Two vectors have magnitudes 3 unit and 4 unit respectively. What shoul

1.	Two vectors have magnitudes 3 unit and 4 unit respectively. What should be the angle betweenthem if the magnitude of the resultant is (a) 1 unit, (b) 5 unit and (c) 7 unit.
Answer» Solution:- $: \implies \sf \: magnitude \: of \: \vec{a} \: \|a\| = 3unit \:$ $: \implies \sf \: magnitude \: of \: \vec{<klux>B</klux>} \: \|b\| = 4unit \:$ $\sf \: \to formula$ $\boxed{\sf \: R = \sqrt{ {A }^{2} + {B}^{2} + 2A B \cos \theta } }$ *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* $\sf \: R = \sqrt{ {A }^{2} + {B}^{2} + 2A B \cos \theta }$ $\sf \: 7 = \sqrt{ {3}^{2} + {4}^{2} + 2 \times 3 \times 4 \cos \theta }$ $\sf \: 49= {9}^{} + {16}^{} + 24 \cos \theta$ $\sf \: 49 = 25 + 24 \cos \theta$ $\sf \: 24 = 24 \cos \theta$ $\cos\theta = 1$ $\theta = 0 \degree$