Numerical Problem No. 2.3 (Vectors and Equilibrium) Physics HSSC - I (Class -11) (Punjab Curriculum and Text Board, Lahore) With Verified Answers

What is the unit vector in the direction of the vector A=4 i +3 j ?

Given:

\to A

$\vec {A}$ = 4

ˆ i

$\hat {i}$ + 3

ˆ j

$\hat {j}$

To Find:

Unit Vector =

ˆ A

$\hat {A}$ = ?

Solution:

By definition: A vector dived by its magnitude is called a unit vector. Mathematically it is written as

ˆ A

$\hat {A}$ =

\frac{\to A}{\to A}

$\ frac {vec {A}}{vec {A}}$ ----eqn (1)

as given

\to A

$\vec {A}$ = 4

ˆ i

$\hat {i}$ + 3

ˆ j

$\hat {j}$ Here x-component = 4 and y-component = 3. So, to find its magnitude |

\to A

$\vec {A}$ | we have

\to A

$\vec {A}$ | =

\sqrt{x^{2} + y^{2}}

$\sqrt {x^2 + y^2}$

by putting values

A =

\sqrt{4^{2} + 3^{2}}

$\sqrt {4^2 + 3^2}$

A =

\sqrt{16 + 9}

$\sqrt {16 + 9}$

A =

\sqrt{25}

$\sqrt {25}$

A = 5

so eqn (1)

ˆ A

$\hat {A}$ =

\frac{4 ˆ i + 3 ˆ j}{5}

$\ frac {4 \hat {i} + 3 \hat {j} }{5}$ ----- Ans

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Numerical Problem 2.3
⇑

List of 11th Physics Numerical Problems of Chapter - 2

⇑

Numerical Problem 2.2 ⇚ 11th Physics Main Content List ⇛ Numerical Problem 2.4

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