̄̄̄1 1

¡2 3

¯̄̄̄~i ¡

¯̄̄̄1 11 3

¯̄̄̄~j +

¯̄̄̄1 11 ¡2

¯̄̄̄~k

= 5~i ¡ 2~j ¡ 3~k.

To …nd the equation of the line, we also need a point on the line, i.e., onboth planes. So we look for one solution to the system

x+ y + z = 1

x ¡ 2y + 3z = 1.

10

This system has in…nite many solutions (why). Since we only need onesolution, we set z = 0 to reduce the system to

x+ y = 1

x ¡ 2y = 1.Subtracting the second equation from the …rst, we …nd

3y = 0 =) y = 0

x = 1.

So P (1, 0, 0) 2 l. The equation of the line, in parametric form, is

x = 1 + 5t

y = ¡2tz = ¡3t.

Solution #2: Another way to …nd the equation of this line is to solve thesystem

x+ y + z = 1

x ¡ 2y + 3z = 1directly in terms of z. In other words, we choose z as parameter. To this end,we subtract the second equation from the …rst one to get

3y ¡ 2z = 0 =) y =2

3z.

Substituting this into plane π1 :

x+

µ2

3z

¶+ z = 1 =) x = 1¡ 5

3z,

we obtain the equation of the line

x = 1¡ 5

3z

y =2

3z

z = z.

11

Note that z is basically a free variable. If we set z = ¡3t, this becomes

x = 1 + 5t

y = ¡2tz = ¡3t

which is identical to what we got earlier.(b) The angle between two planes is the same as the angle between their

normal vectors. So

cos θ =~n1 ¢ ~n2j~n1j j~n2j =

h1, 1, 1i ¢ h1, ¡2, 3ijh1, 1, 1ij jh1, ¡2, 3ij =

1¡ 2 + 3p3p1 + 4 + 9

=2p3p14

θ = arccos

µ2p3p14

¶= 1. 257 1(rad) = 1. 257 1

180

π(deg) = 720.

Practical advice in …nding equations of lines or planes:Regardless what information a problem provides, we always look for a

point and a direction. That would be su¢cient to solve the problem. Moreprecisely

(1) If we want equations of a line, then we look for one point on the lineand a vector PARALLEL to the line.

(2) If we want an equation of a plane, then we look for one point on theplane and a vector PERPENDICULAR to the plane. Cross product may beused to create a vector perpendicular given two vectors.

12

P0

P1

Line l

V

dist

C) Distance between points, lines and planes:Let S and T be two sets of points. Then

dist (S, T ) = min fdist (P, Q) j P 2 S, Q 2 Tg .

In other words, the distance between two sets is de…ned as the smallestdistance between two points from di¤erent sets.

1) Distance between a point P1 (x1, y1, z1) and the line l:

x = x0 + at

y = y0 + bt

z = z0 + ct.

Pick a point on the line, say P0 (x0, y0, z0) , and a direction (unit vector)of the line

~v =1p

a2 + b2 + c2ha, b, ci .

Then the cross product ³¡¡!P0P1

´£ ~v

13

by de…nition, has the length

dist (P1, l) =¯̄̄³¡¡!

P0P1´

£ ~v¯̄̄=

¯̄̄¡¡!P0P1

¯̄̄sin θ.

Example 4.8. Find the distance from P1 (1, ¡2, 1) to the line l :

x = 1 + 2t

y = 2¡ 3tz = 4t.

Solution. P0 (1, 2, 0) is a point on l, and

¡¡!P0P1 = h1, ¡2, 1i ¡ h1, 2, 0i = h0, ¡4, 1i .

A unit direction of the line is

~v =1p29

h2, ¡3, 4i

So

³¡¡!P0P1

´£ ~v =

1p29

¯̄̄̄¯̄ ~i ~j ~k0 ¡4 12 ¡3 4

¯̄̄̄¯̄

=1p29

µ¯̄̄̄ ¡4 1¡3 4

¯̄̄̄~i ¡

¯̄̄̄0 12 4

¯̄̄̄~j +

¯̄̄̄0 ¡42 ¡3

¯̄̄̄~k

¶=

1p29

³¡13~i+ 2~j + 8~k

´,

and

dist (P1, l) =¯̄̄³¡¡!

P0P1

´£ ~v

¯̄̄=

p132 + 4 + 64p

29=

r237

29= 2. 86.

(2) Distance from a point P1 (x1, y1, z1) to a plane π : Ax+By+Cz+D =0.

Pick any point P0 (x0, y0, z0) on the plane, i.e., (x0, y0, z0) solves

Ax0 +By0 + Cz0 +D = 0,

14

P0

Plane π

N

P1

N

then the distance is the absolute value of the dot product of¡¡!P0P1 and normal

direction~n =

1pA2 +B2 + C2

hA, B, Ci

dist (P0, π) =¯̄̄³¡¡!

P0P1´

¢ ~n¯̄̄

=

¯̄̄̄hx1 ¡ x0, y1 ¡ y0, z1 ¡ z0i ¢ hA, B, CipA2 +B2 + C2

¯̄̄̄=

¯̄̄̄A (x1 ¡ x0) +B (y1 ¡ y0) + C (z1 ¡ z0)p

A2 +B2 + C2

¯̄̄̄=

¯̄̄̄Ax1 +By1 + Cz1 ¡ (Ax0 +By0 + Cz0)p

A2 +B2 + C2

¯̄̄̄=

¯̄̄̄Ax1 +By1 + Cz1 +Dp

A2 +B2 + C2

¯̄̄̄.

Example 4.9. Find the distance from P (1, 2, 3) to the planeπ : 2x ¡y + 3z = 4.

Solution: We rewrite plane π in the standard form as

2x ¡ y + 3z ¡ 4 = 0.

15

So

dist (P, π) =

¯̄̄̄Ax1 +By1 + Cz1 +Dp

A2 +B2 + C2

¯̄̄̄=

¯̄̄̄2x1 ¡ y1 + 3z1 ¡ 4p

22 + 12 + 42

¯̄̄̄=

¯̄̄̄2¡ 2 + 9¡ 4p22 + 12 + 42

¯̄̄̄= 1. 09.

(3) Distance between two lines.We …rst …ne the plane π containing line l1 and being parallel to l2. If

~v1 and ~v2 are directions of l1 and l2,respectively. Then,

~n = ~v1 £ ~v2

is a normal vector to the plane π .One may pick any point P0 on l1 and thisnormal vector to obtain the equation of π . Pick any point P1 on line l2,and

dist (l1, l2) = dist (P1, π) (A)

Another approach is to use projection. Pick one point from each line, say P02 l1, P1 2 l2. Then,

dist (l1, l2) =¯̄̄Proj~n

³¡¡!P0P1

´¯̄̄=

¯̄̄³¡¡!P0P1

´¢ ~n

¯̄̄j~nj (B)

Example 4.10. Consider two skewed lines in Example 9.5.3.

l1 : x = 1 + t, y = ¡2 + 3t, z = 4¡ t

l2 : x = 2t, y = 3 + t, z = ¡3 + 4t.Find their distance.

Solution. We …rst use method (A). The normal to the plane π containingl1 and parallel to l2 is

~n = ~v1 £ ~v2 =

~n = ~v1 £ ~v2 =

¯̄̄̄¯̄ ~i ~j ~k1 3 ¡12 1 4

¯̄̄̄¯̄ = ¯̄̄̄

3 ¡11 4

¯̄̄̄~i ¡

¯̄̄̄1 ¡12 4

¯̄̄̄~j +

¯̄̄̄1 32 1

¯̄̄̄~k

= 13~i ¡ 6~j ¡ 5~k.

16

P0

P1

Line l1

V1

dist

Line l2

V2

N

17

Obviously, P0 (1, ¡2, 4) 2 l1. So equation of π is

13 (x ¡ 1)¡ 6 (y + 2)¡ 5 (z ¡ 4) = 0.Since P1 (0, 3, ¡3) 2 l2, formula (A) and the distance formula lead to

dist (l1, l2) = dist (P1, π)

=

¯̄̄̄Ax1 +By1 + Cz1 +Dp

A2 +B2 + C2

¯̄̄̄=

¯̄̄̄13 (0¡ 1)¡ 6 (3 + 2)¡ 5 (¡3¡ 4)p

132 + 36 + 25

¯̄̄̄= 0.527 5

Let now solve the same problem using formula (B). Note that

¡¡!P0P1 = h0, 3, ¡3i ¡ h1, ¡2, 4i = h¡1, 5, ¡7i

~n = 13~i ¡ 6~j ¡ 5~k.

So

dist (l1, l2) =

¯̄̄³¡¡!P0P1

´¢ ~n

¯̄̄j~nj =

¯̄̄̄¡1 ¢ 13 + 5 ¢ (¡6) + (¡7) ¢ (¡5)p132 + 36 + 25

¯̄̄̄= 0.527 5.

(4) Distance between a line l and a plane π .Pick any point P0 on the line and the distance dist (l, π ) between line l

and plane π is the distance from P0 to plane π :

dist (l, π ) = dist (P0, π ) , P0 2 l.

Homework:

1. Determine whether each statement is true or false.

(a) Two lines parallel to a third line are parallel.

(b) Two lines perpendicular to a third line are parallel.

(c) Two planes parallel to a third plane are parallel.

(d) Two planes perpendicular to a third plane are parallel.

18

(e) Two lines parallel to a plane are parallel.

(f) Two lines perpendicular to a plane are parallel.

(g) Two planes parallel to a line are parallel.

(h) Two planes perpendicular to a line are parallel.

(i) Two planes are either intersect or are parallel.

(j) Two lines are either intersect or are parallel.

(k) A plane and a line are either intersect or are parallel.

2. Find parametric equation and symmetric equation of line.

(a) The line through the point (1, 0,¡3) and parallel to h2, ¡4, 5i .

(b) The line through the point the origin and parallel to the line x =2t, y = 1¡ t, z = 4 + 3t.

(c) The line through (1, 1, 6) and perpendicular to the plane x+3y+z = 5.

(d) The line through (2, 1, 1) and perpendicular to~i+~j+~k and~i+2~k.

(e) The line of intersection of the planes x+2y+z = 1 and x+y = 0.

3. Find equation of plane.

(a) The plane through (1, 0, 1) , (0, 1, 1), and (1, 1, 0) .

(b) The plane through (5, 1, 0) and parallel to two lines x = 2+t, y =¡2t, z = 1 and x = t, y = 2¡ t, z = 2¡ 3t.

(c) The plane through (¡1, 2, 1) and contains the line of intersectionof two planes x+ y ¡ z = 2 and 2x ¡ y + 3z = 1.

(d) The plane through (¡1, 2, 1) and perpendicular to the line of in-tersection of two planes x+ y ¡ z = 2 and 2x ¡ y + 3z = 1.

(e) The plane that passes through the line of intersection of the planesx ¡ z = 1 and y + 2z = 3, and is perpendicular to the planex+ y ¡ 2z = 1.

4. Determine whether two lines are parallel, skew, or intersecting. If theyintersect, …nd the point of intersection.

19

(a) L1 : x = 1+2t, y = 3t, z = 2¡ t; L2 : x = ¡1+ s, y = 4+s, z =1 + 3s.

(b) L1 :x ¡ 22

=y ¡ 32

=z ¡ 2¡1 ; L2 :

x ¡ 21

=y ¡ 6¡1 =

z + 2

3

5. (Optional) Find distance.

(a) Distance from (1, 0, ¡1) to the line x = 5¡ t, y = 3t, z = 1+ 2t.

(b) Distance from (3, ¡2, 7) to the plane 4x ¡ 6y + z = 5.

(c) Distance between two planes 3x+6y¡9z = 4 and x+2y¡3z = 2.

20