VECTORS CHAPTERS 6-9
UNIT 3
LINES
Lesson
Homework
Booklet page 6
3
Objectives
 Define direction vector for a line, normal for a line
 Equation of a line in two space
 Vector equation of a line in two space
 Parametric equations of a line in two space
 Symmetric equation of a line in two space
 Scalar (Cartesian) equation of a line in two space
 Finding equations of lines in two space given clues
 Changing from one form of a line to another
 Recognizing equivalent equations of lines in two space
 Vector equation of a line in three space
 Parametric equations of a line in three space
 Symmetric equation of a line in three space
 Scalar (Cartesian) equation of a line in three space is
impossible
 Finding equations of lines in three space given clues
 Changing from one form of a line to another
 Recognizing equivalent equations of lines in three space
 Intersection of lines in two and three space
4

Booklet page 16
1
2



Unit 3 - Lines
Distance between a point and a line and parallel lines in
two and three space
Distance between skew lines in three space
Review
Test
Booklet page 9
Booklet page 13
Review Sheet page 17
.
Day 1: Lines in Two Space
Definition: A direction vector for a line is a vector parallel to the line. A line

d2
d
with direction vector  d1 , d 2  has slope m  d1 .
RECALL: Parallel (collinear) vectors can be written as scalar multiples of each
other.
To get the equation of a line, we need two points (because two points make a
direction vector) or a point and a direction vector.
Example – Sketch the line that passes through P(5,1) and has a direction vector

d   3,3 .

P
d
P0 ( x0 , y0 )
P ( x, y )
kd

p0
p

d  d1 , d 2 
Suppose we have a line  , with
point P0 ( x0 , y0 ) and direction

d  d1 , d 2 
Now suppose P ( x, y ) is a general
point on  . The point P0 has
position vector OP0  p0 and the
point P has position vector
OP  p
We can write P0 P  k d , where
k   , so that k d is any vector
collinear with d .
Unit 3 - Lines
Page 2 of 18.
By the triangle law, we have
OP  OP0  k d
( x, y)  ( x0 , y0 )  k (d1 , d 2 )
p  p0  k d
These equations are called the vector equations of a line in the plane (there are an


infinite number of representations). You could use any d and any p0 . The scalar
k is called a parameter.
If we equate the components, we get
 x  x0  kd1

 y  y0  kd2



which is called the parametric equations of a line in two space.
If we then solve for the parameter, we get the symmetric equation of a line in two
space.
x  x0  kd1
k
x  x0
d1

y  y0  kd 2
k
y  y0
d2
x  x0 y  y0

d1
d2
Example – Determine the vector equation for the line that is perpendicular to
p  (4,1)  k (3, 2), k  , and passes through point P(6,5).
{ p  (6,5)  k (2,3), k  }
Unit 3 - Lines
Page 3 of 18.

Definition: A normal vector for a line is a vector n  n1,n2  which is

perpendicular to the line. A line with direction vector d  d1 , d 2  has normal

n  d2 ,d1  .
n


d  d1 , d 2 
n   n1 , n2 

P0 P  n

P0 P  n  0
n
P ( x, y )
x  x0 , y  y0   n1 , n2   0
n1  x  x0   n2  y  y0   0
n1 x  n2 y   n1 x0  n2 y0   0
P0 ( x0 , y0 )

p0
p
n1 x  n2 y  C  0
Ax  By  C  0
n   n1 , n2 
So this last equation is the standard equation of a line which is now called the

scalar equation of a line (or Cartesian) in two space. Note that n   A, B
Ax  By  C  0
Unit 3 - Lines
Page 4 of 18.
Example – Determine vector and parametric equations for the line containing
points P(-1,5) and Q(6,11). Determine three other points on this line. Where
does this line cross the x-axis?
{ p  (1,5)  k (7,6), k   }
{answers vary} {(  41 ,0)}
6

Ex. Given y  x  3 which is a line with d  4,1 and P0 0,3
1
4
So the vector equation is x, y   0,3  t 4,1, t  
 x  4t
And the parametric equations are  y  3  t

And the symmetric equation is
x y3

4
1
4 y  12  x
Solving for y gives y  1 x  3 which is the scalar equation of a line in two
4
space. Right back to where we started.


:
r
Ex. Given 1 1  3,4  k 2,1 and  2 : r2   9,8  m 6,3 . Do these
{ 1 el  2 }
represent the same line?

Ex. Find the scalar equation of the line with n  2,5 , passing through A(1,3).
{2x-5y+13=0}
Unit 3 - Lines
Page 5 of 18.
Homework:
 x  3  2t
1. Given 
, find a scalar equation for the line.
 y  4  t
2. Given 2x-3y+6=0, find a vector equation for the line.
3.
{ x  2y  5  0 }
{ x, y    3,0  t 3,2 }
Convert each of the following equations to the requested form.

a) r  2,2  t  2,5 to scalar form.
b) 2 x  y  6  0 to vector form.
{5x+2y-6=0}
{(x,y)=(3,0)+k(1,2)}
4. Find a direction vector for a line which is:
a) perpendicular to 3x  7 y  21  0 . {(3,7)}
b) parallel to x, y    1,0  k 3,8
{(3,-8)}
5. Find the symmetric equation of the line through P 1,5 with slope
7
5
.
 x 1 y  5



7 
 5

6. Find the scalar equation of the line with normal n   3,1 passing through P4,3
{3x-y-9=0}
7. Find the vector equation of the line perpendicular to 2 x  3 y  9  0 and passing through the y
 x  4  k
intercept of 
.
 y  3  2k
{(x,y)=(0,-5)+t(2,-3)}
8. Find the scalar equation of the line perpendicular to 3x  y  7  0 passing through P2,3 .
{x+3y+7=0}
9. Do these represent the same line?
a)
 1 : x, y    1,3  t  2,4
 2 : 2x  y  1  0
{identical}
b)
 1 : x, y    1,4  t  1,4
 2 : 4x  y  3  0
{parallel}
Section 8.1 Page 433-434 #2, 4, 5, 9, 10, 12
Section 8.2 Page 443-444 #1, 3, 4, 6, 7, 13
Unit 3 - Lines
Page 6 of 18.
Day 2: Lines in R3
Definition: A direction vector for a line is a vector parallel to the line. A line with direction vector

d  d1 , d 2  has slope m 
P(x,y,z)

p
d2
d1
P0 x0 , y0 , z0 
.
Line


kd 
p0
O

d  d1, d2 , d3 

P0 P  kd
x  x0 , y  y0 , z  z0   k d1 , d 2 , d3 
x, y, z   x0 , y0 , z0   k d1 , d 2 , d3 

OP  OP0  kd

 
p  p0  kd
Unit 3 - Lines
Page 7 of 18.
These last three equations are called the vector equation of a line in space (there are an infinite number of
representations). You could use any
If we equate the
components, we get
the parametric
equations
of a line in three
space.

d
and any
 x  x0  kd1

 y  y 0  kd2
 z  z  kd
0
3


p0 . The scalar k is called a parameter.
Then solve for the
parameter k to get
the symmetric
equation of a line
in three space


x  x0 y  y0 z  z0


d1
d2
d3

Definition: A normal vector for a line is a vector n  n1 , n2 , n3 which is perpendicular to the line.
So …. there is no scalar equation of a line (or Cartesian) in three space because there is no unique normal for a
line in 3-space.
Ex 1. Find vector, parametric, and symmetric equations of the line that passes through P1 2,4,0 and
P2 5,0,7 . Does the point Q(-4,12,-14) lie on that line?
Ex 2. Given

1 : r1  3,4,3  k 2,1,5 and  2 : r2   9,8,6  m 6,3,15 . Do these represent the
{ 1 el  2 }
same line?
Ex 3. Find symmetric equations of the line through A(1,2,3) and B(2,-1,3).
Ex 4. Do the equations
1 :
x  5 y  4 z 1


2
3
5
and
2 :
x 1 y 1 z  3


4
6
10
represent the
same line?
Unit 3 - Lines
Page 8 of 18.
Homework:
1. Convert each of the following equations to the requested form.
x8 y 9 2 z


to vector form.
5
2
1
 x  3  4t

b)  y  t
to vector form.
 z  2  2t

a)
x, y, z    8,9,2  t 5,2,1
{ x, y, z    3,0,2  t 4,1,2 }
 x  1  2k

 y  4  2k
 z  5  2k

c) x, y, z    1,4,5  k 2,2,2 to parametric form.
2. Find the parametric equations of the line through A 3,2,3 and is perpendicular to both  1 and  2
 x  3  t
x4
z 3

 y2
where  1 :
and  2 : x, y, z    1,1,5  k  1,2,3
y  2  t
5
4
3  t

3. Give the coordinates of three points on the line x, y, z   1,1,2  k 3,1,1
{many answers}
4. Find the symmetric equation of the line through the origin parallel to the line through A(4,3,1) and
x y
z
B(-2,-4,3).
{  
}
6 7 2
5. For each of the following pairs of lines, determine whether they are identical, parallel or neither.


a) r1  1,0,3  t 3,6,3 and r2  2,2,5  m2,4,2
{parallel}


b) r1  2,1,4  t 3,0,6 and r2   3,0,1  m2,0,2
{neither}


c) r1  1,1,1  t 6,2,0 and r2   5,3,1  m 9,3,0
{identical}
6. Which of the following points lies on the line x  2t , y  3  t , z  1  t ?
P(2,4,2)
Q(-2,2,1)
R(4,5,2)
S(6,6,2)
{P}
7. Find parametric equations of the line:
x 1 y  2

 z  3 and passes through the origin.{ x  3t , y  2t , z  t }
a) that is parallel to the line
3
2
b) that passes through Q(6,-4,5)and is parallel to the y axis.
{ x  6, y  4  t , z  5 }

c) that has a z intercept of -3 and direction vector d  1,3,6  .
{ x  t , y  3t , z  3  6t }
8. Find a vector equation of the line through A(2,0,-3) and B(-3,2,-2)
{ x, y, z   2,0,3  k 5,2,1 }
Section 8.3 Page 449-450 #4, 5, 7, 9, 14
Unit 3 - Lines
Page 9 of 18.
Day 3: Intersection of Lines in Two and Three Space
Finding the point of intersection of two lines is also referred to as solving a system of linear
equations.
The solution can be classified in the following manner:
System of
equations
Consistent
(solution(s))
Inconsistent (no
solution)
R2: lines are || but not
the same
3
R : lines are || but not
same or lines are
skew
Inconsistent
parallel
Independent
(one solution)
in R2: lines are not ||
in R3: lines are not || or
skew
Consistent
Dependent
(  ’ly many
solutions)
in R2 and in R3,
lines are || and
equal
1,  2
skew
Unit 3 - Lines
Page 10 of 18.
Examples: Find and classify the intersection of the following lines
 1 : ( x, y, z )  (2,0,3)  t (5,1,3)
Example a)  : ( x, y, z )  (5,8,6)  s (1,2,3)
2
x  2  5t
x 5s
1 : y  t
 2 : y  8  2s
z  3  3t
z  6  3s
set coordinates equal to each other
 2  5t  5  s
5t  s  7
1
t  8  2s
 3  3t  6  3s
t  2s  8
3t  3s  3
2
2 1 +2:

11t  22
t2
sub t  2 and s  3 into
1
-5 2 :
3
11s  33
s  3
3
LS  3(2)  3(3)
69
 3
 RS
 point of intersection:
x  2  5(2)  8
y2
z  3  3(2)  3
The point is (8,2,3) and the system is consistent and independent.
1 : x  2 y  3
1
Example b)  : 3 x  6 y  9
2
2
3
1
2
- :
0x+0y = 0
This is always true.  lines are equal and have
 ’ly many points of intersection.
The system is consistent and dependent.
Unit 3 - Lines
Page 11 of 18.
For solution, set one coordinate equal to a variable and solve for the other coordinate in terms of
this variable.
Let y  a , sub into
1
x  2a  3
x  3  2a
solution, (3  2a, a )
a 
 1 : r  (2,1,0)  t (1,2,3)
Example c)  : x  1  y  1  z  2
2
2
1
1
x  2t
 1 : y  1  2t
sub into
z  3t
2
x 1 y 1 z  2


2
1
1
2  t  1  1  2t  1

2
1
1
t
5
 1  2t  1  3t  2

1
1
t 0
t' s are not equal  no solution
d1  (1,2,3)
d 2  (2,1,1)
d1  m d 2
Unit 3 - Lines
 skew lines
Page 12 of 18.
Homework:
1. Find the intersection of the following lines. Classify the system as consistent or inconsistent, if possible
dependent or independent
 : 5 x  2 y  25  0
a) 1
 2 : 5x  2 y  5  0
c)
e)
g)
i)
b)
 1 : x, y    1,3  t  2,4
 2 : 2x  y  1  0
 1 : x, y    1,4  t  1,4
 2 : 4x  y  3  0
 1 : 2 x  3 y  30  0
 2 : 2x  y  3  0
 x  13  3k

d)  1 : x, y, z    1,3,5  t 1,2,6 and  2 :  y  8  k
 z  1  3k

 x  13  3k

f)  1 : x, y, z    1,3,5  t 1,2,6 and  2 :  y  8  k
 z  8  3k

 x  18  3k
 x  5  2t
h)  1 : 
and  2 : 
 y  2  2k
y  4  t
 2 : x  2 y  13  0
 x  1  3k

 1 :  y  1  4k and  2
 z  2k

 1 : 2 x  y  14  0
 x  1  2t

:  y  3t
 z  7  t

j)  1 : x, y, z   2,1,0  t 1,1,1 and  2 : x, y, z   3,0,1  k 2,3,1
Answers:
a) system is inconsistent, lines are parallel
b) system is inconsistent, lines are parallel
c) system is consistent and dependent, solution is (0.5  0.5a, a ), a  
d) system is inconsistent, lines are skewed
e) system is inconsistent, lines are parallel
f) system is consistent and independent, solution is (2, -3, 23)
g) system is consistent and independent, solution is (3,8)
h) system is consistent and independent, solution is (3,8)
i) system is consistent and independent, solution is (5, 9, -4)
j) system is inconsistent, lines are skewed
Unit 3 - Lines
Page 13 of 18.
Day 4:
Distances for Lines in Two Space and Three Space
Recall:
u
 
 u v
Pr ojv u  
v
v
Pr ojv u
TWO SPACE
a) Finding the distance between a point and a line in two space:

Find the magnitude of the projection of the vector between the point and a point on the line onto the normal
for the line. This will be the distance between the point and the line.
Distance
n
Pr ojn PP0
P0
In two space, a line has a normal and the
distance is equal to the magnitude of the
projection of PP0 onto n where
P0 is a given point and P is a point on the line
(you pick)
P
dist  Pr ojn PP0

n  PP0


n
This works in R2 only! Why?
OR
 Find the magnitude of the projection of the vector between the point and a point on the line onto the
direction vector for the line. Use this and Pythagorean’s Theorem to get the distance between the point and
the line.
Pythagorean Theorem:
P0
2
dist  Pr ojd PP0  PP0
2
2
2
2
distance
P
Pr ojd PP0
dist  PP0  Pr ojd PP0
2
This works in R2 and R3. Why??
b) Finding the distance between parallel lines in two space:

Select a point on one line and find the distance between a point and a line as outlined above.
Unit 3 - Lines
Page 14 of 18.
THREE SPACE
a) Finding the distance between a point and a line in three space:


Find the magnitude of the projection of the vector between the point and a point on the line onto the
direction vector for the line. Use this and Pythagorean’s Theorem to get the distance between the point and
the line.
OR
Find the magnitude of the cross product of the vector between the point and a point on the line with the
direction vector for the line. Divide this number by the magnitude of the direction vector. This will give you
the distance between the point and the line. Below is the derivation of this formula:
P0
PP0

distance
P
d
Using trig. we have,
sin  
We also have
PP0  d  PP0 d sin  eˆ
PP0  d  PP0 d sin 
dist
PP0
PP0 sin  
dist  PP0 sin 
PP0  d
 dist 
d
PP0  d
d
b) Finding the distance between parallel lines in three space:
 Select a point on one line and find the distance between a point and a line as outlined above.
c) Finding the distance between skew lines in three space:
 Find the magnitude of the projection of the vector between the point on one line and a point on the other line
onto the vector which is perpendicular to both lines.
The shortest distance occurs on a line that
is  to both 1 and 2 .
 The direction vector for this line is
1
P1
d1  d 2
given by
d1  d 2
 dist  Pr ojd d P1 P2
1
distance
P2
2
Unit 3 - Lines

2
d  d  P P
1
2
d1  d 2
1 2
Page 15 of 18.
Examples: Find the distance from:
1)
Point (-3,7) to the line 2 x  y  5  0
2)
Point (1,-1,2) to the line
3)
{2.68}
: ( x, y, z )  (3,5, 3)  k (1,1, 1)
1
: ( x, y, z )  (1, 3, 2)  k (1, 1, 2)
2
: ( x, y, z )  (3,1, 4)  t (3,1, 4)
Homework:
1. Find the distance between the parallel lines:
 1 : 5 x  2 y  25  0
30 29
b)
.
{
}
29
 2 : 5x  2 y  5  0
{2.94}
{1.83}
b)
 1 : 2 x  y  14  0
 2 : 2x  y  3  0
2. Find the distance between:
a) the point A 6,5,3 and the line x, y, z   6,1,3  t 5,3,3
b) the lines

 1 : r  1,6,2  t 1,2,5

 2 : r  3,4,9  k  2,7,1
{
11 5
}
5
{2.76}
{0.387}
c) the point A 1,8,4 and the line x, y, z   3,4,0  t  2,7,3
{8.53}

 1 : r  3,5,2  t  3,1,4
d) the lines

 2 : r  3,3,1  k 2,3,9
{6.4}

 1 : r  4,2,4  t  3,1,4
e) the lines

 2 : r  3,7,1  k 6,2,8
{9}
f) the lines

 1 : r  7,1,3  t  1,4,2

 2 : r   3,6,5  k  2,8,4
{9.84}
Section 9.1 Page 496-498 #8, 9, 11, 18
Section 9.5 Page 540-541 #1, 2, 4, 8, 10
Section 9.6 Page 549-550 #7, 8
Unit 3 - Lines
Page 16 of 18.
MCV 4U Lines Review
1. Convert each of the following equations to the requested form.
x8 y 9 5 z



a)
to vector form.
b) r  2,1  t  2,7 to scalar form.
4
2
1
 x  3  4t

c)  y  t
to symmetric form.
d) x, y, z    1,4,9  k 5,2,2 to parametric form.
 z  2  2t

e) 3x  5 y  15  0 to vector form.
Answers: a)
c)
x, y, z    8,9,5  k 4,2,1, kR
b) 7x + 2y – 12 = 0
x3
y
z2


4
1  2
d) x  1  5k , y  4  2k , z  9  2k
e) x, y, z   0.  3  k 5,3, kR
2. Find a normal vector for a line which is:
a) parallel to 5 x  6 y  15  0 .
b) perpendicular to x, y    1,0  k 3,8
Ans: a) (5, - 6) b) (3, - 8)
3. Find the symmetric equation of the line through P 3,5 with slope
9
4
.
Ans:
x3 y 5

4
9

4. Find the scalar equation of the line with direction vector d   3,1 passing through P 4,7 .
Ans: x + 3y – 17 = 0
5. Find the requested version of the line described in each of the following:
a) Vector equation of the line parallel to 4 x  3 y  12  0 and passing through the y intercept of
 x  4  2k
.

 y  3  5k
b) Scalar equation of the line perpendicular to 2 x  3 y  8  0 passing through P2,3 .
c) Parametric equations of the line through A 3,2,5 and is perpendicular to both  1 and  2 where
x4
z 3
1 :
 y2
 2 : x, y, z    1,1,5  k  1,2,3
3
2
Answers: a) x, y   0,7  k 3,4
b) 3x + 2y = 0
c) x = - 3 – t, y = 2 – 7t, z = 5 – 5t
6. Find the intersection of the following lines. Then classify them as an inconsistent or consistent, dependent or
independent system.
 x  13  3k
 1 : x, y    1,3  t  2,4

a)
b)  1 : x, y, z    1,3,5  t 1,2,6 and  2 :  y  8  k
 2 : 2x  y  1  0
 z  1  3k

Unit 3 - Lines
Page 17 of 18.
x  2 y 1 z
 x  3  2u
 x  1 t




1
1
1
c)
d)  1 :  y  1  2t and  2 :  y  5  4u
x  3 y 1 z
 z  5  6u
 z  1  3t
2 :
 


2
3
1
answers: (t , 1  2t ), t   , { system is inconsistent }, { ( 3, 0, 1)}, { (a, 2a - 1, 4 - 3a)}
1 :


7. Show that the lines r  (4, 7,  1)  t (4, 8,  4) and r  (1, 5, 4)  u (1, 2, 3) intersect at right angles and find
the point of intersection.
{ ( 2, 3, 1)}
8. Find the distance between:
 : 5 x  2 y  25  0
a) The point (3, 7) and the line 2 x  3 y  7
b) lines: 1
 2 : 5x  2 y  5  0
c) the point A 6,5,3 and the line x, y, z   6,1,3  t 5,3,3
 x  4t

 1 : r  1,6,2  t 1,2,5
x 1 y  4


, z  1 and  2 :  y  1  2t
d) the lines
e) lines:  1 :

2
1
 2 : r  3,4,9  k  2,7,1
z6

Answers:
{ 6.10 units}, { 5.57 units}, { 2.76 units}, { 0.39 units}, { 7.01 units}
Unit 3 - Lines
Page 18 of 18.