1. A vector A makes an angle of 20° and B makes an angle of 110° with the X-axis. The magnitudes of these vectors are 3 m and 4 m respectively. Find the resultant.
Answer: It is clear that angle between vectors A and B, θ = 110° -20° = 90°. Hence magnitude of the resultant = √(A2+B2) = √(32+42) =√(9+16) =√25 = 5 m.
If α be the angle between resultant and A, then tanα=B/A= 4/3, that gives α = 53.13°. So the resultant makes an angle 20°+53.13° =73.13° with the X-axis.
2. Let A and B be the two vectors of magnitude 10 units each. If they are inclined to X-axis at an angle 30° and 60° respectively, find the resultant.
If α be the angle between resultant and A, then tanα=B/A= 4/3, that gives α = 53.13°. So the resultant makes an angle 20°+53.13° =73.13° with the X-axis.
2. Let A and B be the two vectors of magnitude 10 units each. If they are inclined to X-axis at an angle 30° and 60° respectively, find the resultant.
Answer: Angle between the vectors θ= 60°-30° =30°. Hence magnitude of the resultant = √(A²+B²+2ABcosθ) = √(10²+10²+2.10.10.cos30°) =√(200+200.√3/2) =10√(2+√3) = 19.32 units
Since the vectors have equal magnitudes so the resultant bisects the angle θ between them. Angle between resultant and vector A =30°/2 =15°. So angle between resultant and X-axis =15°+30° =45°.
Since the vectors have equal magnitudes so the resultant bisects the angle θ between them. Angle between resultant and vector A =30°/2 =15°. So angle between resultant and X-axis =15°+30° =45°.
3. Add vectors A, B and C each having magnitude of 100 units and inclined to the X-axis at an angle 45°, 135° and 315° respectively.
Answer: For clarity. let us draw the given vectors as below,
It is clear that out of the three vectors B and C are equal in magnitude but opposite in direction, so their sum B + C = 0. Hence sum of the three vectors A+B+C = A which has magnitude of 100 units and inclined at 45° to X-axis.
4. Let a=4i+3j and b=3i+4j. Find the magnitudes of (a) a,(b) b,(c) a+band (d) a-b.
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| Refer Question - 3 |
4. Let a=4i+3j and b=3i+4j. Find the magnitudes of (a) a,(b) b,(c) a+band (d) a-b.
Answer: (a) Magnitude of a = √(4²+3²) =√25 =5,
(b)Magnitude of b = √(3²+4²) =√25 =5,
a+b =7i+7j and a-b =i-j
(c) Magnitude of a+b = √(7²+7²) =√(49+49) =7√2
(d) Magnitude of a-b = √(1²+1²) =√2
5. Refer to figure below. Find (a) the magnitude,(b) x and y components and (c) the angle with the X-axis of the resultant of OA, BC and DE.
Answer: Let us resolve these vectors along x and y axes first.
Magnitude of x-component of OA = OA. cos 30° =2*√3/2 =√3 m
Magnitude of x-component of BC = BC. cos 120° =1.5 *(-1/2) =-0.75 m
Magnitude of x-component of DE = DE. cos 270° =0 m
(angles of cosine are taken from the positive direction of x-axis)
Magnitude of y-component of OA = OA. sin 30° =2*1/2 =1 m
Magnitude of y-component of BC = BC. sin 120° =1.5 *(√3/2) =0.75√3
Magnitude of y-component of DE = DE. sin 270° =-1 m
(b)Now x-component of the resultant = Sum of x-components of the vectors =√3-0.75+0 =1.73-0.75 =0.98 m and y-component of the resultant = Sum of y-components of the vectors =1+0.75√3-1 =1.30 m
(a) Magnitude of resultant =√(square of x-comp.+square of y-comp) =√(0.98²+1.3²) =√(0.96+1.69) =1.6 m
(c) angle with X-axis of resultant is given by tanθ =Magnitude of y-comp/Magnitude of x-comp = 1.30/0.98 =1.32, so θ=tan-1(1.32).
6. Two vectors have magnitudes 3 units and 4 units respectively. What should be the angle between them if the magnitude of the resultant is (a) 1 unit, (b) 5 units and (c) 7 units.
Answer: Magnitudes of vectors =a =3 units, b =4 units, let θ be angle between them and c= magnitude of resultant.
We know that c²=a²+b²+2abCosθ,
So Cosθ=(c²-a²-b²)/2ab,
(a) If c =1, Cosθ =(1-9-16)/24 =-1, So θ =180°
(b) If c =5, Cosθ =(25-9-16)/24 =0, So θ =90°
(c) If c =7, Cosθ =(49-9-16)/24 =1, So θ =0°
7. A spy report about a suspected car reads as follows. "The car moved 2.00 km towards east, made a perpendicular left turn, ran for 500 m, made a perpendicular right turn, ran for 4.00 km and stopped". Find the displacement of the car.
Answer: Let us take east as X-axis and north as Y-axis and draw the displacements of the car as in figure below,
Taking unit vectors in X and Y directions as i and j, the given displacements can be written as,
OA = 2i, AB =0.5j and BC =4i
Hence the resultant displacement vector OC = OA+AB+BC=2i,+0.5j+4i =6i+0.5j
Magnitude of the displacement = √(6²+0.5²) =√36.25 = 6.02 km
Angle between resultant displacement and east is given by tanθ =(0.5/6) =1/12,
So θ=tan-1(1/12)
8. A carrom board (4 ft x 4 ft square) has the queen at the center. The queen, hit by the striker moves to the front edge, rebounds and goes in the hole behind the striking line. Find the magnitude of displacement of the queen (a) from the center to the front edge, (b) from the front edge to the hole and (c) from the center to the hole.
Answer: Let us draw the sketch,
O is the center of the carrom board. P is the point on the front edge where queen strikes. PA is the rebound path. We must understand that like the reflection of light OP and PA will make equal angle θ with the normal at P. From the figure PQ=OQ tan θ =2tan θ and
DP=DA tanθ =4 tanθ,
DQ =DP+PQ = 2 tanθ + 4 tanθ =6 tan θ = 2,
tan θ =1/3
Now Magnitude of displacement of the queen from.
(a) the center to the front edge OP = √(OQ²+PQ²) =√(4+4 tan²θ) =2√(1+1/9) =2√(10/9) =⅔√10 ft
(b) the front edge to hole =PA =√(PD²+DA²) =√(16 tan²θ+16) =(4√10)/3 ft
(c) the center to the hole OA =√(2²+2²) =√8 =2√2 ft
9. A mosquito net over a 7 ft x 4 ft bed is three ft high. The net has a hole at one corner of the bed through which a mosquito enters the net. It flies and sits at the diagonally opposite upper corner of the net. (a) Find the magnitude of the displacement of the mosquito. (b) Taking the hole as the origin, the length of the bed as the X-axis, its width as the Y-axis and vertically up as the Z-axis, write the components of the displacement vector.
Answer: Let us see this problem in the following figure:-
(a) The mosquito net is a cuboid of dimension 7 ft x 4 ft x 3 ft and the magnitude of displacement vector OA of the mosquito is the length of the diagonal of the cuboid which is = √(7²+4²+3²) =√74 ft.
(b) From the figure magnitudes of the components of displacement vector are 7 ft, 4 ft and 3 ft along X,Y and Z-axes respectively.
10. Suppose a is a vector of magnitude 4.5 units due north. What is the vector (a) 3a, (b) -4a ?
Answer: (a) Vector 3a is such a vector which magnitude is three times that of magnitude of vector a =3x4.5 =13.5 units and has the same direction, due north.
(b) The vector -4a is due south with magnitude 4 times that of a =4x4.5 =18 units.
11. Two vectors have magnitudes 2 m and 3 m. The angle between them is 60°. Find (a) The scalar product of the two vectors, (b) The magnitude of their vector product.
Answer: Let the magnitudes of two vectors be, a=2 m and b=3 m
θ= angle between the vectors =60°.
(a) The scalar product of the twovectors is given by abCosθ
= 2.3.Cos60° = 3 m².
(b) The magnitude of their vector product is = abSinθ =2.3.Sin60° =2.3.√3/2 =3√3 m².
12. Let A1A2A3A4A5A6A1 be a regular hexagon. Write the x- components of the vectors represented by the six sides taken in order. Use the fact that the resultant of these six vectors is zero, to prove that
cos0+cosπ/3+cos2π/3+cos3π/3+cos4π/3+cos5π/3=0.
Use the known cosine values to verify the result.
Answer: Let 'a' be the length of each side of the regular hexagon and ibe the unit vector along x-axis.
X=components of the vector A1A2 =aCos0° i=a i
vector A2A3 =aCos60° i =0.5a i
vector A3A4 =aCos120° i=-0.5a i
vector A4A5 =aCos180° i=-a i
vector A5A6=aCos240° i=-0.5a i
vector A6A1=aCos300° i=0.5a i
Since resultant of these vectors are zero hence sum of magnitudes of x-components of these vectors will also be zero ie.
aCos0°+aCos60°+aCos120°+aCos180°+aCos240°+aCos300°=0
writing the angles in radian and dividing both sides by a we get
Cos0+Cosπ/3+Cos2π/3+Cos3π/3+Cos4π/3+Cos5π/3=0
13. Let a=2i+3j+4k and b=3i+4j+5k. Find the angle between them.
Answer: Magnitudes a=√(2²+3²+4²) =√29, b=√(3²+4²+5²) =√50
Dot product of these vectors a.b =(2i+3j+4k).(3i+4j+5k) =2.3+3.4+4.5 =6+12+20 =38
But the value of dot product is also given by abCosθ
So, abCosθ=38
Cosθ=38/ab=38/(√29.√50) =38/√1450
θ=Cos-1(38/√1450)
14. Prove that A.(AxB)=0.
Answer: A.(AxB) = A.(ABsinθ)k =(ABsinθ)A.k = 0
(Where A and B are magnitudes and θ angle between them. k is the unit vector perpendicular to both A and B. Since A and k are perpendicular hence A.k=0)
15. If A=2i+3j+4k and B=4i+3j+2k. Find AxB.
Answer: AxB=(3.2-4.3)i+(4.4-2.2)j+(2.3-3.4)k = -6i+12j-6k
16. If A,B,C are mutually perpendicular, show that Cx(AxB)=0. Is the converse true?
Answer: Let i, j and k be the unit vectors along A, B and C.
Now Cx(AxB)= Ck x (ABsin90°)k =(ABC) k x k =0
(Since k x k =0)
Let us see the converse, given (AxB)xC=0,
if θ be angle between A and B and k unit vector perpendicular to bothA an B
=> ABsinθ k x C = 0
=> ABsinθ.1.Csinß.u =0 (ß is angle between k and C and u the unit vector along perpendicular to both k and C)
=> This condition gives either θ =0 or ß=0. So they are not mutually perpendicular and the converse is not true.
17. A particle moves on a given straight line with a constant speed v. At a certain time it is at a point P on its straight line path. O is a fixed point. Show that OP x v is independent of the position P.
Answer:
OP x v = OP.v.sin(180°-θ) k (k unit vector perpendicular to plain containing OP and v, and 180°-θ is the angle between them)
=>OP x v = OP.v.sinθ k = v.(OP.sinθ) k =(v.OQ) k
This expression is independent of θ or OP and constant because OQ is the perpendicular distance between line and point O. So OP x v does not depend on the position P.
18. The force on a charged particle due to electric and magnetic fields is given by F=qE+qvxB. Suppose E is along the X-axis and B along the Y-axis. In what direction and with what minimum speed v should a positively charged particle be sent so that the net force on it is zero?
Answer: Given that F=qE+qvxB=0,
=> qvxB=-qE.
=> -vB i = -E i (If v is along z-axis θ=90° for v and B)
=> v=E/B
19. Give an example for which A.B=C.B but A ≠ C .
Answer: A.B=C.B
=> AB cosθ = CB cosß
=> A cosθ = C cosß
=> cosθ = (C/A) cosß - (necessary condition)
Let C/A=0.5 and ß=60° then cosθ = 0.5 x 0.5 =0.25, This gives θ =75.5°
So if A=2C, angle between B & C, ß=60° and angle between B & A, θ =75.5° then A.B=C.B ; is an example.
(b)Magnitude of b = √(3²+4²) =√25 =5,
a+b =7i+7j and a-b =i-j
(c) Magnitude of a+b = √(7²+7²) =√(49+49) =7√2
(d) Magnitude of a-b = √(1²+1²) =√2
5. Refer to figure below. Find (a) the magnitude,(b) x and y components and (c) the angle with the X-axis of the resultant of OA, BC and DE.
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| Fig for Q-5 |
Answer: Let us resolve these vectors along x and y axes first.
Magnitude of x-component of OA = OA. cos 30° =2*√3/2 =√3 m
Magnitude of x-component of BC = BC. cos 120° =1.5 *(-1/2) =-0.75 m
Magnitude of x-component of DE = DE. cos 270° =0 m
(angles of cosine are taken from the positive direction of x-axis)
Magnitude of y-component of OA = OA. sin 30° =2*1/2 =1 m
Magnitude of y-component of BC = BC. sin 120° =1.5 *(√3/2) =0.75√3
Magnitude of y-component of DE = DE. sin 270° =-1 m
(b)Now x-component of the resultant = Sum of x-components of the vectors =√3-0.75+0 =1.73-0.75 =0.98 m and y-component of the resultant = Sum of y-components of the vectors =1+0.75√3-1 =1.30 m
(a) Magnitude of resultant =√(square of x-comp.+square of y-comp) =√(0.98²+1.3²) =√(0.96+1.69) =1.6 m
(c) angle with X-axis of resultant is given by tanθ =Magnitude of y-comp/Magnitude of x-comp = 1.30/0.98 =1.32, so θ=tan-1(1.32).
6. Two vectors have magnitudes 3 units and 4 units respectively. What should be the angle between them if the magnitude of the resultant is (a) 1 unit, (b) 5 units and (c) 7 units.
Answer: Magnitudes of vectors =a =3 units, b =4 units, let θ be angle between them and c= magnitude of resultant.
We know that c²=a²+b²+2abCosθ,
So Cosθ=(c²-a²-b²)/2ab,
(a) If c =1, Cosθ =(1-9-16)/24 =-1, So θ =180°
(b) If c =5, Cosθ =(25-9-16)/24 =0, So θ =90°
(c) If c =7, Cosθ =(49-9-16)/24 =1, So θ =0°
7. A spy report about a suspected car reads as follows. "The car moved 2.00 km towards east, made a perpendicular left turn, ran for 500 m, made a perpendicular right turn, ran for 4.00 km and stopped". Find the displacement of the car.
Answer: Let us take east as X-axis and north as Y-axis and draw the displacements of the car as in figure below,
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| Figure for question no.-7 |
OA = 2i, AB =0.5j and BC =4i
Hence the resultant displacement vector OC = OA+AB+BC=2i,+0.5j+4i =6i+0.5j
Magnitude of the displacement = √(6²+0.5²) =√36.25 = 6.02 km
Angle between resultant displacement and east is given by tanθ =(0.5/6) =1/12,
So θ=tan-1(1/12)
8. A carrom board (4 ft x 4 ft square) has the queen at the center. The queen, hit by the striker moves to the front edge, rebounds and goes in the hole behind the striking line. Find the magnitude of displacement of the queen (a) from the center to the front edge, (b) from the front edge to the hole and (c) from the center to the hole.
Answer: Let us draw the sketch,
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| Fig. Q no.- 8 |
DP=DA tanθ =4 tanθ,
DQ =DP+PQ = 2 tanθ + 4 tanθ =6 tan θ = 2,
tan θ =1/3
Now Magnitude of displacement of the queen from.
(a) the center to the front edge OP = √(OQ²+PQ²) =√(4+4 tan²θ) =2√(1+1/9) =2√(10/9) =⅔√10 ft
(b) the front edge to hole =PA =√(PD²+DA²) =√(16 tan²θ+16) =(4√10)/3 ft
(c) the center to the hole OA =√(2²+2²) =√8 =2√2 ft
9. A mosquito net over a 7 ft x 4 ft bed is three ft high. The net has a hole at one corner of the bed through which a mosquito enters the net. It flies and sits at the diagonally opposite upper corner of the net. (a) Find the magnitude of the displacement of the mosquito. (b) Taking the hole as the origin, the length of the bed as the X-axis, its width as the Y-axis and vertically up as the Z-axis, write the components of the displacement vector.
Answer: Let us see this problem in the following figure:-
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| Figure for Question no -9 |
(b) From the figure magnitudes of the components of displacement vector are 7 ft, 4 ft and 3 ft along X,Y and Z-axes respectively.
10. Suppose a is a vector of magnitude 4.5 units due north. What is the vector (a) 3a, (b) -4a ?
Answer: (a) Vector 3a is such a vector which magnitude is three times that of magnitude of vector a =3x4.5 =13.5 units and has the same direction, due north.
(b) The vector -4a is due south with magnitude 4 times that of a =4x4.5 =18 units.
11. Two vectors have magnitudes 2 m and 3 m. The angle between them is 60°. Find (a) The scalar product of the two vectors, (b) The magnitude of their vector product.
Answer: Let the magnitudes of two vectors be, a=2 m and b=3 m
θ= angle between the vectors =60°.
(a) The scalar product of the twovectors is given by abCosθ
= 2.3.Cos60° = 3 m².
(b) The magnitude of their vector product is = abSinθ =2.3.Sin60° =2.3.√3/2 =3√3 m².
12. Let A1A2A3A4A5A6A1 be a regular hexagon. Write the x- components of the vectors represented by the six sides taken in order. Use the fact that the resultant of these six vectors is zero, to prove that
cos0+cosπ/3+cos2π/3+cos3π/3+cos4π/3+cos5π/3=0.
Use the known cosine values to verify the result.
Answer: Let 'a' be the length of each side of the regular hexagon and ibe the unit vector along x-axis.
X=components of the vector A1A2 =aCos0° i=a i
vector A2A3 =aCos60° i =0.5a i
vector A3A4 =aCos120° i=-0.5a i
vector A4A5 =aCos180° i=-a i
vector A5A6=aCos240° i=-0.5a i
vector A6A1=aCos300° i=0.5a i
Since resultant of these vectors are zero hence sum of magnitudes of x-components of these vectors will also be zero ie.
aCos0°+aCos60°+aCos120°+aCos180°+aCos240°+aCos300°=0
writing the angles in radian and dividing both sides by a we get
Cos0+Cosπ/3+Cos2π/3+Cos3π/3+Cos4π/3+Cos5π/3=0
13. Let a=2i+3j+4k and b=3i+4j+5k. Find the angle between them.
Answer: Magnitudes a=√(2²+3²+4²) =√29, b=√(3²+4²+5²) =√50
Dot product of these vectors a.b =(2i+3j+4k).(3i+4j+5k) =2.3+3.4+4.5 =6+12+20 =38
But the value of dot product is also given by abCosθ
So, abCosθ=38
Cosθ=38/ab=38/(√29.√50) =38/√1450
θ=Cos-1(38/√1450)
14. Prove that A.(AxB)=0.
Answer: A.(AxB) = A.(ABsinθ)k =(ABsinθ)A.k = 0
(Where A and B are magnitudes and θ angle between them. k is the unit vector perpendicular to both A and B. Since A and k are perpendicular hence A.k=0)
15. If A=2i+3j+4k and B=4i+3j+2k. Find AxB.
Answer: AxB=(3.2-4.3)i+(4.4-2.2)j+(2.3-3.4)k = -6i+12j-6k
16. If A,B,C are mutually perpendicular, show that Cx(AxB)=0. Is the converse true?
Answer: Let i, j and k be the unit vectors along A, B and C.
Now Cx(AxB)= Ck x (ABsin90°)k =(ABC) k x k =0
(Since k x k =0)
Let us see the converse, given (AxB)xC=0,
if θ be angle between A and B and k unit vector perpendicular to bothA an B
=> ABsinθ k x C = 0
=> ABsinθ.1.Csinß.u =0 (ß is angle between k and C and u the unit vector along perpendicular to both k and C)
=> This condition gives either θ =0 or ß=0. So they are not mutually perpendicular and the converse is not true.
17. A particle moves on a given straight line with a constant speed v. At a certain time it is at a point P on its straight line path. O is a fixed point. Show that OP x v is independent of the position P.
Answer:
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| Fig. Q no.-17 |
OP x v = OP.v.sin(180°-θ) k (k unit vector perpendicular to plain containing OP and v, and 180°-θ is the angle between them)
=>OP x v = OP.v.sinθ k = v.(OP.sinθ) k =(v.OQ) k
This expression is independent of θ or OP and constant because OQ is the perpendicular distance between line and point O. So OP x v does not depend on the position P.
18. The force on a charged particle due to electric and magnetic fields is given by F=qE+qvxB. Suppose E is along the X-axis and B along the Y-axis. In what direction and with what minimum speed v should a positively charged particle be sent so that the net force on it is zero?
Answer: Given that F=qE+qvxB=0,
=> qvxB=-qE.
=> -vB i = -E i (If v is along z-axis θ=90° for v and B)
=> v=E/B
19. Give an example for which A.B=C.B but A ≠ C .
Answer: A.B=C.B
=> AB cosθ = CB cosß
=> A cosθ = C cosß
=> cosθ = (C/A) cosß - (necessary condition)
Let C/A=0.5 and ß=60° then cosθ = 0.5 x 0.5 =0.25, This gives θ =75.5°
So if A=2C, angle between B & C, ß=60° and angle between B & A, θ =75.5° then A.B=C.B ; is an example.
