Question

Let $\overline{a}=x\overline{i}-2\overline{j}+3\overline{k}$, $\overline{b}=-2\overline{i}+x\overline{j}-\overline{k}$ and $\overline{c}=7\overline{i}-2\overline{j}+x\overline{k}$. If $x=x_0$ is the point of the local maxima of $f(x)=\overline{a}\cdot(\overline{b}\times\overline{c})$, then at $x=x_0$, $\overline{a}\cdot\overline{b}+\overline{b}\cdot\overline{c}+\overline{c}\cdot\overline{a}=$

• A. \(-30 \)
• B. \(-22 \)
• C. \(-4 \)
• D. \(-14 \)

Hint:

For scalar triple products defined by variables, the determinant method is more efficient than calculating cross products. Always verify critical points using the second derivative test ($f''(x) < 0$ for local maxima).

Answer 1

The Correct Option is B

Concept: The scalar triple product $\overline{a}\cdot(\overline{b}\times\overline{c})$ is given by the determinant of the matrix formed by the components of vectors $\overline{a}$, $\overline{b}$, and $\overline{c}$. \[ f(x) = \begin{vmatrix} x & -2 & 3 \\ -2 & x & -1 \\ 7 & -2 & x \end{vmatrix} \]

Step 1: Expand the determinant to find the polynomial \(f(x)\). \[ f(x) = x(x^2 - 2) - (-2)(-2x + 7) + 3(4 - 7x) = x^3 - 27x + 26 \]

Step 2: Find the critical points by setting the derivative \(f'(x) = 0\). \[ f'(x) = 3x^2 - 27 = 0 \implies x^2 = 9 \implies x = \pm 3 \]

Step 3: Identify the local maxima using the second derivative test \(f''(x) = 6x\). At $x = -3$, $f''(-3) = -18 < 0$, confirming $x_0 = -3$ is the point of local maxima.

Step 4: Evaluate the vectors \(\overline{a}, \overline{b}, \overline{c}\) at \(x = -3\). For $x = -3$: $\overline{a} = -3\overline{i}-2\overline{j}+3\overline{k}$, $\overline{b} = -2\overline{i}-3\overline{j}-\overline{k}$, $\overline{c} = 7\overline{i}-2\overline{j}-3\overline{k}$.

Step 5: Calculate the sum of the dot products \(\overline{a}\cdot\overline{b}+\overline{b}\cdot\overline{c}+\overline{c}\cdot\overline{a}\).

• $\overline{a}\cdot\overline{b} = (-3)(-2) + (-2)(-3) + (3)(-1) = 9$

• $\overline{b}\cdot\overline{c} = (-2)(7) + (-3)(-2) + (-1)(-3) = -5$

• $\overline{c}\cdot\overline{a} = (7)(-3) + (-2)(-2) + (-3)(3) = -26$
Sum $= 9 - 5 - 26 = -22$.