Compute the dot product of the vectors \( \mathbf{u} \) and \( \mathbf{v} \), and find the angle between the vectors. \( \mathbf{u}=\langle-12,0,5\rangle \) and \( \mathbf{v}=\langle 1,2,3\rangle \). \( \mathbf{u} \cdot \mathbf{v}=\square \). (Type an integer or a simplified fraction.)

To compute the dot product of the vectors \(\mathbf{u} = \langle -12, 0, 5 \rangle\) and \(\mathbf{v} = \langle 1, 2, 3 \rangle\), we use the formula: \[ \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \] Substituting in the values: \[ \mathbf{u} \cdot \mathbf{v} = (-12)(1) + (0)(2) + (5)(3) = -12 + 0 + 15 = 3 \] So, \(\mathbf{u} \cdot \mathbf{v} = 3\). Next, to find the angle \(\theta\) between the vectors, we use the formula: \[ \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} \] First, we compute the magnitudes of \(\mathbf{u}\) and \(\mathbf{v}\): \[ |\mathbf{u}| = \sqrt{(-12)^2 + 0^2 + 5^2} = \sqrt{144 + 0 + 25} = \sqrt{169} = 13 \] \[ |\mathbf{v}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \] Now we can substitute the values into the cosine formula: \[ \cos(\theta) = \frac{3}{13 \cdot \sqrt{14}} \] Finally, to find the angle \(\theta\), we take the arccosine: \[ \theta = \cos^{-1} \left( \frac{3}{13 \sqrt{14}} \right) \] So, the dot product is \(3\).