Attempt 1: 10 attempts remaining. Find the value of \( k \) for which the vectors \( \left[\begin{array}{c}-1 \\ -2 \\ -4 \\ -3\end{array}\right] \) and \( \left[\begin{array}{c}4 \\ -5 \\ -4 \\ k\end{array}\right] \) are orthogonal. \( k= \)

To determine the value of \( k \) for which the vectors \[ \mathbf{a} = \left[\begin{array}{c}-1 \\ -2 \\ -4 \\ -3\end{array}\right] \] and \[ \mathbf{b} = \left[\begin{array}{c}4 \\ -5 \\ -4 \\ k\end{array}\right] \] are orthogonal, we need to use the property that two vectors are orthogonal if their dot product is zero. The dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as follows: \[ \mathbf{a} \cdot \mathbf{b} = (-1)(4) + (-2)(-5) + (-4)(-4) + (-3)(k) \] Now, let's compute each term: 1. \( (-1)(4) = -4 \) 2. \( (-2)(-5) = 10 \) 3. \( (-4)(-4) = 16 \) 4. \( (-3)(k) = -3k \) Now, we can combine these results: \[ \mathbf{a} \cdot \mathbf{b} = -4 + 10 + 16 - 3k \] Simplifying this gives: \[ \mathbf{a} \cdot \mathbf{b} = 22 - 3k \] To find the value of \( k \) that makes the vectors orthogonal, we set the dot product equal to zero: \[ 22 - 3k = 0 \] Now, we can solve for \( k \): \[ 3k = 22 \] \[ k = \frac{22}{3} \] Thus, the value of \( k \) for which the vectors are orthogonal is \[ k = \frac{22}{3}. \]