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 -
(i) 1 unit
(ii) 5 unit (111) 7 unit
(1) 180°, 90°, 0º (2) 80°, 70°, 0°
(3) 90°, 170°, 50° (4) None of these​

Answer 1

Correct 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 -(i) 1 unit (ii) 5 unit (iii) 7 unit

(1) 180°, 90°, 0º

(2) 80°, 70°, 0°

(3) 90°, 170°, 50°

(4) None of these

Solution

Let the two given vectors be A and B. Their magnitudes are 3 and 4 units respectively

Using Parallelogram Law of Vectors,

$\sf R = \sqrt{A^2 + B^2 + 2AB cos \alpha}$

Case I

When their resultant is 1

$\longmapsto \: \sf \: 1 = \sqrt{ {3}^{2} + {4}^{2} + 2(3)(4) \cos( \alpha ) } \\ \\ \longmapsto \: \sf \: 1 = 25 + 24 \cos( \alpha ) \\ \\ \longmapsto \ \sf \: 24 \cos( \alpha ) = 1 - 25 \\ \\ \longmapsto \: \sf \: \cos( \alpha ) = - 1 \\ \\ \longmapsto \: \boxed{ \boxed{ \sf{ \alpha = {180}^{ \circ} }}}$

Case II

When the resultant is 5

$\longmapsto \: \sf \: 5 = \sqrt{3 {}^{2} + {4}^{2} + 24 \cos( \alpha ) } \\ \\ \longmapsto \: \sf \: 25 = 25 + 24 \cos( \alpha ) \\ \\ \longmapsto \: \sf \: 24 \cos( \alpha ) = 0 \\ \\ \longmapsto \boxed{ \boxed { \sf \: \: \alpha = {90}^{ \circ} }}$

Case III

When the resultant is 7

$\longmapsto \: \sf \: 7 = \sqrt{25 + 24 \cos( \alpha ) } \\ \\ \longmapsto \: \sf{49 = 25 + 24 \: \cos( \alpha ) } \\ \\ \longmapsto \: \sf \: \cos( \alpha ) = 1 \\ \\ \longmapsto \: \boxed{ \boxed{ \sf{ \alpha = {0}^{ \circ}}}}$

Option (1) is correct

Answer 2

$\huge {\red{\boxed{ \overline{ \underline{ \mid\mathfrak{Que}{\mathrm{st}{ \sf{ion}} \colon\mid}}}}}}$

Two vectors have magnitudes 3 unit and 4 unit respectively. What should be the angle between them if the magnitude of the resultant is -(i) 1 unit (ii) 5 unit (iii) 7 unit

(1) 180°, 90°, 0º

(2) 80°, 70°, 0°

(3) 90°, 170°, 50°

(4) None of these

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$\huge {\red{\boxed{ \overline{ \underline{ \mid\mathfrak{An}{\mathrm{sw}{ \sf{er}} \colon\mid}}}}}}$

We have formula for Resultant of Triangle law vectors :

$\large \star {\boxed{\sf{|AB| \: = \: \sqrt{A^2 \: + \: B^2 \: + \: 2AB \cos \theta}}}}$

______________________

(i) Resultant is 1 unit

$\implies {\sf{1 \: = \: \sqrt{(4)^2 \: + \: (3)^2 \: + \: 2(2)(3) \cos \theta}}} \\ \\ \implies {\sf{1 \: = \: \sqrt{16 \: + \: 9 \: + \: 24 \cos \theta}}} \\ \\ \small{\underline{\pink{\sf{\: \: \: \: \: \: \: \: \: \: Square \: Both \: Sides \: \: \: \: \: \: \: \: \: \: \:}}}} \\ \\ \implies {\sf{1 \: = \: 25 \: + \: 24 \cos \theta}} \\ \\ \implies {\sf{1 \: - \: 25 \: = 24 \cos \theta}} \\ \\ \implies {\sf{\cancel{-24} \: = \: \cancel{24} \cos \theta}} \\ \\ \implies {\sf{\cos \theta \: = \: -1}} \\ \\ \implies {\sf{\cos \theta \: = \: \cos 180}} \\ \\ \large {\underline{\boxed{\sf{\theta \: = \: 180 ^{\circ}}}}}$

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(ii) Resultant is 5 units

$\implies {\sf{5 \: = \: \sqrt{(4)^2 \: + \: (3)^2 \: + \: 2(4)(3) \cos \theta}}} \\ \\ \implies {\sf{5 \: = \: \sqrt{25 \: + \: 24 \cos \theta}}} \\ \\ \implies {\sf{25 \: = \: 25 \: + \: 24 \cos \theta}} \\ \\ \implies {\sf{24 \cos \theta \: = \: 0}} \\ \\ \large {\underline{\boxed{\sf{\theta \: = \: 90^{\circ}}}}}$

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(iii) Resultant is 7 units

$\implies {\sf{7 \: = \: \sqrt{(4)^2 \: + \: (3)^2 \: + \: 2(4)(3) \cos \theta}}} \\ \\ \implies {\sf{49 \: = \: 25 \: + \: 24 \cos \theta}} \\ \\ \implies {\sf{\cos \theta \: = \: 1}} \\ \\ \large {\underline{\boxed{\sf{\theta \: = \: 0^{\circ}}}}}$