Numerical Problem No. 2.9 (Vectors and Equilibrium) Physics HSSC - I (Class -11) (Punjab Curriculum and Text Board, Lahore) With Verified Answers

Show that the three vectors $\hat {i}$ + $\hat {j}$ + $\hat {k}$ , 2 $\hat {i}$ - 3 $\hat {j}$ + $\hat {k}$ and 4 $\hat {i}$ + $\hat {j}$ - 5 $\hat {k}$ are mutually perpendicular.

Given:

Let

Vector 1 is

$\vec {A}$ =

$\hat {i}$ +

$\hat {j}$ +

$\hat {k}$

Vector 2 is

$\vec {B}$ = 2

$\hat {i}$ - 3

$\hat {j}$ +

$\hat {k}$

Vector 3 is

$\vec {C}$ = 4

$\hat {i}$ +

$\hat {j}$ - 5

$\hat {k}$

To Find:

To prove that the given three vectors are mutually perpendicular to each other.

Solution:

We know that if two vectors are mutually perpendicular to each other then their scalar (dot) product is equal to Zero. So, we should have to prove

(i)

$\vec {A}$ .

$\vec {B}$ = 0

(ii)

$\vec {A}$ .

$\vec {C}$ = 0

and

(iii)

$\vec {B}$ .

$\vec {C}$ = 0

(i)

$\vec {A}$ .

$\vec {B}$ = 0

$\vec {A}$ .

$\vec {B}$ = (

$\hat {i}$ +

$\hat {j}$ +

$\hat {k}$ )( 2

$\hat {i}$ - 3

$\hat {j}$ +

$\hat {k}$ )

$\vec {A}$ .

$\vec {B}$ = (1 ｘ 2)

$hat {i}$ .

$hat {i}$ + (1 ｘ -3)

$hat {i}$ .

$hat {j}$ + (1 ｘ 1)

$hat {i}$ .

$hat {k}$ + (1 ｘ 2)

$hat {j}$ .

$hat {i}$ + (1 ｘ -3)

$hat {j}$ .

$hat {j}$ + (1 ｘ 1)

$hat {j}$ .

$hat {k}$ + (1 ｘ 2)

$hat {k}$ .

$hat {i}$ + (1 ｘ -3)

$hat {k}$ .

$hat {j}$ + (1 ｘ 1)

$hat {k}$ .

$hat {k}$

[∴

$hat {i}$ .

$hat {i}$ =

$hat {j}$ .

$hat {j}$ =

$hat {k}$ .

$hat {k}$ = 1

and

$hat {i}$ .

$hat {j}$ =

$hat {j}$ .

$hat {i}$ =

$hat {j}$ .

$hat {k}$ =

$hat {k}$ .

$hat {j}$ =

$hat {k}$ .

$hat {i}$ =

$hat {i}$ .

$hat {k}$ = 0 ]

So by putting these values we have

$\vec {A}$ .

$\vec {B}$ = 2 + 0 + 0 + 0 - 3 + 0 + 0 + 0 + 1

$\vec {A}$ .

$\vec {B}$ = 0 --------Eqn (1)

(ii)

$\vec {A}$ .

$\vec {C}$ = 0

$\vec {A}$ .

$\vec {C}$ = (

$\hat {i}$ +

$\hat {j}$ +

$\hat {k}$ )( 4

$\hat {i}$ +

$\hat {j}$ - 5

$\hat {k}$ )

$\vec {A}$ .

$\vec {C}$ = (1 ｘ 4)

$hat {i}$ .

$hat {i}$ + (1 ｘ 1)

$hat {i}$ .

$hat {j}$ + (1 ｘ -5)

$hat {i}$ .

$hat {k}$ + (1 ｘ 4)

$hat {j}$ .

$hat {i}$ + (1 ｘ 1)

$hat {j}$ .

$hat {j}$ + (1 ｘ -5)

$hat {j}$ .

$hat {k}$ + (1 ｘ 4)

$hat {k}$ .

$hat {i}$ + (1 ｘ 1)

$hat {k}$ .

$hat {j}$ + (1 ｘ -5)

$hat {k}$ .

$hat {k}$

[∴

$hat {i}$ .

$hat {i}$ =

$hat {j}$ .

$hat {j}$ =1 =

$hat {k}$ .

$hat {k}$

and

$hat {i}$ .

$hat {j}$ =

$hat {j}$ .

$hat {i}$ =

$hat {j}$ .

$hat {k}$ =

$hat {k}$ .

$hat {j}$ =

$hat {k}$ .

$hat {i}$ =

$hat {i}$ .

$hat {k}$ = 0 ]

So by putting these values we have

$\vec {A}$ .

$\vec {C}$ = 4 + 0 + 0 + 0 + 1 + 0 + 0 + 0 - 5

$\vec {A}$ .

$\vec {C}$ = 0 --------Eqn (2)

(iii)

$\vec {B}$ .

$\vec {C}$ = 0

$\vec {B}$ .

$\vec {C}$ = (2

$\hat {i}$ - 3

$\hat {j}$ +

$\hat {k}$ )( 4

$\hat {i}$ +

$\hat {j}$ - 5

$\hat {k}$ )

$\vec {B}$ .

$\vec {C}$ = (2 ｘ 4)

$hat {i}$ .

$hat {i}$ + (2 ｘ 1)

$hat {i}$ .

$hat {j}$ + (2 ｘ -5)

$hat {i}$ .

$hat {k}$ + (-3 ｘ 4)

$hat {j}$ .

$hat {i}$ + (-3 ｘ 1)

$hat {j}$ .

$hat {j}$ + (-3 ｘ -5)

$hat {j}$ .

$hat {k}$ + (1 ｘ 4)

$hat {k}$ .

$hat {i}$ + (1 ｘ 1)

$hat {k}$ .

$hat {j}$ + (1 ｘ -5)

$hat {k}$ .

$hat {k}$

[∴

$hat {i}$ .

$hat {i}$ =

$hat {j}$ .

$hat {j}$ =1 =

$hat {k}$ .

$hat {k}$

and

$hat {i}$ .

$hat {j}$ =

$hat {j}$ .

$hat {i}$ =

$hat {j}$ .

$hat {k}$ =

$hat {k}$ .

$hat {j}$ =

$hat {k}$ .

$hat {i}$ =

$hat {i}$ .

$hat {k}$ = 0 ]

So by putting these values we have

$\vec {B}$ .

$\vec {C}$ = 8 + 0 + 0 + 0 - 3 + 0 + 0 + 0 - 5

$\vec {B}$ .

$\vec {C}$ = 0 --------Eqn (3)

Eqns (1), (2), and (3) show that the mutual dot product of given vectors

$\vec {A}$ ,

$\vec {B}$ , and

$\vec {C}$ are zero. Hence, it is proved that these vectors are mutually perpendicular to each other.

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Numerical Problem 2.9
⇑

List of 11th Physics Numerical Problems of Chapter - 2

⇑

Numerical Problem 2.8 ⇚ 11th Physics Main Content List ⇛ Numerical Problem 2.10

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