Question

Let $x_0$ be the point of local maxima of $f\left(x\right)=\overrightarrow{a}·(\overrightarrow{b}\times \overrightarrow{c})$ where $\overrightarrow{a}=xi-2j+3k$ and $\overrightarrow{b}=-2i+xj-k$ and $\overrightarrow{c}=7i-2j+xk$. Then the value of $\overrightarrow{a}·\overrightarrow{b}+\overrightarrow{b}·\overrightarrow{c}+\overrightarrow{c}·\overrightarrow{a}$ at$x=x_0$ is:

• A. $-22$
• B. $-4$
• C. $-30$
• D. $14$

Answer 1

The correct option is A

$-22$

Explanation for the correct answer:

$$\begin{gathered} \overrightarrow{a}·(\overrightarrow{b}\times \overrightarrow{c})=\left|\begin{array}{ccc}x& -2& 3\\ -2& x& -1\\ 7& -2& x\end{array}\right| \\ =x(x^2-2)+2(-2x+7)+3(4-7x) \\ =x^3-2x-4x+14+12-21x \end{gathered}$$

$$\begin{gathered} f\left(x\right)=x^3–27x+26 \\ f\text{'}\left(x\right)=3x^2-27 \\ f\text{'}\text{'}\left(x\right)=6x \end{gathered}$$

Now, $$\begin{gathered} x=3,f\text{'}\text{'}\left(3\right)=6\times 3=18>0 \\ x=-3,f\text{'}\text{'}(-3)=6\times 3=-18<0 \end{gathered}$$

Maximum at, $x_0=-3$

$$\begin{gathered} \overrightarrow{a}=(-3,-2,3) \\ \overrightarrow{b}=(-2,-3,-1) \\ \overrightarrow{c}=(7,-2,-3) \end{gathered}$$

So

$$\begin{gathered} \overrightarrow{a}·\overrightarrow{b}+\overrightarrow{b}·\overrightarrow{c}+\overrightarrow{c}·\overrightarrow{a} \\ =6+6-3-14+6+3-21+4-9 \\ =25-47 \\ =-22 \end{gathered}$$

Hence, the correct option is (A)