1. Problem statement: Determine if the vector (1,7,23) can be a vector perpendicular to both $\vec{a} = (2,3,-1)$ and $\vec{b} = (5,-1,-2)$ in problem 7b. 2. Formula: To find a vector perpendicular to both $\vec{a}$ and $\vec{b}$, we use the cross product: $$\vec{w} = \vec{a} \times \vec{b}$$ 3. Calculate the cross product: $$\vec{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & -1 \\ 5 & -1 & -2 \end{vmatrix} = \mathbf{i}(3 \cdot (-2) - (-1) \cdot (-1)) - \mathbf{j}(2 \cdot (-2) - (-1) \cdot 5) + \mathbf{k}(2 \cdot (-1) - 3 \cdot 5)$$ 4. Simplify each component: $$\mathbf{i}(-6 - 1) - \mathbf{j}(-4 + 5) + \mathbf{k}(-2 - 15) = \mathbf{i}(-7) - \mathbf{j}(1) + \mathbf{k}(-17)$$ 5. So the vector is: $$\vec{w} = (-7, -1, -17)$$ 6. Check if (1,7,23) is a scalar multiple of $\vec{w}$: Check ratios: $$\frac{1}{-7} = -\frac{1}{7}, \quad \frac{7}{-1} = -7, \quad \frac{23}{-17} \approx -1.35$$ Since these ratios are not equal, (1,7,23) is not a scalar multiple of $\vec{w}$. 7. Conclusion: The vector (1,7,23) is not perpendicular to both $\vec{a}$ and $\vec{b}$ in problem 7b. Final answer: No, (1,7,23) cannot be $\vec{w}_1$, $\vec{w}_2$, or $\vec{w}_3$ as a vector perpendicular to both $\vec{a}$ and $\vec{b}$ in 7b.