Problem 40 - Precalculus with Limits

Question

In Exercises \(31-40,\) find the angle \(\theta\) between the vectors. $$ \begin{array}{l}{\mathbf{u}=\cos \left(\frac{\pi}{4}\right) \mathbf{i}+\sin \left(\frac{\pi}{4}\right) \mathbf{j}} \\ {\mathbf{v}=\cos \left(\frac{\pi}{2}\right) \mathbf{i}+\sin \left(\frac{\pi}{2}\right) \mathbf{j}}\end{array} $$

Step-by-Step Solution

Answer

\(\theta = \dfrac{\pi}{2}\) or 90 degrees.

Step 1: Calculate dot product of vectors

The dot product of two vectors is given as \(\mathbf{u} \cdot \mathbf{v} = u_i v_i + u_j v_j\). Substituting the given vectors \(\mathbf{u} = \cos\left(\dfrac{\pi}{4} \right) \mathbf{i} + \sin\left(\dfrac{\pi}{4} \right) \mathbf{j}\) and \(\mathbf{v} = \cos\left (\dfrac{\pi}{2} \right) \mathbf{i} + \sin\left(\dfrac{\pi}{2} \right) \mathbf{j}\) into the formula, we get the dot product as \(u \cdot v = 0\). This is because \(cos(\dfrac{\pi}{2}) = 0\) and \(sin(\dfrac{\pi}{4})\) times \(sin(\dfrac{\pi}{2}) = \dfrac{1}{\sqrt{2}}\).

Step 2: Calculate magnitudes of vectors

The magnitude of a vector is given by \(\sqrt{u_i^2 + u_j^2}\). For \(\mathbf{u}\), \(|\mathbf{u}| = \sqrt{{\cos^2\left(\dfrac{\pi}{4} \right)+ \sin^2\left(\dfrac{\pi}{4} \right)}} = 1\) because \(cos(\dfrac{\pi}{4}) = sin(\dfrac{\pi}{4}) = \dfrac{1}{\sqrt{2}}\), and the square of these values added together equals 1. Similarly, for \(\mathbf{v}\), \( |\mathbf{v}| = \sqrt{{\cos^2\left(\dfrac{\pi}{2} \right)+ \sin^2\left(\dfrac{\pi}{2} \right)}} = 1\).

Step 3: Calculate the angle

The angle between two vectors is given by \(\cos\theta = \dfrac{\mathbf{u} \cdot \mathbf{v}} {|\mathbf{u}|*|\mathbf{v}|}\). Plug all calculated values into this equation, \(\cos\theta = \dfrac{0} {1*1} = 0\). Therefore, the angle \(\theta = \cos^{-1}(0) = \dfrac{\pi}{2}\) or 90 degrees.