Question:
Find the area of the parallelogram whose adjacent sides are determined by the vector $\vec{a}=\hat{i}-\hat{j}+3 \hat{k}$ and $\vec{b}=2 \hat{i}-7 \hat{j}+\hat{k}$.
Solution:
The area of the parallelogram whose adjacent sides are $\vec{a}$ and $\vec{b}$ is $|\vec{a} \times \vec{b}|$.
Adjacent sides are given as:
$\vec{a}=\hat{i}-\hat{j}+3 \hat{k}$ and $\vec{b}=2 \hat{i}-7 \hat{j}+\hat{k}$
$\therefore \vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1\end{array}\right|=\hat{i}(-1+21)-\hat{j}(1-6)+\hat{k}(-7+2)=20 \hat{i}+5 \hat{j}-5 \hat{k}$
$|\vec{a} \times \vec{b}|=\sqrt{20^{2}+5^{2}+5^{2}}=\sqrt{400+25+25}=15 \sqrt{2}$
Hence, the area of the given parallelogram is $15 \sqrt{2}$ square units