Given that `\vec {A}` = 2`\hat {i}` + 3`\hat {j}` and `\vec {B}` = 3`\hat {i}` - 4`\hat {j}` find the magnitude and angle of (a) `\vec {C}` =`\vec {A}`+ `\vec {B}` and (b) `\vec {D}` =3`\vec {A}`- 2`\vec {B}`
(Ans: 5.1, 349° ; 17, 90°)

Given:

`\vec {A}` = 2`\hat {i}` + 3`\hat {j}` 
`\vec {B}` = 3`\hat {i}` - 4`\hat {j}`
`\vec {C}` = `\vec {A}` + `\vec {B}`
`\vec {D}` = 3`\vec {A}` - 2`\vec {B}`


To Find:

(a)
Magnitude of the Vector `\vec {C}` = |`\vec {C}`| =  ?

The angle of the Vector `\vec {C}` = Ó¨ = ?

(b)
Magnitude of the Vector `\vec {D}` = |`\vec {D}`| =  ?
The angle of the Vector `\vec {D}` = Ó¨ = ?


Solution: 

(a)

`\vec {C}` = `\vec {A}` + `\vec {B}`

By putting corresponding values we have

`\vec {C}` = 2`\hat {i}` + 3`\hat {j}` + 3`\hat {i}` - 4`\hat {j}`


`\vec {C}` = 5`\hat {i}` - `\hat {j}`

Here x-component = 5 and y-component = -1 So, to find its magnitude |`\vec {C}`| we have

|`\vec {C}`| = `\sqrt {x^2 + y^2}`

by putting values 

C = `\sqrt {(5)^2 + (-1)^2}`

C = `\sqrt {25 + 1}`

C = `\sqrt {26}`

C = 5.1    --------Ans (1)

Thus the magnitude of the vector `\vec {C}` is 5.1

To find the angle of the vector `\vec {C}` with respect to X-axis, we have the formula

tan Ó¨ = `\frac {y}{x}`

or

Ó¨ =  tan⁻¹ `\frac {y}{x}`

by putting value of x and y


Ó¨ =  tan⁻¹ `\frac {-1}{5}`

Ó¨ =  tan⁻¹ (-0.2)

Ó¨ =  12⁰

As x-component = 5 (+ve) and y-component = -1 (-ve) therefore it lies in the 4th quadrant. so we will subtract the angle Ó¨ from 360⁰ 

Ó¨ =  360⁰  - 12⁰

Ó¨ =  348⁰  --------Ans (2)
 
Thus the Angle of the vector `\vec {C}` with respect to x-axis  is 348⁰ 


(b)

`\vec {D}` = 3`\vec {A}` - 2`\vec {B}`

By putting corresponding values we have

`\vec {D}` = 3(2`\hat {i}` + 3`\hat {j}`) - 2( 3`\hat {i}` - 4`\hat {j}`)

`\vec {D}` = 6`\hat {i}` + 9`\hat {j}`) - 6`\hat {i}` + 8`\hat {j}`)

`\vec {D}` = 0`\hat {i}` + 17 `\hat {j}`

or

`\vec {D}` = 17 `\hat {j}`

Here x-component = 0 and y-component = 17 So, to find its magnitude |`\vec {D}`| we have

|`\vec {D}`| = `\sqrt {x^2 + y^2}`

by putting values 

D = `\sqrt {(0)^2 + (17)^2}`

D = `\sqrt {0 + (17)^2}`

D = `\sqrt {(17)^2}`

D = 17    --------Ans (1)

Thus the magnitude of the vector `\vec {D}` is 17

To find the angle of the vector `\vec {D}` with respect to X-axis, we have the formula

tan Ó¨ = `\frac {y}{x}`

or

Ó¨ =  tan⁻¹ `\frac {y}{x}`

by putting value of x and y


Ó¨ =  tan⁻¹ `\frac {17}{0}`

Ó¨ =  tan⁻¹ (0)

Ó¨ =  90⁰ --------Ans (2)

As x-component = 0 (+ve) and y-component = 17 (+ve) therefore the angle will lie along the Y-axis ie. Ó¨ =  90⁰

************************************
************************************
Numerical Problem 2.6
⇑
List of 11th Physics Numerical Problems of Chapter - 2
⇑



************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.