Question

Three vectors each of magnitude $3\sqrt{1.5}$ units are acting at a point. If the angle between any two vectors is $\frac{\pi}{3}$, then the magnitude of the resultant vector of the three vectors is

• A. $9\sqrt{3}$ units
• B. 9 units
• C. $\sqrt{6}$ units
• D. 3 units

Hint:

Resultant of vectors of equal magnitude with equal angles can be found using vector addition formula and symmetry.

Answer 1

The Correct Option is B

Given each vector magnitude $= 3\sqrt{1.5}$ and angle between each two vectors $= \frac{\pi}{3}$. Let the vectors be $\vec{A}$, $\vec{B}$, $\vec{C}$. The resultant vector magnitude $R$ is given by: \[ R = \sqrt{A^2 + B^2 + C^2 + 2(AB\cos \theta + BC \cos \theta + CA \cos \theta)} \] Since $A = B = C = 3\sqrt{1.5}$ and $\theta = \frac{\pi}{3}$, \[ R = \sqrt{3(3\sqrt{1.5})^2 + 2 \times 3 \times (3\sqrt{1.5})^2 \times \cos \frac{\pi}{3}} = \sqrt{3 \times 13.5 + 2 \times 3 \times 13.5 \times \frac{1}{2}} = \sqrt{40.5 + 40.5} = \sqrt{81} = 9 \]