3D Vectors and Space

BMM 104: ENGINEERING MATHEMATICS I Page 1 of 22

CHAPTER 3: 3-D SPACE VECTORS

Basic Concept of Vector

A vector is a quantity that having a magnitude/length (absolute) and a direction.

→ →
Notation: i. AB or a is vector
→ →

ii. AB or a is modulus/absolute value/length of the vector

→ →
BA is opposite to AB

→ →
BA = − AB
→ →

Meanwhile AB = BA .

Characteristics of vectors


→ → → → → →
1. If AB is parallel to c , then AB = k c OR c = t AB where k and t are the
scalars or parameters.
→ →
2. If k < 0 OR t < 0 , then AB and c are in the opposite directions.

3. If k > 0 OR t > 0 , then they are in the same direction.

Example:

→ →
→ 1 →
i) AB = 2 c or c= AB
2
→

ii) 1→ or
→ →
AB = − d d = −2 AB
2

Addition Law of Vectors
→ → → →
Let c = AB , d = BC . Refer to below diagram.

→ → → → →
AC = AB + BC = c + d

→ →
If OA is the position vector for A and OB is the position vector for B then

→ → →
OA+ AB = OB

→ → → → → →
AB = OB − OA BUT AB ≠ OA −OB

→
Example: Find NS = ?

Components of Vectors

2-D Space


i j
2 basic vectors : ~ and ~

They are also called unit vectors as i =1 and j =1
~ ~

→
Position vector of point A( a ,b ) is given by OA = a ~ + b j .
i
~
→

Absolute/ modulus OA = a 2 + 2
b by Pythagoras Theorem

Example:
→
i) OA = −3 i + 4 j
~ ~
→
OA = 9 +16 =5

3-D Space


o All the x-axes, y-axes are perpendicular to each other.
o There are 3 basic vectors: i , j , k .
~ ~ ~

o They are all unit vectors that parallel to the axes respectively and thus they also
perpendicular to each other.

a 
→  
Position vector of A( a ,b , c ) is OA = a ~ + b j + c k = ( a ,b , c ) =  b  and
i
~
~
c 
 
→

Length is OA = a 2 + 2 + 2
b c

3 
 
→ →

Example: Given OA =  4  . Find OA =?
6 
 

Addition of vectors for 3-D Space
→ →
The sum of two vector OA = ( a1 , a 2 , a 3 ) and OB = ( b1 ,b2 ,b3 ) is the vector formed by
adding the respective component;


→ →
OA+ OB = ( a1 + b1 , a 2 + b2 , a 3 + b3 )

Subtraction of vectors for 3-D Space
→ →
The subtraction of two vectors OA = ( a1 , a 2 , a 3 ) and OB = ( b1 ,b2 ,b3 ) is the vector
formed by adding the respective component;
→ → →
AB = OB − OA

→ →
OB − OA = ( b1 − a1 ,b2 − a 2 ,b3 − a 3 )

Example: Given A( −1,1,4 ) , B ( 8 ,0 ,2 ) and C ( 5 ,− ,11) . Find
2

→
(i) OA
→
(ii) OB
→
(iii) OC
→
(iv) AB
→
(v) AC
→

(vi) AB
→

(v) AC

→
Unit Vector v~
in the Direction of v
→

→
→
v
v=
~
→
→
is a unit vector in the direction of v .
v


 − 1
→  →
Example: Given v =  2  . Find unit vector in the direction of v ?
 3 
 

Dot/ Scalar Product
→ →
Notation for Dot product: a • b ∈ ℜ

→ →
The dot product of a = ( a1 , a 2 , a 3 ) and b = ( b1 ,b2 ,b3 ) is the real number a • b obtained
→ →

by
→ → →→
a• = a
b b cos θ

where θ is the angle between a and b and 0 ≤ θ ≤ π .
→ →

When we measuring angle between two vectors, the vectors must have the same initial
point.

→ → → →
Example: (i) Given a =6 and b =7. Find a • b .
→ →
= and θ =
π → →
(ii) Given a =5 and b 4 . Find a • b .
4


Example:

i • i = i i cos 0
~ ~ ~ ~

i • i = 1 ×1 ×1 = 1
~ ~

Important:

(a) j• j = k • k = i • i = 1
~ ~ ~ ~ ~ ~

(b) i• j = i
~ ~ ~
j cos 90 
~

(c) i• j = 1×1×0 = 0
~ ~

(d) i • j = 0 = i • k = j• k
~ ~ ~ ~ ~ ~

.
→ →
→ →
If a ⊥ b then a • b = a b cos 90  = 0 .
~ ~

Note:


~ ~ → →
(i) If a• b = 0 then a ⊥ b
→ → → →
(ii) a× b ≠ b× a

Properties of the dot product
→ → → →
a) a• b = b• a : Cumulative
→ → → → → →
b •a = b a cos θ=a •b

→
→ → → → → →
b) a •b + c  = a • b + a • c
 
: Distributive
→ →  → → →  →
c) k a • b  = k a  • b = a •k b 
     

Example:
 5   8 
  → 
→  → →
Finding angle between the vectors a =  3  and b =  − 9  given that a • b = −9 .
− 2  11 
   
Formula to compute scalar product

 a1   b1 
→ →    
a • b =  a 2  •  b2 
a  b 
 3  3
→ →
a • b = a1b1 + a 2 b2 + a 3 b3

 3  − 2 
  →→   → → → →
Example: Given a =  4  and b =  6  . Verify a • b = b • a .
 − 1  3 
   

Remark: By using addition law of vectors and the law of cosine, it can be showed that
→ →
a b + b + b = a b cos θ.
a 1 a
1 2 2 3 3

→
→ →
Component of a in the direction of n : Denoted as comp n a →


OQ
cos θ =
OP
OQ = OP cos θ
a cos θ
→

=

→ →
OQ is called component of vector a in the direction of n

→
OQ = Comp→ a
n

= OP cos θ
= a cos θ
→

^
n=1
→ ^
= a cos θ ⋅ n since
~
~

By the definition of product

→ → ^
comp → a = a • n
n ~

Example:

Find
2  1 
→   →
→  
(i) comp → a given a =  1  and n = 1  .
n
1  1 
   


 2  1 
→  →  →  
(ii) comp → a given a =  − 1  and n = 1  .
n
 −7  1 
   

Application of Dot Product : WORK DONE

Work done = Magnitude of force in the direction of motion times the distance it travels
= F cos θPQ
→ →

→ →
W =F •PQ

 1 
→  
Example: A force F =  − 2  causes a body to move from P (1,− ,2 ) to Q (7 ,3 ,6 ) .
1
 3 
 
Find the work done by the force.

Vector/ Cross Product
→ →
Definition: a×b is defined as a vector that

→ → → →
1. a×b is perpendicular to both a and b .

→ → → → → →
( a× b ) • a = 0 and ( a× b ) • b = 0


→ → → →
2. Direction of a×b follows right-handed screw turned from a to b

→ → → →
b× a = − a× b

b sin θ
→→
→ →
3. Modulus of a×b is a

Remarks:

a sin θ
→→
→ →
Modulus of b×a is therefore b

→ → → →
a× b ≠ b× a
→ → → → → → → →
a× b = − b × a or b× a = − a× b


→ →
a×b ∧
=e
Unit Vector : → →
a×b
~

→ → → → ∧ → → ∧
a×b = a×b e = a b sin θ e
~ ~

4. i× i = 0 j× j = 0 k× k = 0
~ ~ ~ ~ ~ ~

i× j = i
~ ~ ~
j sin 90  k =k
~ ~ ~
and ~ ~ ~ ~
( )
j×i = j i sin 90  −k = −k
~ ~

Similarly, j× k = i and k× j = −i
~ ~ ~ ~ ~ ~

k× i = j and i× k = − j
~ ~ ~ ~ ~ ~

→ →
Determinant Formula for a×b


i j k
→ → ~ ~ ~
a× b = a1 a2 a3
b1 b2 b3

a2 a3 a1 a3 a1 a2
= i− j+ k
b2 a3 ~ b1 b3 ~ b1 b2 ~

= ( a 2 b3 − a 3 b2 ) i − ( a1b3 − a 3 b1 ) ~ + ( a1b2 − a 2 b1 ) k
~
j
~

 3 
→ → → → →
→
 → →
→
Example: Find a×b , b×a and verify that a× b = −b× a given a =  − 4  and
 2 
 
 9 
→  
b = −6  .
 2 
 

Applications of Cross Product

1. The moment of a force
→ →
A force F is applied at a point with position vector r to an object causing the object to
rotate around a fixed axis.


As the magnitude of moment of the force at O is

M 0 = F • d = ( Magnitude of force perpendicular to d ) × (Magnitude of displacement)
→ → → →

Thus we have M 0 = F sin θ• r = r×F

Therefore we define the moment of the force about O as the vector
→ → →
M 0 = r×F
→ → → → →

As r×F =r F sin θ= M 0 =M 0
→

The magnitude, M0 , is a measure of the turning effect of the force in unit of Nm.

2 
 →
Example: Calculate the moment about O of the force F =  3  that is applied at the point
1 
 
with position vector 3j. Then calculate its magnitude.

2. Calculate the area of a triangle


By the sine rule:
1 1 → →
Area of ∆ABC = 2 bc sin A = 2 AC× AB
1 1 → →
= 2 ac sin B = 2 BC× BA
1 1 → →
= 2 ab sin C = 2 CB×CA
Example: Find area for a triangle with vertices A(0 ,7 ,1) , B (1,3 ,2 ) and C ( − 2 ,0 ,3 ) .

Equations (Vector, parametric and Cartesian equations) of a line


x  v1 
→ →   →  
Let r ( t ) = OP =  y  and v =  v 2  as a vector that parallel to the line L.
z v 
   3

→ → → →
As P0 P // v thus P0 P = t v , t is a parameter (scalar).

By the addition law of vectors, we obtain
→ → →
OP = OP 0 + P0 P
i.e. the vector equation of a line passing through a fixed point P0 and parallel to a vector
→ → → →
v is r ( t ) = OP 0 + t v .

 x   a   v1 
     
 y  =  b  + t  v2 
 z c v 
     3

 x   a + tv1 
   
 y  =  b + tv 2 
 z   c + tv 
   3 

⇒ x = a + tv1 , y = b + tv 2 , z = c + tv 3 are the parametric equations of L.
x −a y −b z −c
⇒ = = is the Cartesian equation of L.
v1 v2 v3


Example: Find the vector equation of line passes through A( 3,2 ) and B(7 ,5 ) .

Example: Find Cartesian equation of line passes through A( 5 ,− ,3 ) and B ( 2 ,1,− ) .
2 4

Equation (Vector and Cartesian equations) of a plane

 n1   x   n1   a 
       
Vector equation:  n2  •  y  =  n2  •  b 
n   z n   c
 3    3  
Cartesian equation: n1 x + n 2 y + n 3 z = n1 a + n 2 b + n 3 c

Remark: In general, Ax + By + Cz = D OR Ax + By + Cz = 1 is the Cartesian equation of a
A
 
plane with a normal vector  B .
C 
 


Example: A plane contains A(1,0 ,1) , B ( − 2 ,5 ,0 ) and C ( 3 ,1,1) . Find the vector and
Cartesian forms of the equation of the plane.

→ →
AB = −3 i + 5 j − k and AC = 2 i + j
~ ~ ~ ~ ~

Distance From A Point to A Line and to A Plane

Distance From A Point to A Line
→ → →
Distance, d, from a fixed point P to a line: r ( t ) = OA+ t v where A is a point on the line
→
and v is a vector parallel to the line is given by

→ → ∧ → ∧
d = AP sin θ = AP v sin θ = AP×v
~ ~

 1−t 
→  
Example: Find the distance from point ( 4 ,3 ,2 ) to the line L : r ( t ) =  2 + 3t  .
 3 +t 
 

Distance From A Point to A Plane


→ → → →
Distance, D, from a fixed point P to a plane n • r = n • OA where A is a point on the plane
→
and n is a normal vector to the plane is given by

→
→ → ∧ ∧
D = AP cos θ = AP n cos θ = AP •n .
~ ~

Example: Find the distance from point ( 4 ,− ,3 ) to the plane x + 3 y − 6 z = 9 .
3

PROBLEM SET: 3-D SPACE VECTORS

1. Points P , Q and R have coordinates ( 9 ,1,0 ) , ( 8 ,− ,5 ) and C ( 5 ,5 ,7 )
3
respectively. Find

(a) the position vectors of P, Q and R.
→ →
(b) PQ and QR .


→ →

(c) PQ and QR .

2. A triangle has vertices A(1,3 ,2 ) , B ( −1,5 ,9 ) and C ( 2 ,7 ,1) respectively.
Calculate the vectors which represent the sides of the triangle.
→ → → → → →
3. Find a • b and verify that a • b = b • a if

 2  3  1  3 
→   →   →   →  
(a) a = − 5  , b = 2  (b) a = 8  , b = − 2 
 0  0  7   5 
       
4. (a) Find the component of the vector 2 ~ + ~ + 7 k in the direction of the vector
i j
~

i + j+ k .
~ ~ ~

(b) Find the component of the vector 7 ~ + 2~ j − k in the direction of the vector
i
~

i − j+ 2 k .
~ ~ ~

→
5. A force F = 3 i + 7 k causes a body to move from point A(1,1,2 ) to point
~ ~
B (7 ,3 ,5 ) .
Find the work done by the force.

 3   5 
→   →   → →
6. (a) If a =  2  and b =  − 1 , find a×b .
 − 1  − 1
   
→ → → →
(b) Verify that a× b = −b× a .

7. (a) Find the area of the triangle with vertices P (1,2 ,3 ) , Q( 4 ,− ,2 ) and
3
R ( 8 ,1,1) .
→
(b) A force F of magnitude 2 units acts in the same direction of the vector
3 i − 2 j + 4 k . It causes a body to move from point S ( −2 ,− ,− ) to
3 4
~ ~ ~

point T (7 ,6 ,5 ) . Find the work done by the force.

8. Find the vector equation of the line passing through
(a) A( 3 ,2 , ) and B ( −1,2 ,3 ) .
7
→ →
(b) the points with position vectors p = 3 i + 7 k − 2 k and q = −3 ~ + 2 j + 2 k .
i
~
~ ~ ~ ~

Find also the cartesian equation of this line.


1 
 
(c) ( 9 ,1,2 ) and which is parallel to the vector 1  .
1 
 
9. Given A( 9 ,1,1) , B ( 8 ,1,1) and C ( 9 ,0 ,2 ) . Find
(a) the area of the triangle ABC.
(b) the Cartesian equation of the plane containing A, B and C.

 4 −t 
→  
10. (a) Find the distance from point (6 ,3 ,5 ) to the line L : r ( t ) =  − 3 + 2t  .
 2 + 5t 
 
(b) Find the distance from point ( 4 ,− ,6 ) to the plane 2 x − 4 y + 3 z = 8 .
3

 −7 
  →
11. (a) Find the distance from A( 2 ,1,− ) to the plane  5  • r ( t ) = 10.
3
 6 
 
y +4 z −3
(b) Find the distance from point B (6 ,3 ,− ) to the line L :
5 = ,x = 5
3 2
.

ANSWERS FOR PROBLEM SET: 3-D SPACE VECTORS

9   8  5 
→   →   →  
1. (a) OP =  1  , OQ =  − 3  , OR =  5 
0   5  7 
     
 −1  − 3 
  
(b) − 4 , 8  (c) 42 , 77
 5  2 
  
− 2   3   1 
→   →   →  
2. AB =  2  , BC =  2  , AC =  4 
 7  − 8   − 1
     


3. (a) −4 (b) 22

10 3
4. (a) (b)
3 6
5. 39 Joule
6. (a) − 3 i − 2 j − 13 k
~ ~ ~

 3 
1 → 2   90
7. (a) 1106 (b) F = − 2  , W = Joule
2 29   29
 4 
3  − 4 
→    
8. (a) r =  2  + t 0 
7   − 4 
   
 3  − 6 
→     x −3 y −3 z +2
(b) r =  7  + t − 5  , = =
− 2  4  −6 −5 4
   
 9  1
→    
(c) r =  1  + t 1
 2  1
   
1
9. (a) (b) y +z =2
2
845 30
10. (a) ≈ 5.31 (b) ≈ 5.57
30 29
37 1457
11. (a) (b) ≈ 10.59
110 13

3D Vectors and Space

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