FORM SIX MATHEMATICS STUDY NOTES VECTOR ANALYSIS 2

VECTOR ANALYSIS 2

VECTOR ANALYSIS 2

VECTOR ANALYSIS 2

USIBAKI NYUMA>PATA NOTES ZETU KWA HARAKA:INSTALL APP YETU-BOFYA HAPA

 

UNAWEZA JIPATIA NOTES ZETU KWA KUCHANGIA KIASI KIDOGO KABISA:PIGA SIMU:0787237719

 

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 2 = 2 + 2 – 2   cos

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Therefore

2 = 2 + 2 – 2   cos

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

–      Suppose the projection of a onto b

 

i.e

Cos Q =

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Cos Q =

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proj  =  cos Q        ———– i

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Also

.  =    cos Q

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=  cos Q            ———–ii

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Equalizing i and ii as follows;

proj  =  cos Q =

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proj  =

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similary

proj  =

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Where;

Proj  = projection of  onto

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Proj  = projection of  onto

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VECTOR PROJECTION

This is given by

V proj =.

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And

V. proj =.

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Where;

V. proj  = vector projection of  onto

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V. proj  = vector projection of  onto

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03. TO FIND THE WORKDONE

– Consider the diagram below

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Force applied (F) = component of tone

cos Q =

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Cos Q =

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F =  Cos Q

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Also

Distance d =

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d = ….ii

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Hence

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

W.D = F x d

W. D =  cos Q x

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=  cos Q

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= .

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W.D =

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Note

i) Force F in the direction of vector

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Force applied = F.

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ii) Distance in the direction of vector

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Displacement  = d.

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Individual

i)  = 2i +  + 2

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= i +  +

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W.D =

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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.

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 Diagram

 

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= (cos A)  + (sin A)

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= (cos B)  – (sin B)

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Hence

=  cos (A + B)

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but

.  =

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= cos A cos B + -sin A sin 0

= cos A cos B – sin A sin B

Also

=

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=

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=

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= 1

 

=

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=

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=

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= 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

 

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= (cos A)  + (sin A) j

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  = (cos B) I + (sin B)

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Hence

 .   =  cos (A – B)

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But

 .   =  

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     .   = cos A cos B + sin A sin B

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Also

       =

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=

=

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= 1

=

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=

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=

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= 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

 

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          To prove that

< SR = 900

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 .  = 0

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-b +  + –  = 0

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–  +  =

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=  +

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+  + -= 0

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+  =

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=  –

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        Hence

.  = ( + ) ( – )

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.  = 2 – (2

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.  = 22

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.  = 22

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But

=  = radius, r

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.  = 0

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Proved

 

QUESTION

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

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

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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

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

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CROSS (VECTOR) PRODUCT (X or )

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x  =  sin Ø.

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Where

– is the unit vector perpendicular to both vector  and

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=  x

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=

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Hence

x  =   sin Ø.

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1 =

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=   sin Ø

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Therefore

=   sin Ø

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OR

=   sin Ø

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Where

Ø – is the angle between the vector  and

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Again

Suppose the vector

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Hence

+  =

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+  =   – j +

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  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.

 

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=   sin

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Individual

  = (1, 0, 0)

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j = (0, 1, 0)

k = (0, 0, 1)

= (1) (1) (0)

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= 0

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                 = 0

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Hence

i x i = j x  =  x  = 0

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iv) From the definition

 

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x  = k

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Generally

Consider the component vector

 

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For anticlockwise (+ve)

Pg. 4 drawing

 

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i)  x  =

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ii)  x  =

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iii)  x  =

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For clockwise (-ve)

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i)  x  =

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ii)  x  = –

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iii)  x  = –

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THEOREM

From the definition

 

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= 0

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Is the condition for collinear (parallel) vectors

a)     ≠

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= –

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Questions

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

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

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24.  If  = 2 + j + 2k and   = 3 + 2 +

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Find  â‹€

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Box product

-This involves both cross and dot product

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

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.  x  =.  x)

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-This is sometimes called scalar triple product

Note

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

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-Then the vector ,  and  are said to be COMPLANAR

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Question

25.  If  = 2 +  + 2

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= 2 +  and

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= 3 + 2 + k

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Find

a) .  x c

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b)  x.

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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

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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

         

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           = (cos A)  + (sin A)  + o

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           = (cos A)  + (sin A)  + o

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Hence

=  sin (A + B)

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But

=

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           = i  + k

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

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

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           = 2

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= 2

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           =  sin A cos B + cos A sin B

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Also

=

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= 2

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          2 = 2 B

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                          = 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

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= (cos A)  + (sin A)  + o

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           = (cos B)  + (sin B)  + o

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=  =  sin (A – B)

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but

=

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= i  – j  + k

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= 2

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= 2

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= sin A cos B – cos A sin B

Also

= + 0

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=

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=

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= 1

 

=

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=

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=

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= 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

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Area (A) = ½ x base (b) x height (h)

A = ½ bh ……i

Also

Sin  =

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Sin  =

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h =  Sin ….ii

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And

b = …iii

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Substitute …ii and …iii into 1 as follows

A = ½  sin

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A = ½

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USED TO FIND THE AREA OF PARALLELOGRAM

–      Consider the parallelogram below

Pg. 7 drawing

 

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Area = length (l) x height (h)

= A = Lh …i

L = …ii

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Also

Sin Q =

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h =  Sin Q….ii

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Substitute ii and iii into —1 as follows

Area (A) =  sin Q

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A =

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Generally

Area (A) =

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=

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=

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=

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Where

=  –

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=  –

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=  –

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=  –

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USED TO FIND THE VOLUME OF PARALLELOPIPED

–      Consider the diagram below

Pg. 7 drawing

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–      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

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V =

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V =

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V =

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Again, for the sides with position vectors,  and

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Volume (v) =   x

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=

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=

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USED TO FIND THE VOLUME OF A TETRAHEDRON

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

Pg. 8 drawing

 

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

V = 1/3 x ½  x

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V = 1/6  x

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V =

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Therefore

Volume = 1/6

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USED TO FIND THE VECTOR PERPENDICULAR TO THE PLANE

–      Consider the diagram below

Pg. 8 drawing

 

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Where

  = is the vector perpendicular or normal to the plane

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=  x

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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

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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

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COLLINEAR AND COPLANAR VECTORS

1. COLLINEAR VECTOR

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

    Pg. 10 drawing

 

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= μ

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= t

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Where

Æ›, μ and t are scalar

Again  and  to be collinear  x  = 0

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2. COPLANAR VECTOR

These are vectors which lie on the same plane

Eg. Pg. 10 drawing

 

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For the vectors,  and  to be coplanar

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= 0

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= 0

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= 0

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Generally

=   =    = 0

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Question

31.  Given that

= 3 + 4
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32.  Find the value are collinear vectors 2 – + ,  +  2+ 3 and 3 + + 5    are coplanar.

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

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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

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NB

To solve vectors means to put the vectors into linear form

Question

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

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

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= 2 –  +

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= 3 + 2 –     and

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= – + 3 – 2

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

Proving cosine rule using dot product

Consider the triangle ABC above

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Subtopic: Cross Product

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

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Construct a line AH which lies on BC and perpendicular.

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Equating area

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