Key Concepts and Formulas

1. Scalar Triple Product (STP): For three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$, the scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ can be computed as the determinant of the matrix formed by their components: $[\vec{a} \, \vec{b} \, \vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$
2. Local Maxima/Minima (Second Derivative Test): To find local extrema of a function $f(x)$, we find critical points by setting $f'(x) = 0$. The nature of the critical point $x_c$ is determined by the second derivative $f''(x_c)$:
   • $f''(x_c) < 0 \implies$ local maximum.
   • $f''(x_c) > 0 \implies$ local minimum.
3. Dot Product: For vectors $\vec{u} = u_1\widehat{i} + u_2\widehat{j} + u_3\widehat{k}$ and $\vec{v} = v_1\widehat{i} + v_2\widehat{j} + v_3\widehat{k}$, the dot product is $\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3$.

Step-by-Step Solution

1. Expressing $f(x)$ using the Scalar Triple Product

The function $f(x)$ is given by the scalar triple product of vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$. We write out the components of these vectors: $\vec{a} = x\widehat{i} - 2\widehat{j} + 3\widehat{k}$ $\vec{b} = - 2\widehat{i} + x\widehat{j} - \widehat{k}$ $\vec{c} = 7\widehat{i} - 2\widehat{j} + x\widehat{k}$

The scalar triple product $f(x) = \vec{a} \cdot (\vec{b} \times \vec{c})$ is calculated as the determinant of the matrix formed by the components of these vectors: $f(x) = \begin{vmatrix} x & -2 & 3 \\ -2 & x & -1 \\ 7 & -2 & x \end{vmatrix}$ Expanding this determinant along the first row: $f(x) = x \begin{vmatrix} x & -1 \\ -2 & x \end{vmatrix} - (-2) \begin{vmatrix} -2 & -1 \\ 7 & x \end{vmatrix} + 3 \begin{vmatrix} -2 & x \\ 7 & -2 \end{vmatrix}$ $f(x) = x(x^2 - (-1)(-2)) + 2((-2)x - (-1)(7)) + 3((-2)(-2) - x(7))$ $f(x) = x(x^2 - 2) + 2(-2x + 7) + 3(4 - 7x)$ Distributing and simplifying: $f(x) = x^3 - 2x - 4x + 14 + 12 - 21x$ $f(x) = x^3 - 27x + 26$

2. Finding Critical Points of $f(x)$

To find the points of local maxima or minima, we first find the critical points by calculating the first derivative $f'(x)$ and setting it to zero. $f'(x) = \frac{d}{dx}(x^3 - 27x + 26)$ $f'(x) = 3x^2 - 27$ Setting $f'(x) = 0$: $3x^2 - 27 = 0$ $3x^2 = 27$ $x^2 = 9$ $x = \pm 3$ The critical points are $x = 3$ and $x = -3$.

3. Identifying the Local Maximum using the Second Derivative Test

We use the second derivative test to determine which critical point corresponds to a local maximum. First, we compute the second derivative $f''(x)$: $f''(x) = \frac{d}{dx}(3x^2 - 27)$ $f''(x) = 6x$ Now, we evaluate $f''(x)$ at each critical point:

• For $x = 3$: $f''(3) = 6(3) = 18$. Since $f''(3) > 0$, $x=3$ is a point of local minimum.
• For $x = -3$: $f''(-3) = 6(-3) = -18$. Since $f''(-3) < 0$, $x=-3$ is a point of local maximum.

The problem states that $x_0$ is the point of local maxima, so $x_0 = -3$.

4. Evaluating Vectors at $x_0 = -3$

We substitute $x_0 = -3$ into the definitions of $\vec{a}$, $\vec{b}$, and $\vec{c}$:

• $\vec{a} = (-3)\widehat{i} - 2\widehat{j} + 3\widehat{k} = -3\widehat{i} - 2\widehat{j} + 3\widehat{k}$
• $\vec{b} = -2\widehat{i} + (-3)\widehat{j} - \widehat{k} = -2\widehat{i} - 3\widehat{j} - \widehat{k}$
• $\vec{c} = 7\widehat{i} - 2\widehat{j} + (-3)\widehat{k} = 7\widehat{i} - 2\widehat{j} - 3\widehat{k}$

5. Calculating the Required Expression $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$

We compute each dot product individually:

• $\vec{a} \cdot \vec{b} = (-3)(-2) + (-2)(-3) + (3)(-1) = 6 + 6 - 3 = 9$
• $\vec{b} \cdot \vec{c} = (-2)(7) + (-3)(-2) + (-1)(-3) = -14 + 6 + 3 = -5$
• $\vec{c} \cdot \vec{a} = (7)(-3) + (-2)(-2) + (-3)(3) = -21 + 4 - 9 = -26$

Finally, we sum these values: $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = 9 + (-5) + (-26) = 9 - 5 - 26 = 4 - 26 = -22$

Common Mistakes & Tips

• Determinant Calculation Errors: Be extremely careful with signs when expanding determinants, especially the middle term.
• Sign Errors in Derivatives: Ensure correct application of differentiation rules, particularly for polynomial terms.
• Confusing Maxima and Minima: Remember that $f''(x) < 0$ indicates a local maximum and $f''(x) > 0$ indicates a local minimum.

Summary

The problem required us to first express the scalar triple product as a polynomial in $x$, then find its local maximum using differential calculus. We identified the point of local maxima, $x_0 = -3$, by applying the second derivative test. Finally, we substituted this value back into the vector definitions and computed the sum of the dot products $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$.

The final answer is $\boxed{-22}$, which corresponds to option (D).