Calculus and Vectors – How to get an A+
8.3 Vector, Parametric, and Symmetric Equations of a Line in R3
A Vector Equation
The vector equation of the line is:
r r
r
r = r0 + tu , t ∈ R
where:
r
Ö r = OP is the position vector of a generic point
P on the line,
r
Ö r0 = OP0 is the position vector of a specific
point P0 on the line,
r
Ö u is a vector parallel to the line called the
direction vector of the line, and
Ö t is a real number corresponding to the
generic point P .
Ex 1. Find two vector equations of the line L that passes
through the points A(1,2,3) and B(2,−1,0) .
r
If we use the direction vector u = AB = (1,−3,−3) and the point
A(1,2,3) ∈ L , then the vector equation of the line L is:
r
L : r = (1,2,3) + t (1,−3,−3), t ∈ R
r
If we use the direction vector u = BA = (−1,3,3) and the point
B(2,−1,0) ∈ L , then the vector equation of the line L is:
r
L : r = (2,−1,0) + s (−1,3,3), s ∈ R
Ex 2. Find the vector equation of a line L2 that passes
through the origin and is parallel to the line
r
L1 : r = (−2,0,3) + t (−1,0,2), t ∈ R .
r
∴ L2 : r = s (−1,0,2), s ∈ R
B Specific Lines
r
A line is parallel to the x-axis if u = (u x ,0,0), u x ≠ 0 .
In this case, the line is also perpendicular to the
yz-plane.
Ex 3. Find the vector equation of a line that:
a) passes through A(3,−2,0) and is parallel to the y-axis
r
r = (3,−2,0) + t (0,1,0), t ∈ R
b) passes through M (−1,0,4) and is perpendicular to the yzplane
r
r = (−1,0,4) + t (1,0,0), t ∈ R
r
c) passes through P (3,0,0) and is perpendicular to the x-axis
A line with u = (0, u y , u z ), u y ≠ 0, u z ≠ 0 is parallel to r
r = (3,0,0) + t (0, a, b), t ∈ R . At least one of a or b is not 0 .
the yz-plane.
d) passes through the origin and is parallel to the xz-plane
r
r = t (a,0, b), t ∈ R . At least one of a or b is not 0 .
C Parametric Equations
Let rewrite the vector equation of a line:
r r
r
r = r0 + tu , t ∈ R
as:
( x, y, z ) = ( x0 , y 0 , z 0 ) + t (u x , u y , u z ), t ∈ R
The parametric equations of a line in R3 are:
⎧ x = x0 + tu x
⎪
⎨ y = y0 + tu y , t ∈ R
⎪
⎩ z = z 0 + tu z
Ex 4. Find the parametric equations of the line L that passes
through the points A(0,−1,2) and B (1,−1,3) . Describe the line.
r
u = AB = (1,0,1); A(0,−1,2) ∈ L
r
L : r = (0,−1,2) + t (1,0,1), t ∈ R
( x, y, z ) = (0,−1,2) + t (1,0,1)
⎧x = t
⎪
L ∴ ⎨ y = −1
, t∈R
⎪z = 2 + t
⎩
The line is parallel to the xz-plane.
8.3 Vector, Parametric, and Symmetric Equations of a Line in R
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Calculus and Vectors – How to get an A+
D Symmetric Equations
The parametric equations of a line may be written
as:
⎧ x − x0 = tu x
⎪
⎨ y − y 0 = tu y , t ∈ R
⎪
⎩ z − z 0 = tu z
From here, the symmetric equations of the line
are:
x − x0 y − y 0 z − z 0
=
=
ux
uy
uz
u x ≠ 0, u y ≠ 0, u z ≠ 0
Ex 6. Convert the symmetric equations for a line:
x − 2 y +1 z
=
= to the parametric and vector
3
−2 4
equations.
⎧ x − 2 = 3t
x − 2 y +1 z
⎪
=
= = t ⇒ ⎨ y + 1 = −2t
−2 4
3
⎪ z = 4t
⎩
⎧ x = 2 + 3t
⎪
∴ ⎨ y = −1 − 2t , t ∈ R
⎪ z = 4t
⎩
r
∴ r = (2,−1,0) + t (3,−2,4), t ∈ R
Ex 5. Convert the vector equation of the line
r
L : r = (0,1,−3) + t (−1,2,0), t ∈ R to the parametric and
symmetric equations.
( x, y, z ) = (0,1,−3) + t (−1,2,0)
⎧ x = −t
⎪
∴ ⎨ y = 1 + 2t , t ∈ R
⎪ z = −3
⎩
x
y −1 z + 3
=
=
−1
2
0
x
y −1
∴
=
, z = −3
−1
2
Ex 7. For each case, find if the given point lies on the given
line.
r
a) L : r = (1,2,−3) + t (0,1,−2); P(1,4,−7)
(1,4,−7) = (1,2,−3) + t (0,1,−2)
(1,4,−7) − (1,2,−3) = t (0,1,−2)
(0,2,−4) = t (0,1,−2) ⇒ t = 2 ⇒∴ P ∈ L
⎧ x = −2 + 3t
⎪
b) L : ⎨ y = −t
; P (0,1,5)
⎪z = 5
⎩
⎧0 = −2 + 3t ⇒ t = 2 / 3⎫
⎪
⎬ ⇒∴ P ∉ L
⎨1 = −t ⇒ t = −1
⎭
⎪5 = 5 ⇒ true
⎩
x +1 y − 2
z
=
=
; P(−3,3,−3)
−2
1
−3
−3 + 1 3 − 2 −3
=
=
1
−3
−2
1=1=1⇒ P∈ L
r
Ex 7. Consider the line L : r = (3,−2,3) + t (−1,2,−3), t ∈ R . Find
the intersection points between this line and the coordinates
axes and planes.
c) L :
E Intersections
A line intersects the x-axis when y = z = 0 .
⎧x = 3 − t
⎪
L : ⎨ y = −2 + 2t
⎪ z = 3 − 3t
⎩
A line intersects the xy-plane when z = 0 .
x = 0 ⇒ t = 3 ⇒ y = −2 + 2(3) = 4, z = 3 − 3(3) = −6
∴ yz − int = A(0,4,−6) = L ∩ yz − plane
y = 0 ⇒ t = 1 ⇒ x = 3 − 1 = 2, z = 3 − 3(1) = 0
∴ xz − int = B(2,0,0) = L ∩ xz − plane
z = 0 ⇒ t = 1 ⇒ x = 3 − 1 = 2, y = −2 + 2(1) = 0
∴ xy − int = B(2,0,0) = L ∩ xy − plane
∴ x − int = B(2,0,0) = L ∩ x − axis
Note that y-intercept and z-intercept do not exist.
Reading: Nelson Textbook, Pages 445-448
Homework: Nelson Textbook: Page 449 #1abc, 5acf, 6, 8, 9, 12, 13, 14
8.3 Vector, Parametric, and Symmetric Equations of a Line in R
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