Answer to Question #114879 in Linear Algebra for Ayanda Nkabinde

Question

Determine the area of the parallelogram determined by u(1,0) and v(0,1)

Answer 1

We have given that vectors, $\textbf{u} =(1,0)$ and $\textbf{v} =(0,1)$ , Since, the vector cross product of any two vector gives the area of parallelogram. For simplicity let represent the vector in as usual way i.e $\textbf{u}=\hat{i} +0\hat{j} \: \& \: \textbf{v}=0\hat{i} + \hat{j}$ ,Now we can calculate the vector cross product in two way,let's focus on the first method,

$\textbf{u} \times \textbf{v} = ||\textbf{u}||\: ||\textbf{v}|| \sin(\theta) \: \hat{n}$, where $\theta$ is the angle between $\textbf{u \& v}$ , $||\textbf{u}||,||\textbf{v}||$ are the norm or magnitude of vector $\textbf{u \& v}$ ( here, $||\textbf{u}||=||\textbf{v}||=\sqrt{1^2 + 0^2} = 1$ ),and $\hat{n}$ is the direction of the area vector $\textbf{u} \times \textbf{v}$ i.e $\hat{n}$ is perpendicular to the plane of $\textbf{u \& v}$ . Now, we have to find the angle $\theta$ .Since, dot product of $\textbf{u \& v}$ is $\textbf{u}\cdot\textbf{v}=1\cdot0 + 0\cdot1=0 \implies \textbf{u} \bot\textbf{v} \implies \theta = \frac{\pi}{2}$ radian ,thus on calculating the vector cross product of $\textbf{u \& v}$ , we get $||\textbf{u} \times \textbf{v}||=1$. Hence, we are done.