Let vec{a}=hat{i}+alpha hat{j}+3 hat{k} and v | vedclass

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(B) Given vectors are $\vec{a} = \hat{i} + \alpha \hat{j} + 3 \hat{k}$ and $\vec{b} = 3 \hat{i} - \alpha \hat{j} + \hat{k}$.The area of the parallelog

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$2$

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(B) Given vectors are $\vec{a} = \hat{i} + \alpha \hat{j} + 3 \hat{k}$ and $\vec{b} = 3 \hat{i} - \alpha \hat{j} + \hat{k}$.The area of the parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by $|\vec{a} 	imes \vec{b}| = 8 \sqrt{3}$.First,calculate the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & \alpha & 3 \ 3 & -\alpha & 1 \end{vmatrix} = \hat{i}(\alpha - (-3\alpha)) - \hat{j}(1 - 9) + \hat{k}(-\alpha - 3\alpha) = 4\alpha \hat{i} + 8 \hat{j} - 4\alpha \hat{k}$.Now,find the magnitude:$|\vec{a} 	imes \vec{b}| = \sqrt{(4\alpha)^2 + 8^2 + (-4\alpha)^2} = \sqrt{16\alpha^2 + 64 + 16\alpha^2} = \sqrt{32\alpha^2 + 64}$.Equating to the given area:$\sqrt{32\alpha^2 + 64} = 8 \sqrt{3} \Rightarrow 32\alpha^2 + 64 = 64 	imes 3 = 192$.$32\alpha^2 = 192 - 64 = 128 \Rightarrow \alpha^2 = 4$.Finally,calculate the dot product $\vec{a} \cdot \vec{b}$:$\vec{a} \cdot \vec{b} = (1)(3) + (\alpha)(-\alpha) + (3)(1) = 3 - \alpha^2 + 3 = 6 - \alpha^2$.Substituting $\alpha^2 = 4$,we get $\vec{a} \cdot \vec{b} = 6 - 4 = 2$.

Let $\vec{a}=\hat{i}+\alpha \hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}-\alpha \hat{j}+\hat{k}$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a}$ and $\vec{b}$ is $8 \sqrt{3}$ square units,then $\vec{a} \cdot \vec{b}$ is equal to ....... .

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