Find the angle \( \theta \) between the vectors. \[ \begin{array}{l}\mathbf{u}=6 \mathbf{i}+7 \mathbf{j}+\mathbf{k} \\ \mathbf{v}=7 \mathbf{i}-6 \mathbf{j}\end{array} \] \( \theta=\square \) radians

To find the angle \( \theta \) between the vectors \(\mathbf{u}\) and \(\mathbf{v}\), we can use the dot product formula: \[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} \] **Step 1: Compute the Dot Product \(\mathbf{u} \cdot \mathbf{v}\)** \[ \mathbf{u} = 6\mathbf{i} + 7\mathbf{j} + \mathbf{k} \] \[ \mathbf{v} = 7\mathbf{i} - 6\mathbf{j} \] \[ \mathbf{u} \cdot \mathbf{v} = (6)(7) + (7)(-6) + (1)(0) = 42 - 42 + 0 = 0 \] **Step 2: Compute the Magnitudes \(|\mathbf{u}|\) and \(|\mathbf{v}|\)** \[ |\mathbf{u}| = \sqrt{6^2 + 7^2 + 1^2} = \sqrt{36 + 49 + 1} = \sqrt{86} \] \[ |\mathbf{v}| = \sqrt{7^2 + (-6)^2} = \sqrt{49 + 36} = \sqrt{85} \] **Step 3: Compute \(\cos \theta\)** \[ \cos \theta = \frac{0}{\sqrt{86} \cdot \sqrt{85}} = 0 \] **Step 4: Determine \(\theta\)** Since \(\cos \theta = 0\), the angle \(\theta\) is: \[ \theta = \frac{\pi}{2} \text{ radians} \] **Final Answer:** \[ \theta = \frac{\pi}{2} \text{ radians} \]