nishant math assingment
TRANSCRIPT
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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).