Three vectors each of magnitude 3 sqrt{1.5} u | vedclass

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(B) Let the three vectors be $\vec{A}$,$\vec{B}$,and $\vec{C}$. The magnitude of each vector is $A = B = C = 3 \sqrt{1.5} = 3 \sqrt{\frac{3}{2}}$.Sinc

Vedclass · Accepted Answer

$9$ units

Vedclass · Answer

(B) Let the three vectors be $\vec{A}$,$\vec{B}$,and $\vec{C}$. The magnitude of each vector is $A = B = C = 3 \sqrt{1.5} = 3 \sqrt{\frac{3}{2}}$.Since the angle between any two vectors is $\frac{\pi}{3}$ $(60^\circ)$,we can place them in a 3D coordinate system.Let $\vec{A} = A \hat{i}$.Since the angle between $\vec{A}$ and $\vec{B}$ is $60^\circ$,$\vec{B} = A(\cos 60^\circ \hat{i} + \sin 60^\circ \hat{j}) = A(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j})$.For the third vector $\vec{C}$,it must make a $60^\circ$ angle with both $\vec{A}$ and $\vec{B}$. This implies $\vec{C}$ lies in the plane formed by $\vec{A}$ and $\vec{B}$ or out of it. However,in 3D space,three vectors with $60^\circ$ between each pair form the edges of a regular tetrahedron.The resultant vector $\vec{R} = \vec{A} + \vec{B} + \vec{C}$.The magnitude squared is $R^2 = A^2 + B^2 + C^2 + 2(AB \cos 60^\circ + BC \cos 60^\circ + CA \cos 60^\circ)$.$R^2 = 3A^2 + 2(3 \cdot A^2 \cdot \frac{1}{2}) = 3A^2 + 3A^2 = 6A^2$.Given $A = 3 \sqrt{1.5} = 3 \sqrt{\frac{3}{2}}$,then $A^2 = 9 \cdot \frac{3}{2} = 13.5$.$R^2 = 6 	imes 13.5 = 81$.$R = \sqrt{81} = 9$ units.

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

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