FORM SIX MATHEMATICS STUDY NOTES VECTOR ANALYSIS 2

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VECTOR ANALYSIS 2

VECTOR ANALYSIS 2

VECTOR ANALYSIS 2

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 2 = 2 + 2 – 2   cos

Therefore

2 = 2 + 2 – 2   cos

 

02. USED TO FIND THE PROJECTION OF ONE VECTOR ONTO ANOTHER  VECTOR

–      Suppose the projection of a onto b

 

i.e

Cos Q =

Cos Q =

proj  =  cos Q        ———– i

Also

.  =    cos Q

=  cos Q            ———–ii

Equalizing i and ii as follows;

proj  =  cos Q =

proj  =

similary

proj  =

Where;

Proj  = projection of  onto

Proj  = projection of  onto

 

VECTOR PROJECTION

This is given by

V proj =.

And

V. proj =.

Where;

V. proj  = vector projection of  onto

V. proj  = vector projection of  onto

 

03. TO FIND THE WORKDONE

– Consider the diagram below

 

Force applied (F) = component of tone

cos Q =

Cos Q =

F =  Cos Q

Also

Distance d =

d = ….ii

Hence

Work done = Force applied (F) x distance (d)

W.D = F x d

W. D =  cos Q x

=  cos Q

= .

W.D =

Note

i) Force F in the direction of vector

Force applied = F.

ii) Distance in the direction of vector

Displacement  = d.

Individual

i)  = 2i +  + 2

= i +  +

W.D =

 

04. TO PROVE COMPOUND ANGLE FORMULA OF COSINE

i.e cos (A + B) = cos A cos B – sin A sin B

– consider the vector diagram below.

 

 Diagram

 

= (cos A)  + (sin A)

= (cos B)  – (sin B)

Hence

=  cos (A + B)

but

.  =

= cos A cos B + -sin A sin 0

= cos A cos B – sin A sin B

Also

=

=

=

= 1

 

=

=

=

= 1

Therefore

Cos A cos B – sin A sin B = (1)(1) cos (A + B)

Cos (A + B) = cos A cos B – sin A sin B

Proved

Pg. 2 drawing

 

= (cos A)  + (sin A) j

  = (cos B) I + (sin B)

Hence

 .   =  cos (A – B)

But

 .   =  

 

     .   = cos A cos B + sin A sin B

Also

       =

=

=

= 1

=

=

=

= 1

          There
Cos A cos B + sin A sin B = (1) (1) cos (A – B)

Cos (A – B) = cos A cos B + sin A sin B
Proved
06.  TO PROVE THAT AN INSCRIBED ANGLE SUBTENDING A SEMI –   CIRCLE IS A RIGHT ANGLE

– Consider the vector diagram below.

Pg. 2 drawing

 

          To prove that

< SR = 900

 .  = 0

-b +  + –  = 0

–  +  =

=  +

+  + -= 0

+  =

=  –

        Hence

.  = ( + ) ( – )

.  = 2 – (2

.  = 22

.  = 22

But

=  = radius, r

.  = 0

Proved

 

QUESTION

17.  Find the projection of  + 2 – 3 onto  + 2 + 2

18.  Find the vector projection of  onto. If  = 2 +2 +  and  = 3+  + 2

19.  Find the work done of the force of (2i + 3i + k) n s pulling a load (3i + j               k) m

20.  Find the work done of the force of (2i + 3j + k) N is pulling a load a distance of 2m in the direction of 2m in the direction of
= 3 + 2 + 2

21.  Find a vector which has magnitude of 14 in the direction of 2 + 3 +

 

CROSS (VECTOR) PRODUCT (X or )

x  =  sin Ø.

Where

– is the unit vector perpendicular to both vector  and

=  x

=

Hence

x  =   sin Ø.

1 =

=   sin Ø

Therefore

=   sin Ø

OR

=   sin Ø

Where

Ø – is the angle between the vector  and

Again

Suppose the vector

Hence

+  =

 

+  =   – j +

  Note

i)  If you cross two vectors, the product is also the vector.

ii)  Cross (vector) product uses the knowledge of determinant of 3 x 3   matrix.

iii)  From the definition.

 

=   sin

Individual

  = (1, 0, 0)

j = (0, 1, 0)

k = (0, 0, 1)

= (1) (1) (0)

= 0

                 = 0

Hence

i x i = j x  =  x  = 0

 

iv) From the definition

 

 

 

 

x  = k

Generally

Consider the component vector

 

         
For anticlockwise (+ve)

Pg. 4 drawing

 

i)  x  =

ii)  x  =

iii)  x  =

For clockwise (-ve)

i)  x  =

ii)  x  = –

iii)  x  = –

 

THEOREM

From the definition

 

 

 

 

= 0

Is the condition for collinear (parallel) vectors

a)     ≠

= –

 

Questions

22.  If  = 2 + 6 + 3 and =  + 2 + 2. Find the angle between  and

23.  Determine a unit vector perpendicular to  = 2 – 6 – 3 and  = 4 + 3 –

24.  If  = 2 + j + 2k and   = 3 + 2 +

Find  â‹€

 

Box product

-This involves both cross and dot product

Suppose.  x  then start with cross (x) followed by DOT (.)

.  x  =.  x)

-This is sometimes called scalar triple product

Note

-If scalar triple product (box product) of three vectors. and  = 0

-Then the vector ,  and  are said to be COMPLANAR

 

Question

25.  If  = 2 +  + 2

= 2 +  and

= 3 + 2 + k

Find

a) .  x c

b)  x.

 

APPLICATION OF CROSS PRODUCT

          USED TO PROVE SINE RULE

–      Consider the diagram below

 

+  + –  = 0
+  =
Cross by  on both sides of
x  +  x  =  x
0 +  x  =  x
x  +  x

Crossing by  on both sides of eqn …. 1 above
+ =
x  +  +   =  x
x  + =  x
x  =  x c
– ( x   =  x
x  = –  x )
x  =   x

Equation i and ii as follows
x  =  x   =  x
Sin. =  sin. =  sin.
Sin =  Sin =  sin
Dividing the whole eqn by

=  =
=  =
Sine rule

 

USED TO PROVED COMPOUND ANGLE FORMULA OF SINE

Sin (A + B) = sin A cos B + sin B cos A

Consider the vector diagram below

Pg. drawing

         

           = (cos A)  + (sin A)  + o

           = (cos A)  + (sin A)  + o

Hence

=  sin (A + B)

But

=

           = i  + k

           = i (o) – j (o) + k (-cos A sin B – sin A cos B)

           = -k [sin A cos B + cos A sin B]

           = 2

= 2

           =  sin A cos B + cos A sin B

Also

=

= 2

          2 = 2 B

                          = 1

Therefore

Sin A cos B + cos A sin B = (1) (1) sin (A+B)

Sin (A + B) = Sin A cos B + cos A sin B

Proved

 

USED TO DETERMINE/ TO PROVE COMPOUND AND FORMULA OF SINE
Sin (A – B) = sin A cos B – Cos A sin B
–  Consider the vector diagram below

Pg.6 drawing

= (cos A)  + (sin A)  + o

           = (cos B)  + (sin B)  + o

=  =  sin (A – B)

but

=

= i  – j  + k

 

= 2

= 2

= sin A cos B – cos A sin B

Also

= + 0

=

=

= 1

 

=

=

=

= 1

Therefore

Sin A cos B – cosA sin B = (1) (1) sin (A – B)

Sin (A – B) = sin A cos B – cos A sin B

Proved

 

USED TO FIND THE AREA OF THE TRIANGLE

–      Consider the triangle ABC below

Pg. 7 drawing

Area (A) = ½ x base (b) x height (h)

A = ½ bh ……i

Also

Sin  =

Sin  =

h =  Sin ….ii

And

b = …iii

Substitute …ii and …iii into 1 as follows

A = ½  sin

A = ½

 

USED TO FIND THE AREA OF PARALLELOGRAM

–      Consider the parallelogram below

Pg. 7 drawing

 

Area = length (l) x height (h)

= A = Lh …i

L = …ii

Also

Sin Q =

h =  Sin Q….ii

Substitute ii and iii into —1 as follows

Area (A) =  sin Q

A =

Generally

Area (A) =

=

=

=

Where

=  –

=  –

=  –

=  –

 

USED TO FIND THE VOLUME OF PARALLELOPIPED

–      Consider the diagram below

Pg. 7 drawing

–      Suppose P Q R and S are the vertices of the parallelepiped, hence the volume (v) of the parallelepiped is given by:

Volume (v) = base area (A) x height (h)

V =  x

V =

V =

V =

 

Again, for the sides with position vectors,  and

Volume (v) =   x

=

=

 

USED TO FIND THE VOLUME OF A TETRAHEDRON

–      Consider the tetrahedron with vertices P, Q, R and S

Pg. 8 drawing

 

Volume (v) = 1/3 x base area x altitude

V = 1/3 x ½  x

 

V = 1/6  x

 

V =

Therefore

Volume = 1/6

USED TO FIND THE VECTOR PERPENDICULAR TO THE PLANE

–      Consider the diagram below

Pg. 8 drawing

 

Where

  = is the vector perpendicular or normal to the plane

 

=  x

 

Question

26.  Find the area of the triangle ABC whose vertices are A (2, 1, 1) B (3, 2, 1) and C (-2, -4, -1)

27.  The position vector of the points A, B and C are (2, 4, 3), (6, 3, -4) and (7,           5, -5) respectively. Find the angle between   and   and hence the       area of the triangle ABC

28.  Find the area of the parallelogram whose vertices P, Q and R are (2, 1, 1),           (3, 2, 1) (2, 4, 1)

29.  Find the volume of the parallelepiped the edges are A (1, 0, 2) B (2, -1, 3) C (4, 1, 3) and D (1, -1, 1)

30.  Find the volume of the tetrahedron whose sides are   = 2 + ,  =  – 3 ` + , and   – + 2

 

COLLINEAR AND COPLANAR VECTORS

1. COLLINEAR VECTOR

These are vectors having the same slope (re direction).

    Pg. 10 drawing

 

 

= μ

= t

Where

Æ›, μ and t are scalar

Again  and  to be collinear  x  = 0

 

2. COPLANAR VECTOR

These are vectors which lie on the same plane

Eg. Pg. 10 drawing

 

For the vectors,  and  to be coplanar

= 0

= 0

= 0

Generally

=   =    = 0

Question

31.  Given that

= 3 + 4

32.  Find the value are collinear vectors 2 – + ,  +  2+ 3 and 3 + + 5    are coplanar.

33.   Find unit vector in the direction of  = 6 + 3 +  and state its length

LINEAR COMBINATION OF VECTORS

Suppose that  are vectors and, β and γ are real numbers (sealers). Then a vector  =    + β   + γ  is a linear combination of vectors

NB

To solve vectors means to put the vectors into linear form

Question

34.  If  =  + ,    =  –  and  = 3 – 4 resolve  into vectors parallel to

35.   Express the vector r = 10 – 3-  as a linear function of  given that

= 2 –  +

= 3 + 2 –     and

= – + 3 – 2

Note: Required to be placed in a right position
Subtopic: Dot Product

Proving cosine rule using dot product

Consider the triangle ABC above

Subtopic: Cross Product

Proof of sine rule Consider the ∆ABC with sides A,B and C respectively

Construct a line AH which lies on BC and perpendicular.

Equating area

 

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