Section 2.4: Equations of Lines and Planes
[Pages:10]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
(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 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 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 to describe the line using algebra so that one is able to digitize it. Suppose that
the 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??b!e 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 ?P?0!P
P0 is on the line, ?P?0!P cannot possibly be cannot possibly be parallel to ~v: We just
parallel to concluded
the line. that
Therefore,
??! 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) z
Po V
l
O
y
x
So
??!
??!
P0P == ~v () P0P = t~v (for a constant t)
which is equivalent to
x ? x0 = y ? y0 = z ? z0 :
a
b
c
We called these three equation symmetric form of the system of equations
for line l:
If we set
x ? x0 = y ? y0 = z ? z0 = t;
a
b
c
which is equivalent to
x ? x0 = t a
y ? y0 = t b
z ? z0 = t; c
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: This system 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?5 = y+1 = z?3
1
0
?2
(ii) ~r = h5; ?1; 3i + t h1; 0; ?2i
(iii)
x=5+t y = ?1 z = 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 line
l
intersects xy ? Solution. (a)
plane: The line
is
parallel
to
vector
?A!B:
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?2 y?4 z+3 1 = ?5 = 4 :
3
(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 = 0 + 3:
1
?5
4
We solve this system to obtain
x = 2 + 3 = 11
4? ?4
y =4?5
3 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 free variable) that can be chosen arbitrarily. Therefore, the parameter "t" in the equations for line l1 is DIFFERENT from the parameter "t" in the equations for 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 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, (x0; y0; z0; t0; s0) is a solution, then the ...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 from the ...rst equation, i.e.,
7 ? 5t = ?6
=)
t = 11; 5
1+t 8 s = 2 = 5:
We need to verify that the solution, t = 11 ; s = 8; from the ...rst two
5
5
equations (1) & (2), satis...es the third equation (3). So
LH S
of
(3) = 4 ? t
=4?
11 5
=
9 5?
?
8 17
RHS of (3) = ?3 + 4s = ?3 + 4 5 = 5 :
Apparently,
11 8 t= 5 ;s = 5
is not a solution of the entire system (1)-(3). We thus conclude that these two line cannot possibly intersect. Answer: skew lines
5
N
Plane
Po
P(x,y,z)
z
l
O
y
x
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 all having ~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 ? can be determined by a point P0 (x0; y0; z0) on the plane and a vector as its normal vector ~n = hA; B; Ci.
For any point P (x; y; z) ; if this point P is on the plane ? ; then the line segment 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 ?; then ~r = P0P is not perpendicular to ~n: We conclude that
??! P 2 ? (P belongs to ?) () P0P ? ~n = 0;
or (~r ? ?!r0 ) ? ~n = 0:
(Vector Equation)
We call it vector equation of the plane ?:In terms of components,
hx ? x0; y ? y0; z ? z0i ? hA; B; Ci = 0:
We obtain scalar form of equation of plane ? :
A (x ? x0) + B (y ? y0) + C (z ? z0) = 0;
(Scalar Equation)
or Ax + By + Cz + D = 0:
(Linear Equation)
In 3D 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 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;
or 2x + 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 = h5; 2; 0i ? h1; 3; 2i = h4; ?1; ?2i ?!
~v = P Q = h3; ?1; 6i ? h1; 3; 2i = h2; ?4; 4i
7
Q V
P
R
U
The vector
~u ? ~v
= ?????? = ????
ij k
4 ?1 ?2
2 ?4
?1 ?2 ?4 4
????4~i
?????? ?
????
4 2
?2 4
????~j + ????
4 2
?1 ?4
???? ~k
= ?12?~i ? 20~j ? 14~k ? = ?2 6~i + 10~j + 7~k
is perpendicular to both ~u and ~v:Thus,
?
?
~n = 6~i + 10~j + 7~k
is perpendicular to ? : 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 6x + 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 6x + 10y + 7z ? 50 = 0:
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