PROPERTIES OF VECTORS:
A vector has both magnitude and direction.
Velocity, force, angular velocity and Torque are some of the vector quantities.
A vector is graphically represented by an arrow whose length is proportional to the magnitude of the vector and direction parallel to the direction of the vector. In print, a vector quantity is either shown in bold (a or A) or in italic form with an arrow above it (a). In handwritten form, the second method is generally used.
A vector can be displaced parallel to itself side wise, forward or backward without changing the direction. Tail Head =
Two vectors are equal only when they have same direction and magnitude. POSITION VECTOR:
If (x, y) is the coordinates of a point, its position with respect to the origin is represented by the vector r = x i + y j where i and j are unit vectors along x and y directions.
A unit vector has magnitude one. In general the position vector of point P(x2,y2,z2) with respect to point Q(x1 ,y1,z1) is given by rPQ = (x2 - x1) + i (y2 - y1) j + (z2 - z1) k
If the line joining P and Q has magnitude r and makes angles a,b and g with x,y and z axes respectively, (x2 -x1) = r cos a, y2–y1=r cos b and (z2 - z1) = r cos g.
RESOLUTION OF A VECTOR :
The component of a vector A along a direction inclined at an angle q to the direction of the vector is given by A cos Y X Z O B . C sin A A 90 Acos
The components of a vector resolved along two mutually perpendicular directions are Acos and A cos (90 - ) = A sin .
When a vector is resolved into two mutually perpendicular components in the same plane, it can be replaced by the components.
F sin F cos F cos F Sin F
F cos and Fsin must lie on either side of F.. In general any vector
A can be resolved into three mutually perpendicular components along x, y and z axes
A= AAx i +Ay j + AAz k
When Ax = A cos ; AAy = A cos ; AAy = Acos where , and are the angles made by
A with x,y and z axes respectively.
The magnitude of a vector is given by |A| = 2 2 2
Ax Ay Az for a null vector magnitude is Zero VECTOR ADDITION :
When two vectors are added, another vector different from these vectors is obtained C A B B A
Graphically vector addition is done by joining head of one vector to the tail of another vector. The arrow joining the free head and free tail gives the resultant vector.
B A C C B A B A
The tail of A can be joined with the head of B or the head of A can be joined with the tail of B using the parallel displacement property of the vectors.
In component form, if A = AAx i +Ay j +Az k and B = Bx i +By j +Bz k C A B B A
AxBx
i
AyBy
j
AzBz
kThe angle between two vectors is the angle between the arrows when joined at the tails or when joined at the heads. B A A B B A
Since angle can be measured either clock wise or anticlockwise, smaller of the two angles will be generally taken as the angle between the vectors and that value lies between 0° and 180°
Angle between A and B is .
B
A '
If is the angle between two vectors A and B and A and B are their magnitudes, the magnitude of the resultant vector C is given by C = 2 2
A B 2AB cos
In component form C =
Ax Bx
2
AyBy
2
AzBz
2When the resultant of more than two vectors is to be obtained, component form is preferred. Resultant R =
xcomp
2
ycomp
2
zcomp
2R
p Q S
For two vectors Aand
B inclined at q with each other, the angle made by the resultant vector
C with
Acan be found from the following vector addition diagram. let PQ, QR and PR represent the magnitudes of
B ,
A and C respectively and a is the angle made by
Cwith
A. From triangles PRS, PRcos = PQ + QS and PR sin = RS From the triangle QRS, QR cos = QS and QR sin = RS
PR sin = QR sin PR cos = PQ + QR cos Taan = cos QR PQ sin QR = cos B A sin B
Resultant vector concept is used in finding out resultant velocity and resultant force acting on a particle
VECTOR SUBTRACTION:
A B A ( B)
Vector subtraction can be thought of as vector addition between A and B To obtain B, reverse the direction of B.
A C B A B C
can be obtained by joining A and B at the tails and joining the free heads with an arrow from head of B to head of A A B C
The magnitude of C = A B is obtained from |C| = 2 2
A B 2AB cos where is the angle between A and B
. This formula should not be con-fused with the formula for the magnitude of the resultant vector
|C| = 2 2
A B 2AB cos . Remember + for addition and - for subtraction in the magnitude formula. If A = AAx i +Ay j +Az k and B = Bx +By +Bz k C A B =
AxBx
i
AyBy
j
AzBz
kand the magnitude of C, |C| =
AxBx
2
AyBy
2
AzBz
2Vector subtraction is particularly useful in finding out relative displacement, relative velocity and relative acceleration between two moving particles.
If
A
V
is the velocity of particle A and
B
V
is the velocity of particle B, the relative velocity of A with respect to B is given by VAB VA VB
. Similarly relative velocity of B with respect to A is given by
BA B A
V V V
|VAB| = |VBA| =
2 2
A B A B
V V 2V V cos where is the angle between
A V and B V .
MULTIPLICATION OF A VECTOR WITH A CONSTANT:
When a vector is multiplied with a constant, all the components will get multiplied with the same constant. If A A ix A jy A ,z k (A) =
KAx
i
KAy
j
KAz
k and the magnitude also will getmultiplied 'k' times without change in the direction.
SCALAR OR DOT PRODUCT:
A .B
is called the scalar or dot product of A
and B . A .B = AB cos = B.A
. Since this product gives only a number without any direction it called scalar product. i.i A
= j.j = k.k = 1 and i.j = j.i = j.k=k.j=k.i=i.k = 0 If A = AAx i + AAy j + AAzk and = Bxi + By j + Bz k A .B =
A ix A jy A kz
.
B ix B jy B kz
= Ax Bx + Ay By + Az BzThe angle between two vectors can be found using scalar product
cos = A .B AB
Where A is the magnitude of A
and B is the magnitude of B
VECTOR OR CROSS PRODUCT:
A
x B
is called vector or cross product of A
and B
. If is the angle between A
and B , A x B
= AB sin n where n is a unit vector perpendicular to the plane containing A
and B
. Since this product gives another vector of magnitude AB sin along the direction of it is called vector product.
The direction of n is given by right hand screw method. If A
is turned towards B
, the direction of movement of a right hand screw gives the direction of . When B
is turned towards A , the direction of A x B
is obtained which is opposite to the direction of A
x B . A x B = - B x A
i x i = j x j = k x k = 0 as the angle between them is zero and sin = 0 i x j = k; j x k = i; k x i = j and j x i = - k; k x j = - i; i x k = -j
(K A) x B = K (A x B) and K1 A x K2 B = K1 K2 (A x B) If A = AAx i + AAy j + AAz k and = Bx i + By j + Bzk A x B =
A ix A jy A k x B iz
x B jy B kz
= Ax Bx
i x i
+ AAx By
i x j
+ AAx Bz
i x k
+ Ay Bx
j x i
+ AAy By
j x j
+ AAy Bz
j x k
+ Az Bx
k x i
+ AAz By
k x j
+ AAz Bz
k x k
= i
A By zA Bz y
j A B
z x A Bx z
k A B
x yA By x
The cross product of A
and B can be found from the method of determinant also.
A x B = x y z x y z i j k A A A B B B = i A B
y zA Bz y
j A B
z x A Bx z
k A B
x yA By x
While representing vectors on a paper, the following symbols are used. represents a vector going in ward into the paper and
Illustrations Illustration - 1 :
At the centre of the roof of a cubical room of side 4 m, a bulb is hanging. Taking the origin at one of the corners on the floor, find the position vector of the bulb.
Y X Z O B . C
a) with respect to the origin
b) with respect to the corner opposite to the origin on the floor. Solution :
coordinates of bulb B are (2, 4,2) coordinates of corner C are (4,0,4) Position vector of B with respect 0,
BO
r
= (2 - 0) i+ (4 - 0) j + (2 - 0) k = 2i+4 j 2k
Position vector of B with respect to C ,
BC
r
= (2 - 4) i+ (4 - 0) j + (2 - 4) k = - 2i+4 j 2k
Illustration - 2 :
Find the components of vector along the given directions 1,2,3 and 4. Solution :
F1 = Fcos30 = Fcos 330 F2 = Fcos45 = Fcos 315 F3 = Fcos135 = Fcos 225 F4 = Fcos270 = Fcos90
Illustration - 3 :
Three persons are applying forces F1, F2 and F3 on a block of mass M in three mutually perpendicu-lar directions. Neglecting other forces write the force acting on the block in vector form and find its magnitude.
Solution :
Taking F1, F2 and F3 along x,y and z directions
F = Fx i + Fy j + Fz k = F1 i+F2 j+F3 k Magnitude of F = |F| = 2 3 2 2 2 1 F F F Illustration - 4 :
A sugar grain is being pulled by four ants in different directions with the same magnitude of force as shown in the diagram. Find the magnitude and direction of the resultant force acting on the sugar grain.
Solution :
Since all the forces are along the same horizontal plane, they can be resolved in to x and y compo-nents. Taking x along the East and y along the North
Fx = F cos300 - F cos450 - Fcos300 + F cos900 = - F
2 - 0. 7 F
Fy = F cos 600 + F cos450 - Fcos600 - F = F F
2 - 0.3 F FR = Fx2Fy2 = F
0.7
2
0.3
2 Fx R F FyIf is the angle a made by the resultant force with the west Tan = y x F F = 3 7 Illustration - 5 :
A car is moving towards east with a velocity of 10 m/s and a bus is moving towards North West with a velocity of 15 m/s. Find the relative velocity of the bus with respect to the car.
Solution :
Take x along east and y along North. For obtaining relative velocity join at the tails. The arrow joining the free heads towards the bus from the car gives the relative velocity of the bus with respect to the car. BC V s / m VB15 s / m VC10 0 135 VBC = 2 2 B C B C V V 2V V cos = 2 2 0 15 10 2(15)(10) cos135 ~ 23 m/s The direction of BC V
can be found from the vector diagram. If is the angle made by VBCwith the west, from the velocity triangle
VBC cos = VC + VB cos450 cos = 10 15 / 2 23 ~ 41 46
This problem can be solved in component form also.
C B V 10 i ; V = -15 cos45 i + 15 cos45 j = - 10.5 i +10.5 j BC V
5 10. 5 20.BC B C V V V =
10.5 10 i
10.5 j = - 20.5 i + 10.5 j |VBC| =
20.5
2
10.5
2 ~ 23 m/s Tan = 10.520.5 when is the angle made by the relative velocity with the west.
Illustration - 6 :
If
A = 3i + 4j - 6k and
B = 2i - 3j + k . Find the magnitude of 3
A + 2 B Solution : C 3A 2 B = 3 (3i + 4j - 6k) +2 (2i - 3j+k) = i (9 + 4) + j (12 - 6) +k (-18 + 2) = 13i+ 6 j - 16k Magnitude of C =
2 2
2 13 6 16 ~ 21.5It may be noted here that the magnitude of 3A 2 B
is not equal to 3 |A| + 2 |B| Illustration - 7 :
Find the dot product of
A= 3i + 4k and B = -2i+3j-4k and and the angle between them. Solution : B . A = (3i+4k). (-2i + 3j - 4k) = (3) (-2) + (0) (3) + (4) (-4) = - 22 A = 2 2 2 3 0 4 = 5 ; B =
2 2
2 2 3 4 = 29 = 5.4 cos =
5 5.4A.B
= 22 27 Scalar product is particularly useful in finding out the component of one vector along the direction of another vector and in finding out the work and power
Work, w = F ,S . where F
is the constant force and S
is the displacement. Power, P = F .
V . where Vis the constant velocity.. Illustration - 8 :
Find the direction of C
= A
x B
in the following cases. A B B A A B A B Solution : a) Since A and B
are on the plane of the paper, C
must be perpendicular to the paper. When A
is rotated towards B
, the screw moves inward C
is normally inward b) c) d) C C C Illustration - 9 : If A = 3i + 4j - 2k and B= 2i + 3 k find A x B . Solution : A x B =
3i 4 j 2k
x
2i 0 j 3k
= 6
i x i
0 i x j
9 i x k
+ 8 jx i
0 j x j
12 j x k
- 4
k x i
0 j x j
12 k x k
= - 9j - 8 k + 12 i - 4 j = 12 i - 13 j - 8kVector product is particularly useful in finding out angular momentum , torque and other angular parameters
Torque,
= r x F
Illustration : 10
Which of the sets given below may represent the magnitudes of three vectors adding to zero a) 2,4,8 b) 4,8,16 c) 1,2,1 d) 0.5,1,2
Solution :
Zero vector sum is possible only when the three vectors can be represented by three sides of a triangle. Since in a triangle the larger side cannot be more than the sum of two smaller sides, a,b and d are not possible C is the answer
Illustration : 11
The resultant of A
and B
makes an angle with A
and with B , then a) < b) < if A < B c) < if A > B d) ) < if A = B Solution : Tan = Bsin A B cos = 1 ; A cos ec cot B Tan an = A sin B A cos = 1 B cos ec cot A When A < B, Tan > Tan >
C is the answer Illustration : 12 Let C = A + B . Then a) C
is always greater than | A | b) It is possible to have | C| < |A| and | C | | B | c) It is always equal to A + B d) C is never equal to A + B Solution : C = 2 2
A B 2AB cos Cmin = A - B and C max = A+B b is the answer
Illustration : 13
The magnitude of the vector product of two vectors A
and B
a) greater than AB b) equal to AB c) less than AB d) equal to zero Solution : A x B = AB sin n magnitude AB sin is zero when = 0 is AB when = /2
is less than AB when is between 0 and /2 But cannot be more than AB as sin
answer is b,c and d
Illustration : 14
Two vectors have magnitudes 2m and 3m. The angle between them is 600. The magnitudes of the
scalar and vector products are respectively
a) 3 and 3 b) 3 and 3 3 c) 3 3 and 3 d) 3 3 and 3 3 Solution :
Scalar product magnitude = AB cos 600 = 2(3) 1/2 = 3
Vector product magnitude = AB sin600 = 2(3) 3
SINGLE ANSWER TYPE QUESTIONS
1. A ship is travelling due east at 10km/h. A ship heading 300 east of north is always due north from the first
ship. The speed of the second ship in km/h is
A) 20 2 B) 20
2 3
C) 20 D) 20 2
2. A river is flowing from west to east at a speed of 5 m per minute. A man on the south bank of the river, capable of swimming at 10 m per minute in still water wants to swim across the river to a point directly opposite in the shortest time. He should then swim in a direction
A) 600 west of north B) 300 east of north
C) 300 west of north D) 600 east of north
3. Wind is blowing from the south at 10 ms-1 but to a cyclist it appears to be blowing from the east at 10 ms-1.
The cyclist has a velocity
A) 10 iˆ 10jˆ B) 10 iˆ 10jˆ C) 10 iˆ 10jˆ D) 10 iˆ 10jˆ
4. The wind appears to blow from the north to a man moving in the north-east direction. When he doubles his velocity the wind appears to move in the direction cot-1 (2) east of north. The actual direction of the
wind is A) 2 towards east B) 2 towards west C) 2 towards west D) 2 towards east
5. The velocity of a boat in still water is h times less than the velocity of flow of the river (> 1). The angle with the stream direction at which the boat must move to minimise drifting is
A) sin-1 1 B) cot -1 1 C) 1 sin 2 1 D) 1 cot 2 1
6. Two particles, A and B move with constant velocities and A . Initially their radius vectors are B rA and r . For the particles to collide the four vectors must be interrelated asB
A) A B rArB B) A Band rArB C) A B A B A B A B r r r r D) B A A B B A A B r r r r
7. A car A is going north east at 80 kmh-1 and another car B is going south east with a velocity of 60 kmh-1.
The velocity of A relative to B makes an angle with the north equal to
A) tan-1(2/7) B) tan-1(7/2) C) tan-1(7) d) tan-1(1/7)
8. Given that P QR0 . Which of the following statements is true ? A) P Q R B) P Q R
C) p Q R D) P Q R
9. The X and Y components of vector A have numerical values 5 and 6 respectively and that of A B have numerical vales 10 and 9. What is the numerical value of B ?
A) 2 B) 3 C) 4 D) 5
10. If A B C and the magnitudes of A, B and C are 5, 4 and 3 units respectively, the angle between A and C is A) cos-1 3 5 B) cos -1 4 5 C) /2 D) sin-1 3 4 11. If A B is a unit vector along x-axis and A iˆjˆkˆ, then what is B ?
A) ˆikˆ B) ˆikˆ C) ˆ ˆi j kˆ D) ˆikˆ 12. If ABˆ BAˆ, then which of the following relations is wrong ?
A) A = B B) Aˆ Bˆ C) A B D)A B = A Bˆ ˆ 13. Given that both A and B are greater than 1. The magnitude of cannot be
A) equal to AB B) less than AB C) more than AB D) equal to A/B
14. Figure shows three vectors a, b and c, where R is theP QR0 midpoint of PQ. Then which of the following relations is correct ?
A) a b 2c B) a b c C) a b 2c D) a b c
15. Two forces Pand Q acting at a point are such that if Pis reversed, the direction of the resultant is turned through 900, then :
A) P = Q B) P = 2Q
16. The resultant of two forces P and Q is of magnitude P. If P is doubled, the resultant will be inclined to Q at an angle :
A ) 00 B) 300 C) 600 D) 900
17. If A B = 0 also A x C =0, the angle between B and C is :
A) zero B) 450 C) 900 D) 1800
18. Two vectors A and B inclined at an angle have a resultant R which makes an angle with A . If the direction of A and B are interchanged the resultant will have the same :
A) magnitude B) direction
C) magnitude as well as direction D) neither magnitude nor direction
19. In an equilateral triangle ABC, AL, BM and CN are medians. Forces along BC and BA represented by them will have a resultant represented by :
A) 2AL B) 2BM C) 2CN
D) AC
20. The condition under which the vectors
a b
and
a b
are parallel is : A) ab B) a b
C) a b D) a || b
21. Vector a is perpendicular to b . Component of a b along a b will be :
A) zero B) a-b C) 2 2 2 2 a b a b D) 2 2 a b
22. ABCD is a quadrilateral. Force BA, BC, CD and DA act at a point. Their resultant is A) 2 AB B) 2 DA C) 2 BC D) 2 BA
23. If the vectors
ˆ ˆi j kˆ
and 3ˆi form two sides of a triangle the area of the triangle isA) 3 B) 2 3 C) 3
2 D) 3 2
24. If a, b
and c are three unit vectors with a b c = 0 then a b b c c a is equal to : A) zero B) 1 C) -3/2 D) none of these 25. If angle between a and b is /3, then angle between 2 a and -3 b is :
A) /3 B) 2 /3 C) /6 D) 5 /3
the angle between a and b is : A) /4 B) /3 C) cos-1 1 3 D) cos -1 2 7 27. Component of ˆ3i + 4 jˆ perpendicular to ˆ ˆij and in the same plane as that of ˆ3i + 4 jˆ is :
A) 1 ˆ ˆ
j i
2 B)
3 ˆ ˆj i 2 C)
5 ˆ ˆj i 2 D)
7 ˆ ˆj i 2 28. A vector of length l is turned through the angle about its tail. What is the change in the position vector of its head ? A) l cos 2 B) 2l sin 2 C) 2l cos 2 D) l sin 2
29. If a vector A makes angles , and respectively with the X, Y and Z-axes respectively then sin2
+ sin2
+ sin2 =
A) 0 B) 1 C) 2 D) 3
30. Two vectors A and B are suc h that A B C and AA2+ B2 = C2
. If is the angle between positive direction of A and B then the correct statement is :
A) = B) = 2/3 C) = 0 D) = /2
31. Given that P QR = 0. Two out of the three vectors are equal in magnitude. The magnitude of the third vector is 2 times that of the other two. Which of the following can be the angles between these vectors ?
A) 900, 1350, 1350 B) 450, 450, 900
C) 300, 600, 900 D) 450, 900, 1350
32. A force
3iˆ4 jˆ
newton acts on a body and displace it by
3iˆ4 jˆ
metre. The work done by the force is:A) 5J B) 25J C) 10J D) 30J
33. The angles which the vector A3iˆ6 j 2kˆ ˆ makes with the coordinate axes are : A) cos-1 3 7, cos -1 6 7 , and cos-1 2 7 B) cos -1 4 7, cos -1 5 7 , and cos-1 3 7 C) cos-1 3 7 , cos-1 4 7 , and cos-1 1 7 D) none of these
34. The sum of two forces at a point is 16N. If their resultant is normal to the smaller force and has a magnitude of 8N. Then two forces are :
A) 6N, 10N B) 8N, 8N C) 4N, 12N D) 2N, 14N
35. A stationary man observes that the rain is falling vertically downward. When he starts running with a velocity of 12km/h he observes that the rain is falling at an angle 600 with the vertical. The actual velocity
of rain is
A) 12km/h B) 12km/h C) 4km/h D) 10km/h
36. A car A is going north east at 80kmh-1 and another car B is going south east with a velocity of 60kmh-1.
The velocity of A relative to B makes an angle with the north equal to :
A) tan-1 2 7 B) tan -1 7 2 C) tan -1 (7) D) tan-1 1 7
37. Two persons P and Q are standing 54m apart on a long moving belt. Person P rolls a round stone towards person Q with a speed of 9ms-1 with respect to the belt. If the belt is moving with a speed of
4ms-1 in the direction from P to Q, what is the speed of the stone with respect to an observer on a
stationary platform :
A) 4ms-1 B) 5 ms-1 C) 9 ms-1 D) 13 ms-1
38 Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity v and other with a uniform acceleration a. If a is the angle between the lines of motion of two particles then the least value of relative velocity will be at time given by :
A) v a sin B) v acos C) v a tan D) v acot
39. Two boats A and B are moving in a river. Boat A moves normal to the river currents as observed by an observer moving with river currents. Boat B moves normal to the river as observed by the observer on the ground.
A) To a ground observer boat B moves faster than A B) To a ground observer boat A moves faster than B C) To the moving observer boat B moves faster than A D) To the moving observer boat A moves faster than B
40. In the above problem, if both these boats have to cross a river of width b, then A) The boat A will take the least time to cross the river
B) The time taken by boat A will be b/v C) The time taken by boat B will be 2 2
b v u
D) The condition of crossing the river is given by v > u.
Velocity of motor boat with respect to water is 5 3 m/sec. The driver should steer the boat an angle A) 300 w.r.t. the line of destination from starting point
B) 600 w.r.t. normal to the bank
C) 1200w.r.t stream direction D) None of these
16.
17. 18. 19.
20.
21. 22. 23
24. 25.
26.
27. 28.
29. 30.
D
C
A
B
D
C
D
C
C
B
B
A
B
C
D
31
32
33
34
35
36
37
38
39
40
41
A
B
A
A
C
D
D
B
B
AB
C
CD
KEY
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12. 13.
14. 15.
C
B
D
C
D
D
B
D
A
D
D
C
A
A
A
1. To a man running at a speed of 4 m/s towards east, the rain appears to fall making an angle of 300 with
the vertical, when he doubles his speed, the rain again appears to fall making an angle of 300 but now in
the opposite direction to the vertical. The true angle q which the rain makes with the vertical and its speed v are
A) v = 4 3 m/s B) v = 6 m/s C) = 600 D) = 450
2. An observer moving with river stream finds that a boat is moving normal to the river currents with a velocity v, then
A) u = v
B) a ground observer finds that the boat is moving at an angle in the downstream direction C) the velocity observed by the ground observer is u 2 v2
D) the velocity observed by the ground observer is v 2 u2
3. Two boats A and B are moving in a river. Boat A moves normal to the river currents as observed by an observer moving with river currents. Boat B moves normal to the river as observed by the observer on the ground. If both these boats have to cross a river of width b, then
A) the boat A will take the least time to cross the river B) the time taken by A will be
v b
C) the time taken by boat B will be 2 2
u v
b
D) the condition of crossing the river is given by v>u
4. In the above problem to a stationary observer on the ground A) the boat B reaches the exactly opposite point on the bank
B) the boat A reaches a point b v u
distance downstream from the starting point C) the boat A travels a longer distance
D) the boat B travels a longer distance
MULTIPLE ANSWER TYPE QUESTIONS
KEY
1.
2.
3.
4.
A,C A,B A,B A,B
COMPREHENSION TYPE QUESTIONS
PASSAGE I :
When an airplane flies, its total velocity with respect to the ground is VtotalVplaneVwind
where Vplane denotes the plane’s velocity through motionless air and
wind
V
denotes the wind’s veloc-ity. Crucially, all the quantities in this equation are vectors. The magnitude of a velocity vector is often called the “speed”.
Consider an airplane whose speed through motionless air is 100 meters per second (m/s). To reach its destination, the plane must fly east.
The “heading” of a plane is the direction in which the nose of the plane points. So, it is the direction in which the engines propel the plane.
1. If the plane has an eastward heading, and a 20 m/s wind blows towards the southwest, then the plane’s speed is
A) 80 m/s B) more than 80 m/s but less than 100 m/s C) 100 m/s D) more than 100 m/s
2. The pilot maintains an eastward heading while a 20 m/s wind blows northward. The plane’s velocity is deflected from due east by what angle ?
A) sin-1 100 20 - B) cos-1 20 100 C) tan -1 20 100 D) None of these
3. Let denotes the answer to question 2. The plane in question 2 has what speed with respect to the ground ?
A) (100 m/s) sin B) (100 m/s) cos C) 100 m / s
sin D)
100 m / s cos
4. Because the 20 m/s northward wind persists, the pilot adjusts the heading so that the plane’s total velocity is eastward. By what angle does the new heading differ from due east ?
A) sin-1 20 100 B) cos -1 20 100 C) tan-1 20 100 D) None of these
5. Let q denote the answer to question 4. What is the total speed, with respect to the ground, of the plane in question 4 ?
A) (100 m/s) sin B) (100 m/s) cos C) 100 m / s
sin D)
100 m / s cos
1.
2.
3.
4.
5.
B
C
D
A
B
MULTIPLE MATCHING TYPE QUESTIONS
1. Match the following:
List – I List - II
a) cross product e) r x F b) scalar product f) scalar c) angle between two vectors g) work d) unit vector h) £ 1800 i) A x B | A x B | j) magnitude in units k) right hand screw rule 2. Match the following:
List – I List - II
a) Addition of two anti parallel vectors e) triangle law of vector b) A B C 0 f) 1
c) Parallelogram law g) null vector d) cos2 + cos2 + cos2S h)
A B (for A B _
3. Match the following:
List - I List - II a) couple e) joule b) work f) c) torque g) N - m d) F.S h) vector i) ML2 T-2
KEY
4. Match the following:
List - I List - II
a) C x A = B e) cos900
b) ˆ ˆi . j f) B.A= 0 c) (i x j). iˆ ˆ ˆ g) zero
h) C and A are in same plane
KEY
1. a e,k , b f,g , c h , d i,j 2. a g,h , b e , c e, d f
3. a g, h, i , b e,f,i , c g,h,i , d e,i 4. a f, h , b e, g , c e,g
