If the angle between two vectors vec{u} = (a, | vedclass

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(C) Let $\vec{u} = (a, 2)$ and $\vec{v} = (a, -2)$.Given that the angle $	heta$ between $\vec{u}$ and $\vec{v}$ is $\frac{\pi}{3}$.The dot product fo

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$\pm 2 \sqrt{3}$

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(C) Let $\vec{u} = (a, 2)$ and $\vec{v} = (a, -2)$.Given that the angle $	heta$ between $\vec{u}$ and $\vec{v}$ is $\frac{\pi}{3}$.The dot product formula is $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos(	heta)$.$\vec{u} \cdot \vec{v} = (a)(a) + (2)(-2) = a^2 - 4$.$|\vec{u}| = \sqrt{a^2 + 2^2} = \sqrt{a^2 + 4}$.$|\vec{v}| = \sqrt{a^2 + (-2)^2} = \sqrt{a^2 + 4}$.Substituting these into the formula: $a^2 - 4 = \sqrt{a^2 + 4} \cdot \sqrt{a^2 + 4} \cdot \cos(\frac{\pi}{3})$.$a^2 - 4 = (a^2 + 4) \cdot \frac{1}{2}$.$2(a^2 - 4) = a^2 + 4$.$2a^2 - 8 = a^2 + 4$.$a^2 = 12$.$a = \pm \sqrt{12} = \pm 2 \sqrt{3}$.

If the angle between two vectors $\vec{u} = (a, 2)$ and $\vec{v} = (a, -2)$ is $\frac{\pi}{3}$,then find the value of $a$.

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