Question

Find the angle \(\theta\) between the vectors. $$ \begin{array}{l} \mathbf{u}=3 \mathbf{i}+2 \mathbf{j}+\mathbf{k} \\ \mathbf{v}=2 \mathbf{i}-3 \mathbf{j} \end{array} $$

Short answer

The angle \(\theta\) between the vectors \(\mathbf{u}\) and \(\mathbf{v}\) is \(\frac{\pi}{2}\) or \(90^\circ\).

Step-by-step solution

Calculate the dot product

To calculate the dot product of vectors \(\mathbf{u}\) and \(\mathbf{v}\), multiply the corresponding components of the vectors and add them together. We have: \(\mathbf{u} . \mathbf{v} = (3 * 2) + (2 * -3) + (1 * 0) = 6 - 6 = 0\).

Calculate the magnitude of the vectors

The magnitude of a vector is calculated by taking the square root of the sum of the squares of the components. We have: \(||\mathbf{u}|| = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{14}\) and \(||\mathbf{v}|| = \sqrt{2^2 + (-3)^2} = \sqrt{13}\).

Find the angle \(\theta\)

From the given formula, we find \(\cos(\theta) = \frac{\mathbf{u} . \mathbf{v}}{||\mathbf{u}|| ||\mathbf{v}||}\). Substituting in our values, we find \(\frac{0}{\sqrt{14} * \sqrt{13}} = 0\). So, \(\cos(\theta) = 0\). The angle \(\theta\) whose cosine is 0 is \(\frac{\pi}{2}\) or \(90^\circ\).