Calculus and Vectors – How to get an A+
8.1 Vector and Parametric Equations of a Line in R2
A Vector Equation of a Line in R2
Let consider the line L that passes through the
r
point P0 ( x0 , y0 ) and is parallel to the vector u .
The point P( x, y ) is a generic point on the line.
Ex 1. The vector equation of the line L is given by:
r
L : r = (0,1) + t (−1,2), t ∈ R
a) Find a direction vector for this line.
r
∴ u = (−1,2)
b) Find a specific point on this line.
∴ P0 = (0,1)
c) Find the points A and B on this line corresponding to
t = 1 and t = 4 respectively.
r
t = 1 ⇒ r = (0,1) + (−1,2) = (−1,3) ⇒∴ A(−1,3)
r
t = 4 ⇒ r = (0,1) + 4(−1,2) = (−4,9) ⇒∴ B (−4,9)
r
P0 P = tu
r
OP − OP0 = tu
r r
r
r − r0 = tu
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 .
d) Explain what does represent the equation:
r
r = (0,1) + t (−1,2), t ∈ [1,4]
∴ The equation represents the set of points of the line
segment AB , where A(−1,3) and B(−4,9) .
e) Verify if the points M (−2,5) and N (2,3) are or not on the
line. Hint: Try to find a t corresponding to each point.
(−2,5) = (0,1) + t (−1,2) ⇒ (−2,5) − (0,1) = t (−1,2)
(−2,4) = t (−1,2) ⇒ t = 2 ⇒∴ M ∈ L
(2,3) = (0,1) + t (−1,2) ⇒ (2,3) − (0,1) = t (−1,2)
⎧2 = −t ⇒ t = −2
⇒ no solution
(2,2) = t (−1,2) ⇒ ⎨
⎩2 = 2t ⇒ t = 1
∴N ∉L
Note. The vector equation of a line is not unique. It
Ex 2. Find two vector equations of the line L that passes
depends on the specific point P0 and on the
r
through the points A(2,−3) and B(−1,2) .
direction vector u that are used.
r
Let use P0 ≡ A(2,−3) and u = AB = (−3,5) . So:
r
∴ L : r = (2,−3) + t (−3,5), t ∈ R
r
Let use P0 ≡ B (−1,2) and u = 2 BA = 2(3,−5) = (6,−10) . So:
r
∴ L : r = (−1,2) + s (6,−10), s ∈ R
B Parametric Equations of a Line in R2
Let rewrite the vector equation of a line:
r r
r
r = r0 + tu , t ∈ R
as:
( x, y ) = ( x0 , y0 ) + t (u x , u y ), t ∈ R
Split this vector equation into the parametric
equations of a line in R2:
⎧ x = x0 + tu x
⎨
⎩ y = y0 + tu y
t∈R
Ex 3. Convert each vector equation into the parametric
equations.
r
a) r = (1,−3) + t (−2,5), t ∈ R
( x, y ) = (1,−3) + t (−2,5), t ∈ R
⎧ x = 1 − 2t
∴⎨
t∈R
⎩ y = −3 + 5t
r
b) r = (−2,0) + s (0,−3), s ∈ R
( x, y ) = (−2,0) + s (0,−3), s ∈ R
⎧ x = −2
∴⎨
⎩ y = −3s
s∈R
8.1 Vector and Parametric Equations of a Line in R
©2010 Iulia & Teodoru Gugoiu - Page 1 of 2
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Calculus and Vectors – How to get an A+
Ex 4. Convert the parametric equations of each
line into a vector equation.
⎧ x = −2 + 3t
t∈R
a) ⎨
⎩y = 5 − t
r
P0 (−2,5) and u = (3,−1)
r
∴ r = (−2,5) + t (3,−1), t ∈ R
⎧ x = 1 − 3s
b) ⎨
s∈R
⎩y = 2
r
P0 (1,2) and u = (−3,0)
r
∴ r = (1,2) + s (−3,0), s ∈ R
Ex 5. Find the points of intersection between the line given by
r
L : r = (1,2) + t (−1,1), t ∈ R and the coordinate axes.
⎧x = 1 − t
L:⎨
t∈R
⎩y = 2 + t
Let A = L ∩ x − axis (the intersection with the x-axis). Then
⎧x = 1 − t
y A = 0 and ⎨
⇒ x A = 1 − (−2) = 3 ⇒∴ A(3,0)
⎩0 = 2 + t ⇒ t = −2
Let B = L ∩ y − axis (the intersection with the y-axis). Then
C Parallel Lines
r
Two lines L1 and L2 with direction vectors u1 and
r
u 2 are parallel ( L1 || L2 ) if:
r r
u1 || u 2
or, there exists k ∈ R such that:
r
r
u 2 = ku1
or:
r
r r
u1 × u 2 = 0
or scalar components are proportional:
u2 x u2 y
=
=k
u1x u1 y
⎧0 = 1 − t ⇒ t = 1
x B = 0 and ⎨
⇒ y B = 2 + 1 = 3 ⇒∴ B (0,3)
⎩y = 2 + t
r
Ex 6. Consider the line L1 : r = (1,−3) + t (−1,2), t ∈ R . Find the
vector equation of a line L2 , parallel to L1 that passes
through the point M (−1,−13) .
r
Because L1 || L2 , a valid direction vector for L2 is u1 = (−1,2) .
So, a possible vector equation for L2 is:
r
L2 : r = (−1,−13) + s (−1,2), s ∈ R
D Perpendicular Lines
r
Two lines L1 and L2 with direction vectors u1 and
r
u 2 are perpendicular ( L1 ⊥ L2 ) if:
r
r
u 2 ⊥ u1
or:
r
r r
u1 ⋅ u 2 = 0
or:
u1x u 2 x + u1 y u 2 y = 0
Ex 7. Show that L1 ⊥ L2 where:
r
L1 : r = (−2,3) + t (2,−1), t ∈ R
E 2D Perpendicular Vectors
r
Given a 2D vector u = (a, b) , two 2D vectors
r
r
r
perpendicular to u are v = (−b, a) and w = (b,−a ) .
r
Ex 8. Consider the line L1 : r = (0,2) + t (2,−3), t ∈ R . Find the
vector equation of a line L2 , perpendicular to L1 that passes
through the point N (−3,0) .
r
r
r
r
u1 = (2,−3); u 2 ⊥ u1 ⇒ u 2 = (3,2)
r
∴ L2 : r = (−3,0) + s (3,2), s ∈ R
Indeed:
r r
r r
u ⋅ v = (a, b) ⋅ (−b, a) = −ab + ba = 0 ⇒ u ⊥ v
⎧ x = −2s
s∈R
L2 : ⎨
⎩ y = 5 − 4s
r
u1 = (2,−1)
r
u 2 = (−2,−4)
r r
u1 ⋅ u 2 = (2)(−2) + (−1)(−4) = −4 + 4 = 0 ⇒∴ L1 ⊥ L2
F Special Lines
A line parallel to the x-axis has a direction vector
r
in the form u = (u x ,0), u x ≠ 0 .
A line parallel to the y-axis has a direction vector
r
in the form u = (0, u y ), u y ≠ 0 .
Reading: Nelson Textbook, Pages 427-432
Homework: Nelson Textbook: Page 433 #2, 3, 4, 5, 8, 9, 10, 11, 13
8.1 Vector and Parametric Equations of a Line in R
©2010 Iulia & Teodoru Gugoiu - Page 2 of 2
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