267 5 parametric eequations of lines

Equations of Lines
Parametric Equations of Lines

Equations of Lines
We start with a vector motivated representation of
lines by recalling the following associations.

Scalar Multiplication
Given a number λ and a vector v,
λv is the extension (compression)
of the vector v by a factor λ.
Equations of Lines

of the vector v by a factor λ. v 2v
Equations of Lines

–2v –v
Equations of Lines

–2v –v
The Base to Tip Rule
Given two vectors u and v, u + v is the
new vector formed by placing the base
of one vector at the tip of the other.
Equations of Lines

–2v –v
u
v
Equations of Lines

–2v –v
v
u
v
Equations of Lines

–2v –v
v
u
u + v
v
Equations of Lines

–2v –v
v
u
u + v
v
Equations of Lines
Again, we begin with the 2D version
then generalize the results to 3D.

Let D = <a, b> be a vector that indicates direction as
shown.
D=<a,b>
x

shown. Let t be a scalar, then t*D = <ta, tb> is an
extension of D.
D=<a,b>
tD=<ta,tb>
x

we get all the points in the "direction" of D,
i.e. the entire line that coincides with D.
extension of D. As t takes on all real numbers,
D=<a,b>
tD=<ta,tb>
x

Hence x(t) = at, y(t) = bt
is a set of parametric equations
for the line through the origin in
the direction of D = <a, b>.
D=<a,b>
tD=<ta,tb>
x

D=<a,b>
tD=<ta,tb>
P=<c, d>
Suppose P = (c, d) specifies a
base point,
x

base point, using the base-to-tip
addition rule, the vector sums
P + tD = <c, d> + t<a, b>
D=<a,b>
tD=<ta,tb>
P=<c, d>
x

D=<a,b>
tD=<ta,tb>
P=<c, d>
P + tD = <c, d> + t<a, b>
x

D=<a,b>
tD=<ta,tb>
P=<c, d>
<at+c,bt+d>
P + tD = <c, d> + t<a, b>
x

D=<a,b>
tD=<ta,tb>
P=<c, d>
<at+c,bt+d>
P + tD = <c, d> + t<a, b> = <c + ta, d + tb>
viewed as points, are all the points on the line L.
L
x

D=<a,b>P=<c, d>
To summarize, let L be the line passing though the
point P = (c, d) in the direction of D = <a, b>
x

D=<a,b>P=<c, d>
L
point P = (c, d) in the direction of D = <a, b>
x

point P = (c, d) in the direction of D = <a, b> then
L = P+ t*D where t is a number.
D=<a,b>P=<c, d>
L
x

D=<a,b>
tD=<ta,tb>
P=<c, d>
L
x

D=<a,b>
tD=<ta,tb>
P=<c, d>
<at+c,bt+d>
L
x

This is a vector–equation of L.
D=<a,b>
tD=<ta,tb>
P=<c, d>
<at+c,bt+d>
L
x

If (x, y) is a generic point on L then
(x, y) = (c, d) + t*(a, b) = (c + at, d + bt).
D=<a,b>
tD=<ta,tb>
P=<c, d>
<at+c,bt+d>
L
x

Hence x = c + at, y = d + bt is a set of
parametric equations of L.
(x, y) = (c, d) + t*(a, b) = (c + at, d + bt).
D=<a,b>
tD=<ta,tb>
P=<c, d>
<at+c,bt+d>
L
x

Example A. Find a set of
parametric equations for the line L
a. in the direction of D = <2, 1>,
passing through P = (4, 5).
(x, y) = (c, d) + t*(a, b) = (c + at, d + bt).
D=<a,b>
tD=<ta,tb>
P=<c, d>
<at+c,bt+d>
L
x

passing through P = (4, 5). P=(4, 5)
D=<2, 1>
x
(x, y) = (c, d) + t*(a, b) = (c + at, d + bt).
D=<a,b>
tD=<ta,tb>
P=<c, d>
<at+c,bt+d>
L
x

passing through P = (4, 5).
The parametric equations of L are
x(t) = 2t + 4, y(t) = 1t + 5.
P=(4, 5)
D=<2, 1>
( 2t + 4, 1t + 5)
x
(x, y) = (c, d) + t*(a, b) = (c + at, d + bt).
D=<a,b>
tD=<ta,tb>
P=<c, d>
<at+c,bt+d>
L
x

the set of points tD =(at, bt, ct)
form the line through the origin
coinciding with the vector D or
that it’s in the same direction as D.
Let D = <a, b, c> be a vector as shown,
Line Equations R3

coinciding with the vector D or that
it’s in the same direction as D. tD=<at,bt,ct>
D = <a, b, c>
x
y
z
Let D = <a, b, c> be a vector as shown,
Line Equations R3

it’s in the same direction as D.
Let P = (d, e, f) be a point in space,
tD=<at,bt,ct>
P=(d,e,f)
D = <a, b, c>
x
y
z
Let D = <a, b, c> be a vector as shown.
Line Equations R3

tD=<at,bt,ct>
P=(d,e,f)
D = <a, b, c>
of points P + tD = (d + at, e + bt, f + ct ).
P+tD
x
y
then the line containing P,
in the direction of D consists
z
Line Equations R3

tD=<at,bt,ct>
Hence x(t) = at + d, y(t) = bt + e, z(t) = ct + f,
where t is any number, is a set of parametric
equations representing the line through
P = (d, e, f) in the direction D = <a, b, c>.
P=(d,e,f)
D = <a, b, c>
of points P + tD = (d + at, e + bt, f + ct ).
P+tD
x
y
then the line containing P,
in the direction of D consists
z
Line Equations R3

Example B. Find a set of parametric equations for the
line in the same direction as QR where Q = (3, –2, 5),
R = (1, 2, 1) that passes through P = (4, 5, 6).

Q(3,–2,5)
x
y
z
R(1,2,1)
QR=D

The vector QR = R – Q
= <1, 2, 1> – <3, –2, 5>
= <–2, 4, –4> = D is a
vector that gives the
direction of the line.
Q(3,–2,5)
x
y
z
R(1,2,1)
QR=D

= <1, 2, 1> – <3, –2, 5>
= <–2, 4, –4> = D is a
direction of the line. So
x(t) = –2t + 4
y(t) = 4t + 5
z(t) = –4t + 6
is the line through P that has the same direction as QR.
Q(3,–2,5)
x
y
z
R(1,2,1)
QR=D

= <1, 2, 1> – <3, –2, 5>
= <–2, 4, –4> = D is a
x(t) = –2t + 4
y(t) = 4t + 5
z(t) = –4t + 6
Q(3,–2,5)
x
y
z
R(1,2,1)
P(4,5,6)
QR=D
P+t*QR

= <1, 2, 1> – <3, –2, 5>
= <–2, 4, –4> = D is a
x(t) = –2t + 4
y(t) = 4t + 5
z(t) = –4t + 6
Q(3,–2,5)
x
y
z
R(1,2,1)
P(4,5,6)
QR=D
P+t*QR
Definition: Two lines are parallel if they have the same
(or opposite) directional vector.

Let L represent the line in the last example, then we
may write L = P + t*D where D is a vector that gives
the direction of the line, P is a vector whose tip is on
the line L, and t as the variable. This is called the
vector-equation of the line.
Q
x
y
z
R
P
QR = D
L = P+t*D

Q
x
y
z
R
P
QR = D
L = P+t*DRemark: We can't
represent a line in 3D with
a single equation in the
variable x, y and z
because the graph of such
an equation is a surface in
3D space in general.

Q
x
y
z
R
P
QR = D
L = P+t*DRemark: We can't
represent a line in 3D with
a single equation in the
variable x, y and z
because the graph of such
an equation is a surface in
3D space in general.
Another method of setting equations to represent a
line L is to give L as the intersection of two planes.

Let L be the line <t, t, t > where t is any real number,
as shown here.
x
z+
<1,1,1>
y

as shown here.
x
z+
<1,1,1>
y
The parametric equations for L are
x(t) = t, y(t) = t, z(t) = t

as shown here.
or that (t =) x = y = z.
x
z+
<1,1,1>
y
x(t) = t, y(t) = t, z(t) = t

as shown here.
These triple–equation is called the
symmetric equation of the line L.
x
z+
<1,1,1>
y
x(t) = t, y(t) = t, z(t) = t

as shown here.
x
z+
<1,1,1>
y
x(t) = t, y(t) = t, z(t) = t
The symmetric equation actually consists of two
systems of linear equations,

as shown here.
x
z+
<1,1,1>
y
x(t) = t, y(t) = t, z(t) = t
systems of linear equations, in this case
x = y
y = z
x = y
x = z
x = z
y = zA: B: C:

as shown here.
x
z+
<1,1,1>
y
x(t) = t, y(t) = t, z(t) = t
systems of linear equations, in this case
x = y
y = z
x = y
x = z
x = z
y = zA: B: C:
Each system of equations consist of two planes and
L is the intersection of two planes.

x = y
x = z
A:

x = y
y = zB:
x = y
x = z
A:

x = y
y = zB:
x = y
x = z
A:
x = z
y = zC:

267 5 parametric eequations of lines

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267 5 parametric eequations of lines