Nishant Math Assingment

8/14/2019 Nishant Math Assingment

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Engineering MathematicsEE002-3-0 Math assignment Page 1 of 15

Level 0 Asia Pacific Institute of Information Technology 2013

QUESTION 1

(A)SolutionZ= i341

Here ,1a and 34b

We know that

R= 22 ba

= 22 341

= 481

= 49

=7

And tana

b

=1

34

= 34

So,

34tan 1

79.81

lies in first quadrant

So, 79.81

At first we converted into polar form

So z= )sin(cos ir

Z= )79.81sin79.81(cos7 i

Now we square root

According to DeMoivres Theoremnir )sin(cos = rn(cosn nisin )

And


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zk=

n

ki

n

krn

)2(sin

)2(cos

So zk= ]

2

)279.81(sin

2

)279.81([cos7

ki

k

Here k=0,1 [ k=0,1,2,3------,n-1]

Z0= 2.645[ ]2

)0279.81(sin

2

)279.81(cos

i

o

= ]895.40sin895.40[cos645.2 i

= ]655.056.7.0[645.2 i

= 732.199.1 i

= 32

Z1= 2.645[cos ]2

1801279.81sin

2

1801279.81

i

= ]895.220sin895.220[cos645.2 i

= ]655.0756.0[645.2 i

= 732.199.1 i

= 32 i

So, Z0 32 i and Z1 = 32

(B)Solution:-12

12

)1(

)1(

n

n

i

i

We know that

baba xxx and bab

a

xx

x

SO ,

=

)1(

)1(

)1()1(2

2

i

i

iin

n

=n

n

i

iii2

2

)1(

)1)(1()1(

[ because (a+b).(a-b)= (a2-b2) ]

=n

n

i

ii2

222

)1(

)1()1(


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=n

n

i

i2

2

)1(

)11()1(

[because i

2= 1 ]

= 2

n

i

i2

)1(

)1(

= 2

n

i

i

i

i2

)1(

)1(

)1(

)1(

= 2

n

i

i2

22

2

)1(

)1(

= 2

n

ii2

2

11

21

= 2

ni

2

2

121

= 2

ni

2

2

2

= 2 ni 2

= 2 ni 2 [because 24 = (22)2=16]= 2 n1 = 2 [ if n is even no.]

= 2 [ if n is odd no. ]

And its conjugate is= 2 [when n is even no.]

= 2 [when n is odd no.]

( C) SOLUTION:-

23

31 i

Here a= 1 and b= 3

So r= 22 ba

= 22 31 = 31

= 4

r =2

And

tana

b


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=1

3

1

3tan 1

3

2

Convert into polar form

Z= 2( )3

2sin

3

2cos

i

According to DeMoivres Theorem

nir )sin(cos = rn(cosn nisin )

So, z=2

13

)3

2sin

3

2(cos2

i

z = 2

1

3 )3

23sin

3

23(cos2

i

Z = 21

)2sin2(cos8 i

Now we find square root

And zk=

n

ki

n

krn

)2(sin

)2(cos

So, zk=

2

)22(sin

2

)22(cos8

ki

k

Here z= 8,1

Z0= ]2

)022(sin

2

)022([cos8

i

= ]2

2sin

2

2[cos8

i

= ]sin[cos8 i

= ]01[8

= 8

= 2.828

Z1= ]2

)122(sin

2

)122([cos8

i

= ]2

4sin

2

4[cos8

i

= ]2sin2cos[8 i


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= ]01[8

= 8 = 2.828

(D.)

(i) Solution:-

Given Z=i

i

55

3

i

i

55

3

=

i

i

i

i

55

55

55

3

=

22 55

553

i

ii

=2

2

2525

551515

i

iii

= 12525

151015

i

=2525

51015

i

=50

1020 i

=50

10

50

20 i

=55

2 i

Modulus of Z= 22

baz 22

5

1

5

2

z

25

1

25

4z

25

5z =

5

1

55

5

5

5

5

5

So, modulus value of Z=

5

1

(ii)Solution:-

Given Z=3 1

2 1

3 1

2 1

=

2

2

2

3

i

i

=

i

i

2

3


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=i

i

i

i

2

2

2

3

=

222

23

i

ii

= 22

2

2236

iiii

= 14

156

i

=14

55

i

=14

55

i

=5

55 i

=5

5

5

5 i

=1+i

So , z =1 + i

Modulus of Z= 22 baz

Z = 22 11 z

Z = 2z

So, modulus value of Z= 2

QUESTION 2

(A) solution:-

Assume the that vector be .

kzjyix

So, Dot product of a vector can be given in question is as:-

).(

kzjyix )(

kji =4

4 zyx -------------------------------(i)

Again

).(

kzjyix )32(

kji =0

032 zyx ----------------------------(ii)

Again given


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).(

kzjyix )(

kji =2

2 zyx .(iii)

Now we have three equations

After solving the equation (i) and (ii) we get

423 zx -------------------------------------------(iv)Again solving the equation (i) and (iii) we get

622 zx ------------------------------------------(v)

Now solving equation (iv) and (v) we get

105 x

So 25

10x

Now put the value of x in equation (iv)

4223 z 46 zz

462 z

12

2z

1z

Again put the value of x and zin equation (iii)

212 y

132 y

1y

We get

1,1,2 zyx

Now require vector is

kji2

(B.)Solution:-

Given point

A=

kji 423

B =

kji 36

C=

kji 375

D=

kji 622


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AB =

kii 353

kjiBC 241

kjiCD 353

We know that point to be collinear the dot triple product must be zero.

0).(

cba

kjiCDBC 17322)(

Now,

).(

CDBCAB = )17322).(353(

kjikji

=66-15-51

=0

So, given points 3 2 4 ,i j k

kji 375

kji 375 and

kji 622 are coplanar.

(C.)Solution:-

kjiF 3261

Unit vector of F1= 9436

326

kji

=49

326

kji

=7

326

kji

Again

kjiF 6232


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Unit vector of F2 =3649

623

kji

=49

623

kji

=7

623

kji

Net force F =F1+F2

=

76233

73265 kjikji

=7

151030

kji+

7

18`69

kji

=7

33439

kji

Displacement (D) =

kjikji 2234

=

kji 22

Work done = F.D

= (7

33439

kji). (

kji 22 )

=7

66478

=7

148

= 21.1

So Work done=21.1

(D) solution

Given Vectors

kjia


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kjib

,c i j k

Now )22()(

jicb

So ).(

cba =

kji .

ji 22

=2+2

= 4

(E) Given vectors

a = 2 4 5i j k

b = 2 3i j k

Sum of vector )(

ba =

kjikji 32542

=

kji 263

Now unit vector of )(

ba = 222 263

263

kji

=4369

263

kji

=49

263

kJi

=7

263

kji

=7

2

7

6

7

3

kji

So unit vector of7

2

7

6

7

3)(

kji

ba


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QUESTION 3

(A)Solution:-

Given, vertices of triangle A= (-4, 6), B= (2,-2), C= (2, 5)

2,2,4 321 xxx

We know that,

Centroid of triangle=

3,

3

321321 yyyxxx

=

3

526,

3

224

=

3

211,

3

44

=

3

211,

3

44

=

3

211,

3

0

=

3

9,

3

0

= 3,0

So, centroid of triangle is (0,3)

5,2,6 321 yyy


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(B)solution:-

Here normal distance from origin to line P = 4and

angle between normal and positive direction of X-axis 135

so, according to normal form of straight line

pyx sincos

4135sin135cos yx

42

1

2

1 yx

42

yx

24 yx

24 xy

So general form of straight line is 24 xy


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(C.)Solution:-

Here distance from origin to normal p=4

Line cut on x-intercept at 5, then co-ordinate of point= (5, 0)

5

4cos

h

b

2cos1sin =

2

5

41sin

25

161sin

25

9sin

5

3sin

We know that equation of straight line in normal form is

pyx sincos

Now put the value of cos and sin


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45

3

5

4 yx

2034 yx

So , equation of line can be given as 4x+3y=20

(D.)Solution:-

Let the co-ordinate of point on x -axis B (a, 0).

7, 21 xax and

6,0 21 yy

By distance formula

AB= 2122

12 YYXX

= 22 067 a

= 361449 2 aa

= aa 1485 2

Similarly,

3, 21 xax

4,0 21 yy

By distance formula

AB= 2122

12 YYXX

= 22 043 a

= 1669 2 aa

= aa 625 2 If given points are equidistant then distance must be equal.

So,

AB=BC

aa 1485 2 = aa 625 2

Squaring both sides of equation

2222 6251485 aaaa aaaa 6251485

22

aa 6251485


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2585614 aa 608 a

8

60a

5.7a

So, co-ordinate of points on x axis will be(7.5,0).

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