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Section 2.4: Equations of Lines and Planes
An equation of three variable F (x; y; z) = 0 is called an equation of a
surface S if
(x 1 ; y 1 ; z 1 ) 2 S if and only if F (x 1 ; y 1 ; z 1 ) = 0:
For instance,
x
2
2
2 = 1
is the equation of the unit sphere centered at the origin. The graph of a
system 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 and
by G (x; y; z) = 0; respectively, and is usually a curve.
A) Lines in R
3 :
A line l is determined by two elements: one point P 0 on the line l and
a direction ~v of l;i.e., any vector that is parallel to l: The goal here is to
describe the line using algebra so that one is able to digitize it. Suppose that
the coordinate of the point P 0 on the line and a direction ~v are given as:
P 0 (x 0 ; y 0 ; z 0 ) 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
P 0 P be the vector connecting
P 0 and P: If P is located exactly on the line, then
P 0 P 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 P 0 is on the line,
P 0 P cannot possibly be parallel to the line. Therefore, ¡¡! P 0 P cannot possibly be parallel to ~v: We just concluded that
P is on l if and only if
P 0 P is parallel to ~v :
Now
P 0 P = hx; y; zi ¡ hx 0 ; y 0 ; z 0 i = hx ¡ x 0 ; y ¡ y 0 ; z ¡ z 0 i
~v = ha; b; ci :
P(x,y,z)
l
V
z
O
x
y
Po
So ¡¡! P 0 P == ~v ()
P 0 P = t~v (for a constant t)
which is equivalent to
x ¡ x 0
a
y ¡ y 0
b
z ¡ z 0
c
We called these three equation symmetric form of the system of equations
for line l:
If we set x ¡ x 0
a
y ¡ y 0
b
z ¡ z 0
c
= t;
which is equivalent to
x ¡ x 0
a
= t
y ¡ y 0
b
= t
z ¡ z 0
c
= t;
(b) If this line crosses xy ¡ plane somewhere at (x; y; z) ; then z = 0: So
this point (x; y; 0) satis…es the line equation, i.e.,
x ¡ 2
y ¡ 4
We solve this system to obtain
x = 2 +
y = 4 ¡ 5
μ 3
z = 0:
Example 4.3. Given two lines
l 1 : x = 1 + t; y = ¡2 + 3t; z = 4 ¡ t
l 2 : 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 free
variable) that can be chosen arbitrarily. Therefore, the parameter "t" in the
equations for line l 1 is DIFFERENT from the parameter "t" in the equations
for line l 2 : To clarify this issue, we rewrite as
l 1 : x = 1 + t; y = ¡2 + 3t; z = 4 ¡ t
l 2 : x = 2s; y = 3 + s; z = ¡3 + 4s;
and intersection of these two lines consists of solutions of the following system
of 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 this
system has a solution. If, for instance, (x 0 ; y 0 ; z 0 ; t 0 ; s 0 ) is a solution, then
the …rst three components, (x 0 ; y 0 ; z 0 ) is a point of intersection.
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 from
the …rst equation, i.e.,
7 ¡ 5 t = ¡6 =) t =
s =
1 + t
We need to verify that the solution, t =
; s =
; from the …rst two
equations (1) & (2), satis…es the third equation (3). So
LHS of (3) = 4 ¡ t = 4 ¡
RHS of (3) = ¡3 + 4s = ¡3 + 4
μ 8
Apparently,
t =
; s =
is not a solution of the entire system (1)-(3). We thus conclude that these
two line cannot possibly intersect. Answer: skew lines
plane º; then ~r =
P 0 P is not perpendicular to ~n: We conclude that
P 2 º (P belongs to º) ()
P 0 P ¢ ~n = 0;
or
(~r ¡
r 0 ) ¢ ~n = 0: (Vector Equation)
We call it vector equation of the plane º:In terms of components,
hx ¡ x 0 ; y ¡ y 0 ; z ¡ z 0 i ¢ hA; B; Ci = 0:
We obtain scalar form of equation of plane º :
A (x ¡ x 0 ) + B (y ¡ y 0 ) + C (z ¡ z 0 ) = 0; (Scalar Equation)
or
Ax + By + Cz + D = 0: (Linear Equation)
In 3 D spaces, any linear equation as above represents a plane with a normal
vector ~n = hA; B; Ci. (In 2D, any linear equation is a straight line.)
Example 4.4. Find the equation of the plane passing through P 0 (2; 4 ; ¡1)
having a normal vector ~n = h 2 ; 3 ; 4 i :
Solution: A = 2; B = 3; C = 4:The equation is
2 (x ¡ 2) + 3 (y ¡ 4) + 4 (z + 1) = 0;
or
2 x + 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 =
P R = h 5 ; 2 ; 0 i ¡ h 1 ; 3 ; 2 i = h 4 ; ¡ 1 ; ¡ 2 i
~v =
P Q = h 3 ; ¡ 1 ; 6 i ¡ h 1 ; 3 ; 2 i = h 2 ; ¡ 4 ; 4 i
Q
P R
V
U
The vector
~u £ ~v =
i j k
~i ¡
~j +
~k
= ¡ 12 ~i ¡ 20 ~j ¡ 14 ~k
6 ~i + 10~j + 7~k
is perpendicular to both ~u and ~v:Thus,
~n =
6 ~i + 10~j + 7
k
is perpendicular to º : 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
or
6 x + 10y + 7z ¡ 50 = 0:
Note that if we chose Q (3; ¡ 1 ; 6) as the known point, then the equation
would be
6 (x ¡ 3) + 10 (y + 1) + 7 (z ¡ 6) = 0
or
6 x + 10y + 7z ¡ 50 = 0:
n
Plane π 1
Line l
n
Plane π 2
Solution: Plane º 1 and plane º 2 have normal vectors ~n 1 and ~n 2 ;respectively,
as
~n 1 = h 1 ; 1 ; 1 i
~n 2 = h 1 ; ¡ 2 ; 3 i :
The line is on both planes and thus is perpendicular to both normal vectors.
The direction of the line is
~v = ~n 1 £ ~n 2 =
i j k
~i ¡
~j +
~k
= 5~i ¡ 2 ~j ¡ 3 ~k:
To …nd the equation of the line, we also need a point on the line, i.e., on
both planes. So we look for one solution to the system
x + y + z = 1
x ¡ 2 y + 3z = 1:
This system has in…nite many solutions (why). Since we only need one
solution, we set z = 0 to reduce the system to
x + y = 1
x ¡ 2 y = 1:
Subtracting the second equation from the …rst, we …nd
3 y = 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 = ¡ 2 t
z = ¡ 3 t:
Solution #2: Another way to …nd the equation of this line is to solve the
system
x + y + z = 1
x ¡ 2 y + 3z = 1
directly 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
3 y ¡ 2 z = 0 =) y =
z:
Substituting this into plane º 1 :
x +
μ 2
z
z;
we obtain the equation of the line
x = 1 ¡
z
y =
z
z = z:
P 0
P 1
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 T g :
In other words, the distance between two sets is de…ned as the smallest
distance between two points from di¤erent sets.
- Distance between a point P 1 (x 1 ; y 1 ; z 1 ) and the line l:
x = x 0 + at
y = y 0 + bt
z = z 0 + ct:
Pick a point on the line, say P 0 (x 0 ; y 0 ; z 0 ) ; and a direction (unit vector)
of the line
~v =
p a
2
2
2
ha; b; ci :
Then the cross product ³ ¡¡! P 0 P 1
£ ~v
by de…nition, has the length
dist (P 1 ; l) =
P 0 P 1
£ ~v
P 0 P 1
¯ sin μ:
Example 4.8. Find the distance from P 1 (1; ¡ 2 ; 1) to the line l :
x = 1 + 2t
y = 2 ¡ 3 t
z = 4t:
Solution. P 0 (1; 2 ; 0) is a point on l; and
P 0 P 1 = h 1 ; ¡ 2 ; 1 i ¡ h 1 ; 2 ; 0 i = h 0 ; ¡ 4 ; 1 i :
A unit direction of the line is
~v =
p 29
h 2 ; ¡ 3 ; 4 i
So
P 0 P 1
£ ~v =
p 29
~i ~j ~k
p 29
μ¯ ¯ ¯ ¯
~i ¡
~j +
~k
p 29
¡ 13 ~i + 2~j + 8~k
and
dist (P 1 ; l) =
P 0 P 1
£ ~v
p 13
2
r
237
(2) Distance from a point P 1 (x 1 ; y 1 ; z 1 ) to a plane º : Ax+By +Cz +D =
0 :
Pick any point P 0 (x 0 ; y 0 ; z 0 ) on the plane, i.e., (x 0 ; y 0 ; z 0 ) solves
Ax 0 + By 0 + Cz 0 + D = 0;
So
dist (P; º) =
Ax 1 + By 1 + Cz 1 + D p A
2
2
2
2 x 1 ¡ y 1 + 3z 1 ¡ 4 p 2
2
2
2
p 22 + 1^2 + 4^2
(3) Distance between two lines.
We …rst …ne the plane º containing line l 1 and being parallel to l 2 : If
~v 1 and ~v 2 are directions of l 1 and l 2 ;respectively. Then,
~n = ~v 1 £ ~v 2
is a normal vector to the plane º :One may pick any point P 0 on l 1 and this
normal vector to obtain the equation of º : Pick any point P 1 on line l 2 ;and
dist (l 1 ; l 2 ) = dist (P 1 ; º) (A)
Another approach is to use projection. Pick one point from each line, say P 0
2 l 1 ; P 1 2 l 2 : Then,
dist (l 1 ; l 2 ) =
¯Proj~n
P 0 P 1
P 0 P 1
¢ ~n
j~nj
(B)
Example 4.10. Consider two skewed lines in Example 9.5.3.
l 1 : x = 1 + t; y = ¡2 + 3t; z = 4 ¡ t
l 2 : x = 2t; y = 3 + t; z = ¡3 + 4t:
Find their distance.
Solution. We …rst use method (A). The normal to the plane º containing
l 1 and parallel to l 2 is
~n = ~v 1 £ ~v 2 =
~n = ~v 1 £ ~v 2 =
~i ~j ~k
~i ¡
~j +
~k
= 13~i ¡ 6 ~j ¡ 5 ~k:
P (^0)
P 1
Line l (^1)
V 1
dist
Line l 2
V 2
N
(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.
- Find parametric equation and symmetric equation of line.
(a) The line through the point (1; 0 ; ¡3) and parallel to h 2 ; ¡ 4 ; 5 i :
(b) The line through the point the origin and parallel to the line x =
2 t; 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:
- 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 =
¡ 2 t; z = 1 and x = t; y = 2 ¡ t; z = 2 ¡ 3 t:
(c) The plane through (¡ 1 ; 2 ; 1) and contains the line of intersection
of two planes x + y ¡ z = 2 and 2 x ¡ 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 2 x ¡ y + 3z = 1:
(e) The plane that passes through the line of intersection of the planes
x ¡ z = 1 and y + 2z = 3, and is perpendicular to the plane
x + y ¡ 2 z = 1:
- Determine whether two lines are parallel, skew, or intersecting. If they
intersect, …nd the point of intersection.
(a) L 1 : x = 1 + 2t; y = 3t; z = 2 ¡ t; L 2 : x = ¡1 + s; y = 4 + s; z =
1 + 3s:
(b) L 1 :
x ¡ 2
y ¡ 3
z ¡ 2
; L 2 :
x ¡ 2
y ¡ 6
z + 2
- (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 4 x ¡ 6 y + z = 5:
(c) Distance between two planes 3 x+ 6y ¡ 9 z = 4 and x +2y ¡ 3 z = 2:
