Question

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

Answer 1

$\overrightarrow{a}=\hat{i}+\alpha \hat{j}+3\hat{k}$
$\overrightarrow{b}=3\hat{i}-\alpha \hat{j}+\hat{k}$
Area of parallelogram $=|\overrightarrow{a}\times \overrightarrow{b}|$
$=|(\hat{i}+\alpha \hat{j}+3\hat{k})\times (3\hat{i}-\alpha \hat{j}+\hat{k})|$
$\Rightarrow 8\sqrt{3}=|\left(4\alpha \right)\hat{i}+8\hat{j}-\left(4\alpha \right)\hat{k}|$
$\Rightarrow \left(64\right)\left(3\right)=16{\alpha }^2+64+16{\alpha }^2$
$\Rightarrow {\alpha }^2=4$
Now, $\overrightarrow{a}\cdot \overrightarrow{b}=3-{\alpha }^2+3$
$=6-{\alpha }^2$
$=6-4$
$=2$