LECTURE NOTES VECTOR

LECTURE 3 OF 5

Subtopic:
5.4 Applications of

Vectors in Geometry

LEARNING OUTCOMES
At the end of the lesson, students
should be able to:

a) find the equation of a straight line.
b) determine the angle between two

straight lines.

5.4 a) Equation of a Straight Line

vR

Ar
a
O

Suppose that R(x, y , z ) is a point which is
free to move on a line containing a fixed point
A(x1 , y1 , z1 ) .

If v = ai + b j + c k is a direction vector of the

line, it is clear that a line consists precisely
of those points for which the vector AR is

parallel to v , that is

AR = tv for some scalar t

OR − OA = tv

OR = OA + tv

r = a + tv is the vector equation of the

straight line.

It can be written as
xi + y j + zk = (x1i + y1 j + z1 k) + t(a i + b j + c k)

By equating the coefficients of i , j and k ,

we have

x = x1 + ta , y = y1 + tb , z = z1 + tc

is the parametric equations of the straight
line.

From the parametric equations, solving for t ,

t = x − x1 , t = y − y1 , t = z − z1
abc

x − x1 = y − y1 = z − z1
abc

is the Cartesian/ Symmetry equation of the
straight line.

Example 1

Find the vector equation, the parametric
equations and the Cartesian equation of a
straight line that passes through the point
(1, –2, 3) and is parallel to the vector
4i +5j −6k .

Solution
a =i − 2 j +3k
v = 4i +5j −6k

The vector equation is

r = a + tv

r = (i − 2 j + 3k) + t(4i + 5 j − 6k)

The parametric equations are

x =1+ 4t , y = −2 + 5t , z = 3 − 6t

The Cartesian equation is

x −1 = y + 2 = z −3
4 5 −6

Example 2
Find the vector equation of a line that passes
through points A(1, 3, 1) and B(4, -1, 2).

Solution
a =i +3j +k

v = AB

= OB − OA

= (4i − j + 2k) − (i + 3 j + k)
= 3i − 4 j + k

Vector equation of line

r = a + tv

r = (i + 3 j + k) + t(3i − 4 j + k)

5.4 b) Angle Between Two Straight Lines

The angle between two straight lines equals the
angle between two direction vectors of the lines.

Suppose the vector equation of two straight
lines are

r1 = a1 + tv1 r1

r2 = a2 + sv2 

r2

with t, s are any scalar and  is angle

between two straight lines.

The angle between two straight lines is
given by

 v1  v2 

 = cos−1 
 v1 v2 

Example 3
Find the angle between two straight lines

r1 = (2i + 2 j − 4k) + t(i + 3 j − 3k)

and r2 = (i + j + k) + s(i + 2 j − 4k)

Solution

v1 = i + 3 j − 3k , v2 = i + 2 j − 4k

v1  v2 = (i + 3 j − 3k)  (i + 2 j − 4k)

= 1+ 6 +12
= 19

v1 = (1)2 + (3)2 + (−3)2 = 19

v2 = (1)2 + (2)2 + (−4)2 = 21

 v1  v2 

 = cos−1 
 v1 v2 

 = cos −1  19 
 
 ( 19)( 21) 

= 18.00

LECTURE 4 OF 5

Subtopic:
5.4 Applications of

Vectors in Geometry

LEARNING OUTCOMES
At the end of the lesson, students
should be able to:

c) find the equation of a plane.

5.4 c) Equation of a Plane

n = ai + b j + ck

R(x, y , z )
A(x1 , y1 , z1 )

Suppose a plane in space has normal vector

n = ai + b j + ck and that it passes through

the fixed point A(x1 , y1 , z1 ) and R(x, y , z )

is any point on the plane.

Then AR is perpendicular to n .

AR  n = 0
(OR − OA)  n = 0

(r − a)  n = 0
rn−an = 0

rn = an

is a vector equation of the plane.

 x   a  x1   a
 y   b  =  y1   b 
 z   c   z1   c 

ax + by + cz = d

is a Cartesian equation of the plane

where d = ax1 + by1 + cz1

The equation of a plane requires a fixed point
on the plane and its normal vector.

Example 1
Find the vector equation of the plane with

normal vector i + 2 j + 3k and containing the

point (-1, 2, 4).

Solution

n =i +2 j +3k , a =−i + 2 j + 4k

Vector equation of the plane

rn = an

r  (i + 2 j + 3k) = (−i + 2 j + 4k)  (i + 2 j + 3k)

= −1+ 4 +12

r  (i + 2 j + 3k) = 15

Example 2

Find the Cartesian equation of the plane
passes through (-5, 1, 2) and perpendicular

to line r = (2i + 2 j − 4k) + t(5i − 3 j + 7 k) .

Solution v = 5i −3j + 7k
n = 5i −3j + 7k
(−5,1, 2) , n = 5i − 3 j + 7 k

Cartesian equation of the plane
5x − 3y + 7z = 5(−5) − 3(1) + 7(2)

5x − 3y + 7z = −14

Example 3

Given the points A(1, 3, 1), B(4, -1, 2),

C(12, 0, 1). Find the equation of the plane

ABC in the Cartesian form. n
Solution
CB
AB = OB − OA A

= (4i − j + 2k) − (i + 3 j + k)

= 3i − 4 j + k

AC = OC − OA

= (12i + k) − (i + 3 j + k)

= 11i − 3 j

n = AB  AC

i jk
= 3 −4 1

11 −3 0

= (0 + 3)i − (0 −11) j + (−9 + 44)k

= 3i +11j + 35k

A(1, 3,1)

Cartesian equation of the plane

3x + 11y + 35z = 3(1) + 11(3) + 35(1)
3x + 11y + 35z = 71

Example 4

Find the Cartesian equation of the plane
passes through A(1,-1,3) and perpendicular

to both planes x − y + 2z = 3 and
2x + y − z = 3 .

Solution
A(1, −1, 3) , n1 =i − j + 2k , n2 = 2i + j − k

n = n1  n2 (a b is perpendicular to both a and b )
ijk

= 1 −1 2
2 1 −1

n = (1− 2)i − (−1− 4) j + (1+ 2)k
= −i + 5 j + 3k

Cartesian equation of the plane

− x + 5y + 3z = −(1) + 5(−1) + 3(3)
− x + 5y + 3z = 3

Example 5

Find the vector equation of the plane that
contains (2, -1, 3) and is
a) parallel to the xy-plane.
b) parallel to the plane with equation

3x – y + z = 2.

Solution

a) a = 2i − j +3k , n = k

Vector equation of the plane n =k

rn = an y
x
r  k = (2i − j + 3k)  (k)

rk =3

b) a = 2i − j +3k , n = 3i − j + k n = 3i − j +k

Vector equation of the plane

rn = an

r  (3i − j + k) = (2i − j + 3k)  (3i − j + k)

= 6+1+3
r  (3i − j + k) = 10

Example 6

Given the point P(-4, 2, -3) and the straight line

L : x + 2 = y = z +1 . Find a Cartesian
4 −3 5

equation of the plane containing the point P
and the straight line L .

Solution n

P(−4, 2, − 3) , Q(−2, 0, −1) v = 4i−3j+5k Q

P

PQ = OQ − OP L: x + 2 = y = z +1
4 −3 5

= (−2i − k) − (−4i + 2 j − 3k)

= 2i − 2 j + 2k

v = 4i−3j+5k

n = v  PQ
i jk

= 4 −3 5
2 −2 2

= (−6 +10)i − (8 −10) j + (−8 + 6)k

= 4i + 2 j − 2k

Cartesian equation of the plane

4x + 2y − 2z = 4(−4) + 2(2) − 2(−3)
4x + 2y − 2z = −6

2x + y − z = −3

LECTURE 5 OF 5

Subtopic:
5.4 Applications of

Vectors in Geometry

LEARNING OUTCOMES
At the end of the lesson, students
should be able to:

d) determine the angle between
i) two planes.
ii) a line and a plane.

e) determine the point of intersection between
a line and a plane.

5.4 d i) Angle Between two Planes

n1  P2

P1 n2



Consider two planes P1 and P2 whose

vector equations are r  n1 = d1 and

r  n2 = d2 .

The angle between two planes P1 and P2
is equal to the angle between two normals

n1 and n2 .

Angle between two planes P1 and P2 is

 n1  n2 
.
 = cos−1 
 n1 n2 

Example 1

Determine the angle between two planes

1 : 2x − 3z = 0 and  2 : − x + 2 y − 3z = 4.

Solution a x + by + cz = d

n1 = 2i − 3k , n2 = −i + 2 j − 3k n = ai + b j + ck

n1  n2 = (2i − 3k)  (−i + 2 j − 3k)
= −2 + 9
=7

n1 = (2)2 + (−3)2 = 13

n2 = (−1)2 + (2)2 + (−3)2 = 14

 n1  n2 

 = cos−1 
 n1 n2 

 = cos−1  7 
 
 ( 13)( 14) 

= 58.70

5.4 d ii) Angle Between a Line and a Plane

v αn



Consider the line r = a + tv and the plane
rn=d .

The angle  between the line and

the normal to the plane is given by

 = cos−1  vn 
 vn  .

Since  and  are complementary angles,
so  = 900 − , where  is angle between line

and plane.

Example 2

Given the straight line L: x+2 = y= z +1
4 −3 5

and the plane  : 2x + y + 3z = 9 . Find an

acute angle between the straight line L

and the plane  .

Solution x − x1 = y − y1 = z − z1
abc
v = 4i − 3 j + 5k , n = 2i + j + 3k
v = ai+bj+ck
v  n = (4i − 3 j + 5k)(2i + j + 3k)

= 8 − 3+15

= 20

v = (4)2 + (−3)2 + (5)2 = 50

n = (2)2 + (1)2 + (3)2 = 14

 = cos−1  vn 
 vn 

 = cos−1  20 
 
 ( 50)( 14) 

= 40.90

 = 90 − 40.90

= 49.10

The angle between the straight line and the
plane is 49.10.

5.4 e) Point of Intersection Between a Line
and a Plane

If a line passes through a plane, there must
be a point of intersection.

Let r = a + tv is a line equation and

r  n = d is a plane equation.

We substitute equation of line (in parametric
form) into equation of plane (in Cartesian
form), and solve for the value of t.

Substitute the value of t into parametric
equation of line, can get the intersection
point between line and plane.

Example 3
Given the straight line L : x + 2 = y = z +1

4 −3 5

and the plane  : 2x + y + 2z = 9 . Find an

intersection point between the straight line
L and the plane  .

Solution x − x1 = y − y1 = z − z1
Parametric equation of line L abc
x = −2 + 4t , y = −3t , z = −1+ 5t
x = x1 + ta y = y1 + tb
z = z1 + tc

Substitute equation line L into equation
plane 

2(−2 + 4t) + (−3t) + 2(−1+ 5t) = 9

− 4 + 8t − 3t − 2 +10t = 9
15t = 15

t =1

x = −2 + 4(1) = 2
y = −3(1) = −3

z = −1+ 5(1) = 4

The intersection point is (2, -3, 4)