# Solve the following

Question:

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______.

Solution:

$\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-\alpha \hat{\mathrm{j}}+\hat{\mathrm{k}}$

area of parallelogram $=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=8 \sqrt{3}$

$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & \alpha & 3 \\ 3 & -\alpha & 1\end{array}\right|=\hat{\mathrm{i}}(4 \alpha)-\hat{\mathrm{j}}(-8)+\hat{\mathrm{k}}(-4 \alpha)$

$\therefore|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\sqrt{64+32 \alpha^{2}}=8 \sqrt{3}$

$\Rightarrow 2+\alpha^{2}=6 \Rightarrow \alpha^{2}=4$

$\therefore \vec{a} \cdot \vec{b}=3-\alpha^{2}+3=2$