Question #114879
Determine the area of the parallelogram determined by u(1,0) and v(0,1)
1
Expert's answer
2020-05-11T10:26:21-0400

We have given that vectors, u=(1,0)\textbf{u} =(1,0) and v=(0,1)\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 u=i^+0j^&v=0i^+j^\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,

u×v=uvsin(θ)n^\textbf{u} \times \textbf{v} = ||\textbf{u}||\: ||\textbf{v}|| \sin(\theta) \: \hat{n}, where θ\theta is the angle between u & v\textbf{u \& v} , u,v||\textbf{u}||,||\textbf{v}|| are the norm or magnitude of vector u & v\textbf{u \& v} ( here, u=v=12+02=1||\textbf{u}||=||\textbf{v}||=\sqrt{1^2 + 0^2} = 1 ),and n^\hat{n} is the direction of the area vector u×v\textbf{u} \times \textbf{v} i.e n^\hat{n} is perpendicular to the plane of u & v\textbf{u \& v} . Now, we have to find the angle θ\theta .Since, dot product of u & v\textbf{u \& v} is uv=10+01=0    uv    θ=π2\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 u & v\textbf{u \& v} , we get u×v=1||\textbf{u} \times \textbf{v}||=1. Hence, we are done.


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