Numerical Problem No. 2.6 (Vectors and Equilibrium) Physics HSSC - I (Class -11) (Punjab Curriculum and Text Board, Lahore) With Verified Answers

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}` = Ө = ?

(b)

Magnitude of the Vector `\vec {D}` = |`\vec {D}`| =  ?

The angle of the Vector `\vec {D}` = Ө = ?

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 Ө = `\frac {y}{x}`

or

Ө =  tan⁻¹ `\frac {y}{x}`

by putting value of x and y

Ө =  tan⁻¹ `\frac {-1}{5}`

Ө =  tan⁻¹ (-0.2)

Ө =  12⁰

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

Ө =  360⁰  - 12⁰

Ө =  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 Ө = `\frac {y}{x}`

or

Ө =  tan⁻¹ `\frac {y}{x}`

by putting value of x and y

Ө =  tan⁻¹ `\frac {17}{0}`

Ө =  tan⁻¹ (0)

Ө =  90⁰ --------Ans (2)

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

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Numerical Problem 2.6
⇑

List of 11th Physics Numerical Problems of Chapter - 2

⇑

Numerical Problem 2.5 ⇚   11th Physics Main Content List    ⇛ Numerical Problem 2.7

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