Question

A=5 B=8 units and separated by60degrees find the resultant between two vectors​

Answer 1

GiveN :

• First Vector $\sf{(\overrightarrow A ) = 5}$ units.
• Second Vector $\sf{(\overrightarrow B) = 8}$ units.
• Angle between vectors $\sf{\theta = 60^{\circ}}$

To FinD :

• Resultant

SolutioN :

Used formula for resultant of the vectors (Triangle law or Parallelogram law of vector addition)

• let resultant vector be $\sf{\overrightarrow R}$

$\implies \sf{\overrightarrow{R} = \sqrt{\overrightarrow A^2 + \overrightarrow B^2 + 2 \overrightarrow{AB} \cos \theta}} \\ \\ \\ \implies \rm{\overrightarrow R = \sqrt{5^2 + 8^2 + (2 \times 5 \times 8) \times \cos \theta}} \\ \\ \\ \implies \rm{\overrightarrow R = \sqrt{25 + 64 + 2(40) \cos 60^{\circ}}} \\ \\ \\ \implies \rm{\overrightarrow R = \sqrt{89 + 80 \cos 60^{\circ}}} \\ \\ \\ \implies \rm{\overrightarrow R = \sqrt{89 + 80 \times \dfrac{1}{2}}} \\ \\ \\ \implies \rm{\overrightarrow R = \sqrt{89 + 40}} \\ \\ \\ \implies \rm{\overrightarrow R = \sqrt{129}} \\ \\ \\ \implies \rm{\overrightarrow R = 11.357} \\ \\ \\ \implies \rm{\overrightarrow R \approx 11.3} \\ \\ \\ \underline{\rm{\therefore \: Resultant \: vector \: is \: 11.3}}$