14. Find \( k \) such that the angle between the vectors \( \left(\begin{array}{c}3 \\ -k \\ -1\end{array}\right) \) and \( \left(\begin{array}{c}1 \\ -3 \\ k\end{array}\right) \) is \( \frac{\pi}{3} \).

To find \( k \) such that the angle between the vectors \( \mathbf{u} = \begin{pmatrix} 3 \\ -k \\ -1 \end{pmatrix} \) and \( \mathbf{v} = \begin{pmatrix} 1 \\ -3 \\ k \end{pmatrix} \) is \( \frac{\pi}{3} \), we start by applying the formula for the cosine of the angle between two vectors: \[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} \] Given \( \theta = \frac{\pi}{3} \), we have \( \cos \frac{\pi}{3} = \frac{1}{2} \). First, we calculate the dot product \( \mathbf{u} \cdot \mathbf{v} \): \[ \mathbf{u} \cdot \mathbf{v} = 3 \cdot 1 + (-k)(-3) + (-1)k = 3 + 3k - k = 3 + 2k \] Next, we find the magnitudes \( \|\mathbf{u}\| \) and \( \|\mathbf{v}\| \): \[ \|\mathbf{u}\| = \sqrt{3^2 + (-k)^2 + (-1)^2} = \sqrt{9 + k^2 + 1} = \sqrt{k^2 + 10} \] \[ \|\mathbf{v}\| = \sqrt{1^2 + (-3)^2 + k^2} = \sqrt{1 + 9 + k^2} = \sqrt{k^2 + 10} \] Since \( \|\mathbf{u}\| = \|\mathbf{v}\| = \sqrt{k^2 + 10} \), we can now write the cosine equation using our calculations: \[ \frac{3 + 2k}{\sqrt{k^2 + 10} \cdot \sqrt{k^2 + 10}} = \frac{1}{2} \] This simplifies to: \[ \frac{3 + 2k}{k^2 + 10} = \frac{1}{2} \] Cross-multiplying gives: \[ 2(3 + 2k) = k^2 + 10 \] Expanding and rearranging yields: \[ 6 + 4k = k^2 + 10 \] \[ k^2 - 4k + 4 = 0 \] Factoring the quadratic results in: \[ (k - 2)^2 = 0 \] Thus, \( k - 2 = 0 \), leading to \( k = 2 \). In summary, the value of \( k \) that makes the angle between the two vectors \( \frac{\pi}{3} \) is \[ \boxed{2} \]