Three vectors vec a, vec b, vec c are incline | vedclass

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(B) Let the angles between $\vec a$ and $\vec b$,$\vec b$ and $\vec c$,and $\vec c$ and $\vec a$ be $	heta_1, 	heta_2, 	heta_3$ respectively.The le

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$\frac{\pi}{4}$

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(B) Let the angles between $\vec a$ and $\vec b$,$\vec b$ and $\vec c$,and $\vec c$ and $\vec a$ be $	heta_1, 	heta_2, 	heta_3$ respectively.The lengths of the projections are given by $p_1 = |\vec a| \cos 	heta_1 = 2 \cos 	heta_1$,$p_2 = |\vec b| \cos 	heta_2 = 3 \cos 	heta_2$,and $p_3 = |\vec c| \cos 	heta_3 = 9 \cos 	heta_3$.Since these lengths are in geometric progression,we have $p_2^2 = p_1 p_3$.Substituting the values: $(3 \cos 	heta_2)^2 = (2 \cos 	heta_1)(9 \cos 	heta_3)$.$9 \cos^2 	heta_2 = 18 \cos 	heta_1 \cos 	heta_3$.$\cos^2 	heta_2 = 2 \cos 	heta_1 \cos 	heta_3$.Given $	heta_1 = \frac{5\pi}{12}$ and $	heta_3 = \frac{\pi}{12}$.$\cos^2 	heta_2 = 2 \cos \left(\frac{5\pi}{12}ight) \cos \left(\frac{\pi}{12}ight)$.Using the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:$\cos^2 	heta_2 = \cos \left(\frac{5\pi}{12} + \frac{\pi}{12}ight) + \cos \left(\frac{5\pi}{12} - \frac{\pi}{12}ight)$.$\cos^2 	heta_2 = \cos \left(\frac{6\pi}{12}ight) + \cos \left(\frac{4\pi}{12}ight) = \cos \left(\frac{\pi}{2}ight) + \cos \left(\frac{\pi}{3}ight)$.$\cos^2 	heta_2 = 0 + \frac{1}{2} = \frac{1}{2}$.Since the angle is acute,$\cos 	heta_2 = \frac{1}{\sqrt{2}}$,which implies $	heta_2 = \frac{\pi}{4}$.

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