Given \( \mathbf{v}=-5 \mathbf{i}-\mathbf{j} \) and \( \mathbf{w}=-\mathbf{i}-5 \mathbf{j} \) (a) find the dot product \( \mathbf{v} \cdot \mathbf{w} \); (b) find the angle between \( \mathbf{v} \) and \( \mathbf{w} \). (a) \( \mathbf{v} \cdot \mathbf{w}=\square \) (Simplify your answer.)

To find the dot product \( \mathbf{v} \cdot \mathbf{w} \), we can use the formula: \[ \mathbf{v} \cdot \mathbf{w} = v_1 w_1 + v_2 w_2 \] Substituting in the components of \( \mathbf{v} = (-5, -1) \) and \( \mathbf{w} = (-1, -5) \): \[ \mathbf{v} \cdot \mathbf{w} = (-5)(-1) + (-1)(-5) = 5 + 5 = 10 \] So, the answer is: (a) \( \mathbf{v} \cdot \mathbf{w} = 10 \) Now to find the angle \( \theta \) between \( \mathbf{v} \) and \( \mathbf{w} \), we can use the formula involving the dot product and magnitudes: \[ \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{v}| |\mathbf{w}|} \] First, we need the magnitudes \( |\mathbf{v}| \) and \( |\mathbf{w}| \): \[ |\mathbf{v}| = \sqrt{(-5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \] \[ |\mathbf{w}| = \sqrt{(-1)^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26} \] Now substituting back into the formula for cosine: \[ \cos(\theta) = \frac{10}{\sqrt{26} \cdot \sqrt{26}} = \frac{10}{26} = \frac{5}{13} \] To find the angle \( \theta \): \[ \theta = \cos^{-1}\left(\frac{5}{13}\right) \] So the final answer for (b) is: (b) The angle \( \theta \) between \( \mathbf{v} \) and \( \mathbf{w} \) is \( \cos^{-1}\left(\frac{5}{13}\right) \).