Question

In Exercises \(9-14,\) find the angle \(\theta\) between the vectors. $$ \mathbf{u}=3 \mathbf{i}+\mathbf{j}, \mathbf{v}=-2 \mathbf{i}+4 \mathbf{j} $$

Short answer

\(\theta = \arccos\left(\frac{-2}{\sqrt{10} \times \sqrt{20}}\right)\)

Step-by-step solution

Calculate the dot product

First, calculate the dot product of vector \(\mathbf{u}\) and vector \(\mathbf{v}\). The dot product is given by the sum of the products of the corresponding entries of the two sequences of numbers. Therefore, \(\mathbf{u} \cdot \mathbf{v} = (3*-2) + (1*4) = -6 + 4 = -2.\

Calculate the magnitudes

Next, calculate the magnitude (or length) of each vector. The magnitude of a vector \(\mathbf{a} = ai + bj\) is given by \(\sqrt{(a^2 + b^2)}\). Therefore, \(||\mathbf{u}|| = \sqrt{(3^2 + 1^2)} = \sqrt{10}\) and \(||\mathbf{v}|| = \sqrt{(-2^2 + 4^2)} = \sqrt{20}.

Calculate the angle

Now calculate the angle theta. Using the dot product formula rearranged, \(\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \times ||\mathbf{v}||}\). Substituting given values, \(\cos(\theta) = \frac{-2}{\sqrt{10} \times \sqrt{20}}\). Next, take the arccos of each side to solve for theta. Doing this yields a final answer of \(\theta = \arccos\left(\frac{-2}{\sqrt{10} \times \sqrt{20}}\right).