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 .

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 .

Prove that the vectors A → = 3 i ^ - 2 j ^ + k ^ , B → = i ^ - 3 j ^ + 5 k ^ and C → = 2 i ^ + j ^ - 4 k ^ , form a Right-Angled triangle.

The given vectors will constitute a triangle only if one of the given vectors is equal to vector sum of the remaining two vectors.In the given problem, B → + C → = A → So, the vectors do form a triangle.This triangle will be right-angled only if the dot product of two vectors is zero. ( A → . B → ) = ( 3 i ^ - 2 j ^ + k ^ ) . ( i ^ - 3 j ^ + 5 k ^ ) = 3 ( i ^ . i ^ ) + 6 ( j ^ . j ^ ) + 5 ( k ^ . k ^ ) = 3 + 6 + 5 = 14 ( B → . C → ) = ( i ^ - 3 j ^ + 5 k ^ ) . ( 2 i ^ + j ^ - 4 k ^ ) = 2 ( i ^ . i ^ ) - 3 ( j ^ . j ^ ) - 20 ( k ^ . k ^ ) = 2 - 3 - 20 = - 21 ( C → . A → ) = ( 2 i ^ + j ^ - 4 k ^ ) . ( 3 i ^ - 2 j ^ + k ^ ) = 6 ( i ^ . i ^ ) - 2 ( j ^ . j ^ ) - 4 ( k ^ . k ^ ) = 6 - 2 - 4 = 0 Since the dot product of C → and A → is zero. Therefore, it implies that C → is perpendicular to A → .

MEDIUM

11th West Bengal Board

IMPORTANT

Earn 100

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 $| \vec{A} | = | \vec{B} |$ or $| \vec{A} | = | 2 \vec{B} |$ .

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 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

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 4 - Scalars and Vectors>EXERCISE>Q 10

A steady wind blows from west at

100 {kmh}^{- 1}

during the flight of an aircraft from airfield

A

to airfield

B

due north of

A

. The air speed indicator reads

260 {kmh}^{- 1}

during the flight. It takes

1

hour

15 mnt

. for the flight from

A

B

(neglecting take off and landing time). Find the direction in which the pilot steers the aircraft and also the distance

A B

HARD

11th West Bengal Board

IMPORTANT

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

Prove that the vectors

\vec{A} = 3 \hat{i} - 2 \hat{j} + \hat{k}, \vec{B} = \hat{i} - 3 \hat{j} + 5 \hat{k}

and

\vec{C} = 2 \hat{i} + \hat{j} - 4 \hat{k}

, form a Right-Angled triangle.

MEDIUM

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 4 - Scalars and Vectors>EXERCISE>Q 12

\hat{i} - \hat{j} - \hat{k}

and

2 \hat{i} + \hat{j} + \hat{k}

are two vectors represented by the two sides of a triangle. Determine the angle between the two sides and the length of the third side.

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

Important Questions on Scalars and Vectors

Learn and Prepare for any exam you want !

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 $| \vec{A} | = | \vec{B} |$ or $| \vec{A} | = | 2 \vec{B} |$ .