Vector
Physical quantities having magnitude and direction e.g. displacement, velocity gradent.
Polar Vector: vectors having a staring point e.g. displament, force etc.
Axial Vector: vectors that represent rotational effect and act along the axis of rotation. It's diration is given by right hand screw rule. e.g. torque, angular velocity, angular momentum.
Scalar
Physical quantities having magnitude only. e.g. volume, speed, potential etc.
. Necessary condition for being vector: The physical quanties must have both magnitude and direction.
. Sufficient condition for being vector: The physical quantities should obey the rules of vector addition.
Q. Which of the following is a vector quantity.
(a) electric current
(b) electric potential
( c) electric flux
(d) current density
Ans : (d)
Q. Physical quantity having magnitude and direction
(a) may be a vector
(b) must be a vector
( c) is not a vector
Ans : (a)
Q. A scalar quantity
(a) may have magnitude and may have direction.
(b) must have magnitude only but not direction.
( c) may not have magnitude and direction.
Ans : (a)
--> All vectors must have magnitude and direction but all physical quantities having magnitude and direction are not vectors. e.g. electric current, time, pressure.
Unit Vector
A vector divided by its own magnitude. It is unit less and dimensionless.
â = a / ǀ a ǀ
Vector can be represented by vector name with arrowhead or bold letter.
a = a1i + a2j + a3k = (a1,a2,a3)
ǀ a ǀ = √(a12 + a22 + a32)
Q. Two forces 3N and 4N are acting perpendicular onto a body a man 5kg. The value of acceleration generated in a body is
(a) 4 m/s2 (b) )3 m/s2 (c) 2 m/s2 (d) 1 m/s2
ans: (d)
we know, F = √(F12 + F22) = √(32 + 42) = 5 --->> a = F/m = 5/5 = 1 m/s2
Physical quantities having magnitude and direction e.g. displacement, velocity gradent.
Polar Vector: vectors having a staring point e.g. displament, force etc.
Axial Vector: vectors that represent rotational effect and act along the axis of rotation. It's diration is given by right hand screw rule. e.g. torque, angular velocity, angular momentum.
Scalar
Physical quantities having magnitude only. e.g. volume, speed, potential etc.
. Necessary condition for being vector: The physical quanties must have both magnitude and direction.
. Sufficient condition for being vector: The physical quantities should obey the rules of vector addition.
Q. Which of the following is a vector quantity.
(a) electric current
(b) electric potential
( c) electric flux
(d) current density
Ans : (d)
Q. Physical quantity having magnitude and direction
(a) may be a vector
(b) must be a vector
( c) is not a vector
Ans : (a)
Q. A scalar quantity
(a) may have magnitude and may have direction.
(b) must have magnitude only but not direction.
( c) may not have magnitude and direction.
Ans : (a)
--> All vectors must have magnitude and direction but all physical quantities having magnitude and direction are not vectors. e.g. electric current, time, pressure.
Unit Vector
A vector divided by its own magnitude. It is unit less and dimensionless.
â = a / ǀ a ǀ
Vector can be represented by vector name with arrowhead or bold letter.
a = a1i + a2j + a3k = (a1,a2,a3)
ǀ a ǀ = √(a12 + a22 + a32)
Q. Two forces 3N and 4N are acting perpendicular onto a body a man 5kg. The value of acceleration generated in a body is
(a) 4 m/s2 (b) )3 m/s2 (c) 2 m/s2 (d) 1 m/s2
ans: (d)
we know, F = √(F12 + F22) = √(32 + 42) = 5 --->> a = F/m = 5/5 = 1 m/s2
Triangle law of vectors
R = A + B
R = A + B
Parallelogram Law of Vector
R = ǀ A + B ǀ = √(A2 + 2AB cosθ + B2)
tanα = B Sinθ / (A + B cosθ) , with A
tanβ = A Sinθ / (B+ A cosθ) , with B (@ exchage A and B)
If A > B then α < β
If A < B then α > β
If A = B then α = β = θ/2
Note: The larger component makes smaller angle with the resultant.
Special cases of resultant
R = ǀ A + B ǀ = √(A2 + 2AB cosθ + B2)
--> If θ = 0o, then R = A + B = Rmax
--> If θ = 90o, then R = √(A2 + B2)
--> If θ = 180o, then R = A - B or B - A = Rmin
Note: so, the limit of resultant of magnitude of A and B lies A-B ≤ R ≤ A+B
R = ǀ A + B ǀ = √(A2 + 2AB cosθ + B2)
tanα = B Sinθ / (A + B cosθ) , with A
tanβ = A Sinθ / (B+ A cosθ) , with B (@ exchage A and B)
If A > B then α < β
If A < B then α > β
If A = B then α = β = θ/2
Note: The larger component makes smaller angle with the resultant.
Special cases of resultant
R = ǀ A + B ǀ = √(A2 + 2AB cosθ + B2)
--> If θ = 0o, then R = A + B = Rmax
--> If θ = 90o, then R = √(A2 + B2)
--> If θ = 180o, then R = A - B or B - A = Rmin
Note: so, the limit of resultant of magnitude of A and B lies A-B ≤ R ≤ A+B
Q. Which of the following can not be resultant of 20 and 30 vectors
(a) 50 (b) 35 (c) 15 (d) 8
Ans: (d)
Hint: Result remains between the range 30-20 ≤ R ≤ 30+20
Q. The magnitude of sum of two vectors is equal to the difference of their magnitude. The angle between these vectors will be
(a) 0o (b) 45o (c) 90o (d) 180o
Ans: (d)
Hint: ǀ a + b ǀ = ǀ a ǀ - ǀ b ǀ
or, a2 + 2ab cosθ + b2 = a2 - 2ab + b2
or, 2ab (1 + cosθ) = 0
or, cosθ = -1
=> θ = 180o
(a) 50 (b) 35 (c) 15 (d) 8
Ans: (d)
Hint: Result remains between the range 30-20 ≤ R ≤ 30+20
Q. The magnitude of sum of two vectors is equal to the difference of their magnitude. The angle between these vectors will be
(a) 0o (b) 45o (c) 90o (d) 180o
Ans: (d)
Hint: ǀ a + b ǀ = ǀ a ǀ - ǀ b ǀ
or, a2 + 2ab cosθ + b2 = a2 - 2ab + b2
or, 2ab (1 + cosθ) = 0
or, cosθ = -1
=> θ = 180o
Q. The magnitude of sum of two vectors is equal to the magnitude of their difference. The angle between these vectors will be
(a) 0o (b) 45o (c) 90o (d) 180o
Ans: (c)
Hint: ǀ a + b ǀ = ǀ a - b ǀ
or, a2 + 2ab cosθ + b2 = a2 - 2ab cosθ + b2
or, 4ab cosθ = 0
or, 4cosθ = 0
=> θ = 90o
Q. The sets of vectors are acting on a body, whose resultant can not be zero.
(a)10, 10, 10 (b) 10, 10, 20 (c) 10, 20, 20 (d) 10, 20, 40
Ans: (d)
Hint: To form a triangle, sum of two smaller sides should be ≥ larger side. (the sides 1, 2,5 can not form a triangle)
Note: A particle is travelling with a velocity v along certain direction If it turns θ, then the change in its velocity
ΔV2 = v12 + 2v1v2 cos(180-θ) + v22
= v2 + 2v2 cos(180-θ) + v2
= 2v2 (1 + cosθ)
= 4 v2 sin2 θ/2
=>ΔV = 2 v sinθ/2 (remember)
Q. A particle is moving eastward with a velocity of 5m/s. In 10s the velocity changes to 5m/s northward. What is average acceleration in this case.
(a) 1 m/s2 (b) √2 m/s2 (c) 1/√2 m/s2 (d) 2√2 m/s2
ans: (c)
Hint: Δv = 2v sin θ/2 = 2 * 5 * sin45o = 5√2
a = Δv/Δt = 5√2/10 = 1/√2
Q. A particle is revolving in a circle at constant speed of 10 m/s. Find the change in velocity in turning through 60o.
(a) 0 (b) 10√3 (c) 10 (d) 5√3
Ans: (c)
Hint: use formula Δv = 2v sin θ/2
Product of two vectors
Dot Product / scalar Product
a.b = a b cosθ
a.b = (a1i + a2j + a3k).( b1i + b2j + b3k) = a1b1 + a2b2 + a3b3
Cases:
when θ = 0, a.b = ab (max)
(a) 0o (b) 45o (c) 90o (d) 180o
Ans: (c)
Hint: ǀ a + b ǀ = ǀ a - b ǀ
or, a2 + 2ab cosθ + b2 = a2 - 2ab cosθ + b2
or, 4ab cosθ = 0
or, 4cosθ = 0
=> θ = 90o
Q. The sets of vectors are acting on a body, whose resultant can not be zero.
(a)10, 10, 10 (b) 10, 10, 20 (c) 10, 20, 20 (d) 10, 20, 40
Ans: (d)
Hint: To form a triangle, sum of two smaller sides should be ≥ larger side. (the sides 1, 2,5 can not form a triangle)
Note: A particle is travelling with a velocity v along certain direction If it turns θ, then the change in its velocity
ΔV2 = v12 + 2v1v2 cos(180-θ) + v22
= v2 + 2v2 cos(180-θ) + v2
= 2v2 (1 + cosθ)
= 4 v2 sin2 θ/2
=>ΔV = 2 v sinθ/2 (remember)
Q. A particle is moving eastward with a velocity of 5m/s. In 10s the velocity changes to 5m/s northward. What is average acceleration in this case.
(a) 1 m/s2 (b) √2 m/s2 (c) 1/√2 m/s2 (d) 2√2 m/s2
ans: (c)
Hint: Δv = 2v sin θ/2 = 2 * 5 * sin45o = 5√2
a = Δv/Δt = 5√2/10 = 1/√2
Q. A particle is revolving in a circle at constant speed of 10 m/s. Find the change in velocity in turning through 60o.
(a) 0 (b) 10√3 (c) 10 (d) 5√3
Ans: (c)
Hint: use formula Δv = 2v sin θ/2
Product of two vectors
Dot Product / scalar Product
a.b = a b cosθ
a.b = (a1i + a2j + a3k).( b1i + b2j + b3k) = a1b1 + a2b2 + a3b3
Cases:
when θ = 0, a.b = ab (max)
when θ = 90o, a.b = 0 (condition of perpendicularity)
when θ = 180o, a.b = -ab(min)
i.i =j.j = k.k = 1
Limit of product: -ab ≤ a.b ≤ ab
when θ = 180o, a.b = -ab(min)
i.i =j.j = k.k = 1
Limit of product: -ab ≤ a.b ≤ ab
Cross of product / vector product
a×b = ab sinθ n
n is unit vector perpendicular to the plane of a and b .
a×b = ab sinθ = (a1i + a2j + a3k) × ( b1i + b2j + b3k)
Cases:
when θ = 0o or 180o , a×b = 0 (null vector) (condition of parallelism)
when θ = 90o, a×b = ab ( max)
Limit of product: 0 ≤ a×b ≤ ab
Q. If the product of two vectors was found to be negative , then this product must be
(a) dot product
(b) cross product
(c) neither dot nor cross
(d) none
Ans: (a)
Note: Dot product may be negative , zero , +ve but cross product is always +ve or equal to zero in magnitude.
Q. If sum of vectors perpendicular to their vector difference, then these vectors
(a) must be equal (b) must be equal in magnitude (c)must be perpendicular (d) must be parallel
ans: (b)
Hint: (a+b).(a-b) = 0
a2 - a.b + a.b +b2 = 0
=> a = b
Angle between two vector
a.b = ab cosθ => cosθ = a.b / ab
ǀ a×b ǀ = ab sinθ => sinθ = ǀa×bǀ / ab
tanθ = ( ǀa×bǀ) / ( a.b)
Q. The dot product of two vectors is equal to the magnitude of their cross product . The angle between those vectors should be
(a) 30o (b) 45o (c) 60o (d) 90o
Ans: (b)
Hint: tanθ = ( ǀa×bǀ) / ( a.b)
Q. A person moves 30 m north, then 20 m east and finally 30√2 m south - west. What is his displacement from the original position?
s = s1 + s2 + s3
s1 = 30 j, s2 = 20 i
s3 = -30√2 sin 45o i - 30√2 sin 45o j
= -30 i - 30 j
so, s = 30 j + 20 i - 30 i - 30 j
= - 10 j
Resultant displacement = 10 m west.
a×b = ab sinθ n
n is unit vector perpendicular to the plane of a and b .
a×b = ab sinθ = (a1i + a2j + a3k) × ( b1i + b2j + b3k)
Cases:
when θ = 0o or 180o , a×b = 0 (null vector) (condition of parallelism)
when θ = 90o, a×b = ab ( max)
Limit of product: 0 ≤ a×b ≤ ab
Q. If the product of two vectors was found to be negative , then this product must be
(a) dot product
(b) cross product
(c) neither dot nor cross
(d) none
Ans: (a)
Note: Dot product may be negative , zero , +ve but cross product is always +ve or equal to zero in magnitude.
Q. If sum of vectors perpendicular to their vector difference, then these vectors
(a) must be equal (b) must be equal in magnitude (c)must be perpendicular (d) must be parallel
ans: (b)
Hint: (a+b).(a-b) = 0
a2 - a.b + a.b +b2 = 0
=> a = b
Angle between two vector
a.b = ab cosθ => cosθ = a.b / ab
ǀ a×b ǀ = ab sinθ => sinθ = ǀa×bǀ / ab
tanθ = ( ǀa×bǀ) / ( a.b)
Q. The dot product of two vectors is equal to the magnitude of their cross product . The angle between those vectors should be
(a) 30o (b) 45o (c) 60o (d) 90o
Ans: (b)
Hint: tanθ = ( ǀa×bǀ) / ( a.b)
Q. A person moves 30 m north, then 20 m east and finally 30√2 m south - west. What is his displacement from the original position?
s = s1 + s2 + s3
s1 = 30 j, s2 = 20 i
s3 = -30√2 sin 45o i - 30√2 sin 45o j
= -30 i - 30 j
so, s = 30 j + 20 i - 30 i - 30 j
= - 10 j
Resultant displacement = 10 m west.
Q. If r = (2i + 2j) m and F = (3i + 2j) N. Find out whether the torque is clock wise or anti clock wise.
T = r × F
= i (0 - 0) - j (0 - 0) + k (4 - 6)
= - 2k
Here n = (- 2k) /(ǀ -2kǀ) = (- 2k) /2 = -k => clock wise.
If n = +ve =>anti close wise
If n = -ve => close wise
Lami's theorem
P/ sin β = Q /sin γ = R /sinα
Q. A mass of 1 kg is attached to the middle of a rope, which is being pulled from both ends in the opposite directions. Take g= 10 m/s2, what is the minimum pull required to completely straighten the rope.
(a) 5N (b) 10N (c) 20N (d) much larger than 20N
Ans: (d)
To strengthen the rope
γ = 180o , α = β = 90o
T1 / sinα = T2 / sin β = mg / sin γ
or T1 = T2 = mg sin 90o / sin 180o = ∞
Q. When forces F1 , F2 , F3 are acting on a particle of mass m such that F2 and F3 are mutually perpendicular , then the particle remains stationary. If the force F1 is now removed, then acceleration of the particle is
(a) F1/m (b) F2 F3/m F1 (c) (F2 -F1)/m (d) F2 /m
Ans: (a)
F3 + F2 = F1 (to be stationary)
Resultant of ǀF3 + F2ǀ = F1
Acceleration = F1/ m
Note: All the gradients (e.g. temperature gradient, velocity gradient) are the vector quantities. Small angle 'dθ' is a vector quantity but the bigger angles are not vector quantities.
T = r × F
= i (0 - 0) - j (0 - 0) + k (4 - 6)
= - 2k
Here n = (- 2k) /(ǀ -2kǀ) = (- 2k) /2 = -k => clock wise.
If n = +ve =>anti close wise
If n = -ve => close wise
Lami's theorem
P/ sin β = Q /sin γ = R /sinα
Q. A mass of 1 kg is attached to the middle of a rope, which is being pulled from both ends in the opposite directions. Take g= 10 m/s2, what is the minimum pull required to completely straighten the rope.
(a) 5N (b) 10N (c) 20N (d) much larger than 20N
Ans: (d)
To strengthen the rope
γ = 180o , α = β = 90o
T1 / sinα = T2 / sin β = mg / sin γ
or T1 = T2 = mg sin 90o / sin 180o = ∞
Q. When forces F1 , F2 , F3 are acting on a particle of mass m such that F2 and F3 are mutually perpendicular , then the particle remains stationary. If the force F1 is now removed, then acceleration of the particle is
(a) F1/m (b) F2 F3/m F1 (c) (F2 -F1)/m (d) F2 /m
Ans: (a)
F3 + F2 = F1 (to be stationary)
Resultant of ǀF3 + F2ǀ = F1
Acceleration = F1/ m
Note: All the gradients (e.g. temperature gradient, velocity gradient) are the vector quantities. Small angle 'dθ' is a vector quantity but the bigger angles are not vector quantities.
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