Question

Find the area of the parallelogram spanned by vectors v and w such that |\v|| = ||w|| = 2 and v . w = 1

Answer 1

$$\vec{v}\cdot\vec{w}=\|\vec{v}\|\|\vec{w}\|\cos\theta\\1=4\cos\theta\\\cos\theta=\frac14$$Now recall the area of the parallelogram is given by \(\|\vec{v}\times\vec{w}\|=\|\vec{v}\|\|\vec{w}\|\sin\theta=4\sin\theta\). Given that \(\cos\theta=\frac14\) we conclude \(\sin\theta=\sqrt{1-(1/4)^2}=\frac{\sqrt{15}}4\)

Answer 2

Hence our area is just \(4\times\sqrt{15}/4=\sqrt{15}\)