When two vectors P → and Q → are inclined at an angle θ and act on a body, the resultant is ( 2 k + 1 ) P 2 + Q 2 . When the vectors are inclined at an angle 90 ° - θ , the resultant is ( 2 k - 1 ) p 2 + Q 2 . Prove that tan θ = ( k - 1 ) / ( k + 1 ) .

Given that the resultant of two vectors P , Q inclined at an angle. θ is R = 2 k + 1 P 2 + Q 2 And when their angle is 90 - θ , then their resultant is R ' = 2 k - 1 P 2 + Q 2 . By the law of parallelogram of vectors, the resultant of two vectors is given by R 2 = P 2 + Q 2 + 2 PQ cos θ In the first case, we have R 2 = 2 k + 1 2 P 2 + Q 2 = P 2 + Q 2 + 2 PQ cosθ ...............(1) And in the second case R ' 2 = 2 k - 1 2 P 2 + Q 2 = P 2 + Q 2 + 2 PQsinθ ..............(2) [ as cos ( 90 - θ ) = sin θ ] Rearranging we have R 2 = 2 k + 1 2 P 2 + Q 2 - P 2 + Q 2 = 2 PQ cosθ ..............(3), and R ' 2 = 2 k - 1 2 P 2 + Q 2 - P 2 + Q 2 = 2 PQ sin θ ................(4) Dividing Eqn. (4) by (3) and Simplifying, we get 2 k - 1 2 - 1 P 2 + Q 2 2 k + 1 2 - 1 P 2 + Q 2 = tan θ Which gives tan θ = k - 1 k + 1

The resultant of two vectors A → and B → acting at a point is equal to 3 B and makes an angle 30 ° with the direction of A → . Prove that | A → | = | B → | or | A → | = | 2 B → | .

Given the resultant of the vectors A → , B → = R → = 3 B . The direction of the resultant with A → = 30 ° . Applying the formula of the parallelogram of vectors, we have, 3 B 2 = A 2 + B 2 + 2 A B cos θ or 2 B 2 = A 2 + 2 AB cosθ ...................(1) The angle between the resultant R → and A → = 30 ° . sin 30 = 1 2 = B sin θ R = B sinθ B 3 or sin θ = 3 2 and cos θ = 1 2 . Now from equation (1), 2 B 2 = A 2 + AB or A 2 + AB - 2 B 2 = 0 . A 2 + 2 AB - AB - 2 B 2 = 0 or A ( A + 2 B ) - B ( A + 2 B ) = 0 . Or A + 2 B ( A - B ) = 0 . That gives A = B and A = 2 B .

To a person moving due west in a train with a velocity 40 kmh - 1 , a car appears to move due north with a velocity of 40 3 kmh - 1 . What is the actual velocity and direction of motion of the car?

Velocity of the train due west V t = 40 km / h . Velocity of the car with respect to the train V c = 40 3 k m / h due north. The actual velocity of the car V A = ? Refer to the figurw , the man in the train is travelling on the west ditection along AB with 40 km / hr . Car is apparently travelling in the the north direction along A O with a speed of 40 3 k m p h . But it it is actually travelling due AC with a velocity of V A . Completing the parallelogram of vectors, we have , V A = V t 2 + V c 2 . V A = ( 40 3 ) 2 + ( 40 ) 2 or 4799 + 1600 = 6399 . 7 = 80 k m p h . Vc = 80 k m / h . And its direction, T a n θ = V t V c = 40 40 3 = 1 3 o r 30 ° W e s t o f n o r t h .

MEDIUM

11th West Bengal Board

IMPORTANT

Earn 100

What is the angle between two equal vectors so that their resultant is one-fourth of one of the vectors?

Important Questions on Scalars and Vectors

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 4 - Scalars and Vectors>EXERCISE>Q 3

When two vectors

\vec{P}

and

\vec{Q}

are inclined at an angle

θ

and act on a body, the resultant is

(2 k + 1) \sqrt{P^{2} + Q^{2}}

. When the vectors are inclined at an angle

(90 ° - θ)

, the resultant is

(2 k - 1) \sqrt{p^{2} + Q^{2}}

. Prove that

\tan θ = (k - 1) / (k + 1)

MEDIUM

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 4 - Scalars and Vectors>EXERCISE>Q 4

\vec{P}, \vec{Q}

and

\vec{R}

are three coplanar vectors. Show by geometric construction that

\vec{P} + (\vec{Q} - \vec{R}) = \vec{Q} - (\vec{R} - \vec{P})

MEDIUM

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 4 - Scalars and Vectors>EXERCISE>Q 5

The resultant of two vectors

\vec{A}

and

\vec{B}

acting at a point is equal to

\sqrt{3} B

and makes an angle

30 °

with the direction of

\vec{A}

. Prove that

| \vec{A} | = | \vec{B} |

| \vec{A} | = | 2 \vec{B} |

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 4 - Scalars and Vectors>EXERCISE>Q 6

To a person moving due west in a train with a velocity

40 {kmh}^{- 1}

, a car appears to move due north with a velocity of

40 \sqrt{3} {kmh}^{- 1}

. What is the actual velocity and direction of motion of the car?

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 4 - Scalars and Vectors>EXERCISE>Q 7

A boat and a motor launch started from the same point in two different directions inclined to each other at an angle

α

. The speed of the motor launch is

m

times that of the boat. To the boatman, the launch appears to move at an angle of

45 °

with the directions of the boat. Show that

\sin 2 α = \frac{m^{2} - 1}{m^{2}}

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 4 - Scalars and Vectors>EXERCISE>Q 8

A boatman can row directly across a river in time

t_{1}

and an equal distance downstream in time

t_{2}

. If

u

be the speed of the boat in still water and

v

that of the river, show that

t_{1} : t_{2} = {(u + v) / (u - v)}^{1 / 2}

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 4 - Scalars and Vectors>EXERCISE>Q 9

A ship is sailing due east at

10 {kmh}^{- 1}

. At

9 AM

it is

30 km

south-west of another ship

B

. If

B

travels at

15 {kmh}^{- 1}

so as to meet

A

in the least possible time, calculate the direction in which

B

must travel

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 4 - Scalars and Vectors>EXERCISE>Q 9

A ship is sailing due east at

10 {kmh}^{- 1}

. At

9 AM

it is

30 km

south-west of another ship

B

. If

B

travels at

15 {kmh}^{- 1}

so as to meet

A

in the least possible time, calculate the time when

B

meets

A

What is the angle between two equal vectors so that their resultant is one-fourth of one of the vectors?

Important Questions on Scalars and Vectors

Learn and Prepare for any exam you want !