Question

Two vectors have magnitudes 3 unit and 4 unit respectively. What should be the angle between
them if the magnitude of the resultant is (a) 1 unit, (b) 5 unit and (c) 7 unit.
​

Answer 1

Solution:-

$: \implies \sf \: magnitude \: of \: \vec{a} \: |a| = 3unit \:$

$: \implies \sf \: magnitude \: of \: \vec{b} \: |b| = 4unit \:$

$\sf \: \to formula$

$\boxed{\sf \: R = \sqrt{ {A }^{2} + {B}^{2} + 2A B \cos \theta } }$

Case :- 1 R = 1 unit

A = 3 unit and B = 4 unit

Using this

$\sf \: R = \sqrt{ {A }^{2} + {B}^{2} + 2A B \cos \theta }$

$\sf \: 1 = \sqrt{ {3}^{2} + {4}^{2} + 2 \times 3 \times 4 \cos \theta }$

$\sf \: 1 = {9}^{} + {16}^{} + 24 \cos \theta$

$\sf \: 1 = {25}^{} {}^{} + 24 \cos \theta$

$\sf \: 1 - 25= {}^{} {}^{} 24 \cos \theta$

$\sf \: - 24= {}^{} {}^{} 24 \cos \theta$

$\sf \cos \theta = - 1$

$\theta \: = 180 \degree$

Case :- 2 R = 5 unit

Case :- 2 R = 5 unit A = 3 unit and B = 4 unit

Case :- 2 R = 5 unit A = 3 unit and B = 4 unit Using this

$\sf \: R = \sqrt{ {A }^{2} + {B}^{2} + 2A B \cos \theta }$

$\sf \: 5 = \sqrt{ {3}^{2} + {4}^{2} + 2 \times 3 \times 4 \cos \theta }$

$\sf \: 25= {9}^{} + {16}^{} + 24 \cos \theta$

$\sf\: 25 = 25 + 24 \cos \theta$

$\sf0 = 24 \cos \theta$

$\cos \theta = 0$

$\rm \: \theta = 90 \degree$

Case :- 3 R = 7 unit

unit A = 3 unit and B = 4 unit

unit A = 3 unit and B = 4 unit Using this

$\sf \: R = \sqrt{ {A }^{2} + {B}^{2} + 2A B \cos \theta }$

$\sf \: 7 = \sqrt{ {3}^{2} + {4}^{2} + 2 \times 3 \times 4 \cos \theta }$

$\sf \: 49= {9}^{} + {16}^{} + 24 \cos \theta$

$\sf \: 49 = 25 + 24 \cos \theta$

$\sf \: 24 = 24 \cos \theta$

$\cos\theta = 1$

$\theta = 0 \degree$

Answer 2

Answer:

$Solution:-</p><p></p><p>: \implies \sf \: magnitude \: of \: \vec{a} \: |a| = 3unit \::⟹magnitudeofa∣a∣=3unit</p><p></p><p>: \implies \sf \: magnitude \: of \: \vec{b} \: |b| = 4unit \::⟹magnitudeofb∣b∣=4unit</p><p></p><p>\sf \: \to formula→formula</p><p></p><p>\boxed{\sf \: R = \sqrt{ {A }^{2} + {B}^{2} + 2A B \cos \theta } }R=A2+B2+2ABcosθ</p><p></p><p>Case :- 1 R = 1 unit</p><p></p><p>A = 3 unit and B = 4 unit</p><p></p><p>Using this</p><p></p><p>\sf \: R = \sqrt{ {A }^{2} + {B}^{2} + 2A B \cos \theta }R=A2+B2+2ABcosθ</p><p></p><p>\sf \: 1 = \sqrt{ {3}^{2} + {4}^{2} + 2 \times 3 \times 4 \cos \theta }1=32+42+2×3×4cosθ</p><p></p><p>\sf \: 1 = {9}^{} + {16}^{} + 24 \cos \theta1=9+16+24cosθ</p><p></p><p>\sf \: 1 = {25}^{} {}^{} + 24 \cos \theta1=25+24cosθ</p><p></p><p>\sf \: 1 - 25= {}^{} {}^{} 24 \cos \theta1−25=24cosθ</p><p></p><p>\sf \: - 24= {}^{} {}^{} 24 \cos \theta−24=24cosθ</p><p></p><p>\sf \cos \theta = - 1cosθ=−1</p><p></p><p>\theta \: = 180 \degreeθ=180°</p><p></p><p>Case :- 2 R = 5 unit</p><p></p><p>Case :- 2 R = 5 unit A = 3 unit and B = 4 unit</p><p></p><p>Case :- 2 R = 5 unit A = 3 unit and B = 4 unit Using this</p><p></p><p>\sf \: R = \sqrt{ {A }^{2} + {B}^{2} + 2A B \cos \theta }R=A2+B2+2ABcosθ</p><p></p><p>\sf \: 5 = \sqrt{ {3}^{2} + {4}^{2} + 2 \times 3 \times 4 \cos \theta }5=32+42+2×3×4cosθ</p><p></p><p>\sf \: 25= {9}^{} + {16}^{} + 24 \cos \theta25=9+16+24cosθ</p><p></p><p>\sf\: 25 = 25 + 24 \cos \theta25=25+24cosθ</p><p></p><p>\sf0 = 24 \cos \theta0=24cosθ</p><p></p><p>\cos \theta = 0cosθ=0</p><p></p><p>\rm \: \theta = 90 \degreeθ=90°</p><p></p><p>Case :- 3 R = 7 unit</p><p></p><p>unit A = 3 unit and B = 4 unit</p><p></p><p>unit A = 3 unit and B = 4 unit Using this</p><p></p><p>\sf \: R = \sqrt{ {A }^{2} + {B}^{2} + 2A B \cos \theta }R=A2+B2+2ABcosθ</p><p></p><p>\sf \: 7 = \sqrt{ {3}^{2} + {4}^{2} + 2 \times 3 \times 4 \cos \theta }7=32+42+2×3×4cosθ</p><p></p><p>\sf \: 49= {9}^{} + {16}^{} + 24 \cos \theta49=9+16+24cosθ</p><p></p><p>\sf \: 49 = 25 + 24 \cos \theta49=25+24cosθ</p><p></p><p>\sf \: 24 = 24 \cos \theta24=24cosθ</p><p></p><p>\cos\theta = 1cosθ=1</p><p></p><p>\theta = 0 \degreeθ=0°</p><p></p><p>$