User:Eml4500.f08.Lulz.js/HW2

Derivation of $$\vec{\tilde{i}}\cdot\vec{j}$$

The definition of $$\vec{\tilde{i}}$$ given in the lecture notes is: $$\vec{\tilde{i}}=\cos\theta^{(e)}\vec{i}+\sin\theta^{(e)}\vec{j}$$

Taking the dot product of $$\vec{\tilde{i}}$$ and $$\vec{j}$$ yields: $$\vec{\tilde{i}}\cdot\vec{j}=\cos\theta^{(e)}\vec{i}\cdot\vec{j}+\sin\theta^{(e)}\vec{j}\cdot\vec{j}$$

By definition, $$\vec{i}\cdot\vec{j}$$ is equal to 0, and $$\vec{j}\cdot\vec{j}$$ is equal to 1. Using these relationships reduces the above equation to: $$\vec{\tilde{i}}\cdot\vec{j}=\sin\theta^{(e)}$$

Previously, the direction cosine $$m^{\left(e\right)}$$ was defined as $$ \cos \left(\frac{\pi }{2}-\theta ^{(e)}\right)$$, which is equivalent to $$\sin\theta^{\left(e\right)}$$. Thus the derived expression for $$\vec{\tilde{i}}\cdot\vec{j}$$ is: $$\vec{\tilde{i}}\cdot\vec{j}=m^{(e)}$$