# Prove the following

Question:

Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-\alpha \hat{\mathrm{j}}+\hat{\mathrm{k}}$. 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 $=|\vec{a} \times \vec{b}|$

$=\mid(\hat{i}+\alpha \hat{j}+3 \hat{k}) \times(3 \hat{i}-\alpha \hat{j}+\hat{k})$

$8 \sqrt{3}=|(4 \alpha) \hat{i}+8 \hat{j}-(4 \alpha) \hat{k}|$

$(64)(3)=16 \alpha^{2}+64+16 \alpha^{2}$

$(64)(3)=32 \alpha^{2}+64$

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

$\alpha^{2}=4$

$\therefore \quad \vec{a}=\hat{i}+\alpha \hat{j}+3 \hat{k}$

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

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

$=6-\alpha^{2}$

$=6-4$

$=2$