Problem 94 - Precalculus with Limits

Question

PROOF Prove that if \(\mathbf{u}\) is a unit vector and \(\theta\) is the angle between \(\mathbf{u}\) and \(\mathbf{j},\) then $$ \mathbf{u}=\cos \left(\frac{\pi}{2}-\theta\right) \mathbf{i}+\sin \left(\frac{\pi}{2}-\theta\right) \mathbf{j} $$

Step-by-Step Solution

Answer

The unit vector \(\mathbf{u}\) can be given by a unique combination of the basis vectors \(\mathbf{i}\) and \(\mathbf{j}\). With the angle \(\theta\)between \(\mathbf{u}\) and \(\mathbf{i}\), we can relate the vector components with the cosine and sine of this angle respectively. Then, by using the identities for transformations of these trigonometric functions, we arrive at the statement asked to prove.

Step 1: Express the Vector in terms of basis

Express the unit vector \(\mathbf{u}\) in terms of the basis vectors \(\mathbf{i}\) and \(\mathbf{j}\) . The unit vector \(\mathbf{u}\) in two dimensional plane can be expressed as \(\mathbf{u}= u_1\mathbf{i}+u_2\mathbf{j}\).

Step 2: Use the angle between vector and basis vector

The angle \(\theta\) is between the vector \(\mathbf{u}\) and \(\mathbf{i}\), so we can write the components \(u_1\) and \(u_2\) in terms of \(\theta\). Therefore we get \(u_1= \cos(\theta)\) and \(u_2= \sin(\theta)\).

Step 3: Transform the angle

The problem wants us to write our unit vector in terms of \((\frac{\pi}{2}-\theta)\). We therefore use the relations \(\cos(\frac{\pi}{2}-\theta)=\sin(\theta)\) and \(\sin(\frac{\pi}{2}-\theta)=\cos(\theta)\). When substituting the transformed angles into our representation from step 2 we get: \(\mathbf{u}= \cos\left(\dfrac{\pi}{2} - \theta \right)\mathbf{i} + \sin\left(\dfrac{\pi}{2} - \theta \right)\mathbf{j}\). This gives us the desired expression.