Numerical Problem No. 2.14 (Vectors and Equilibrium) Physics HSSC - I (Class -11) (Punjab Curriculum and Text Board, Lahore) With Verified Answers

The magnitude of dot and cross products of two vectors are 6√3 and 6 respectively. Find the angle between the vectors 

(Ans: 30°) 

Given:

Let the two vectors are `\vec {A}` and `\vec {B} 

`\vec {A}` . `\vec {B}` = 6 `\sqrt {3}`

`\vec {A}` ｘ `\vec {B}` = 6

To Find:

Angle between the vectors = θ = ?

Solution: 

As we have

`\vec {A}` . `\vec {B}` = A B cos θ = 6 `\sqrt {3}`

or

A B cos θ =    6 `\sqrt {3}`   ---------Eqn(1)

And 

`\vec {A}` ｘ `\vec {B}` = A B sin θ = 6 

or

A B sin θ = 6    --------- Eqn(2)

Now dividing Eqn(2) by Eqn (1)  

`\frac {A B sin θ}{A B cos θ}`  = `\frac {6}{6 sqrt {3}}`

`\frac {sin θ}{cos θ}`  = `\frac {1}{ sqrt {3}}`

 [∴  `\frac {sin θ}{cos θ}`= tan θ ]  So,

tan θ = `\frac {1}{ sqrt {3}}`

θ = tan⁻¹ ( `\frac {1}{ sqrt {3}}` )

θ = 30⁰ ------Ans

Thus the angle between two vectors is 30⁰

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Numerical Problem 2.14
⇑

List of 11th Physics Numerical Problems of Chapter - 2

⇑

Numerical Problem 2.13  ⇚   11th Physics Main Content List    ⇛ Numerical Problem 2.15

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