Numerical Problem No. 2.7 (Vectors and Equilibrium) Physics HSSC - I (Class -11) (Punjab Curriculum and Text Board, Lahore) With Verified Answers

Find the angle between the two vectors, $\to A$ $\vec {A}$ = 5 $ˆ i$ $\hat {i}$ + $ˆ j$ $\hat {j}$ and $\to B$ $\vec {B}$ = 2 $ˆ i$ $\hat {i}$ + 4 $ˆ j$ $\hat {j}$ (Ans: 52°)

Given:

\to A

$\vec {A}$ = 5

ˆ i

$\hat {i}$ +

ˆ j

$\hat {j}$

\to B

$\vec {B}$ = 2

ˆ i

$\hat {i}$ + 4

ˆ j

$\hat {j}$

To Find:

Angle between the two vectors

\to A

$\vec {A}$ and

\to B

$\vec {B}$ = Ө = ?

Solution:

We will solve this numerical by taking the dot product of the two vectors

\to A

$\vec {A}$ and

\to B

$\vec {B}$ given as

\to A

$\vec {A}$ .

\to B

$\vec {B}$ = |

\to A

$\vec {A}$ | |

\to B

$\vec {B}$ | cos Ө

cos Ө =

\frac{\to A . \to B}{∣ ∣ \to A ∣ ∣ ∣ ∣ \to B ∣ ∣}

$\frac {vec {A} . vec {B}}{|vec A| |vec B|}$

cos Ө =

\frac{(5 ˆ i + ˆ j) . (2 ˆ i + 4 ˆ j)}{(\sqrt{A x^{2} + A y^{2}}) (\sqrt{B x^{2} + B y^{2}})}

$\frac {(5 hat {i} + hat {j}) . (2hat {i} + 4 hat {j})}{(sqrt {Ax^2 + Ay^2}) (sqrt {Bx^2 + By^2})}$

cos Ө =

$\frac {(5 ｘ 2) hat {i} . hat {i} + (5 ｘ 4) hat {i} . hat {j} + (1 ｘ 2) hat {j} . hat {i} + (1 ｘ 4) hat {j} . hat {j}}{(sqrt {5^2 + 1^2}) (sqrt {2^2 + 4^2})}$

cos Ө =

$\frac {10 + 4}{sqrt {25 + 1} ｘ sqrt {4 + 16}}$

cos Ө =

$\frac {14}{sqrt {26} ｘ sqrt {20}}$

cos Ө =

$\frac {14}{sqrt {26 ｘ 20}}$

cos Ө =

$\frac {14}{sqrt {520}}$

cos Ө =

$\frac {14}{22.804}$

cos Ө = 0.61

Ө = cos⁻¹ (0.61)

Ө = 52⁰

Thus, the angle Ө between the two vectors

$\vec {A}$ and

$\vec {B}$ is 52⁰

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Numerical Problem 2.7
⇑

List of 11th Physics Numerical Problems of Chapter - 2

⇑

Numerical Problem 2.6 ⇚ 11th Physics Main Content List ⇛ Numerical Problem 2.8

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