# Prove the following

Question:

Let $x_{0}$ be the point of local maxima of $f(x)=\vec{a} \cdot(\vec{b} \times \vec{c})$, where

$\vec{a}=x \hat{i}-2 \hat{j}+3 \hat{k}, \quad \vec{b}=-2 \hat{i}+x \hat{j}-\hat{k}$ and $\vec{c}=7 \hat{i}-2 \hat{j}+x \hat{k}$.

Then the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ at $x=x_{0}$ is :

1. $-4$

2. $-30$

3. 14

4. $-22$

Correct Option: , 4

Solution:

It is given that

$f(x)=\bar{a} \cdot(\bar{b} \times \bar{c})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ x & -2 & 3 \\ -2 & x & -1 \\ 7 & -2 & x\end{array}\right|=x^{3}-27 x+26$

$\Rightarrow f(x)=x^{3}-27 x+26$

$\Rightarrow f^{\prime}(x)=3 x^{2}-27$

For critical point $f^{\prime}(x)=0$

$\Rightarrow 3 x^{2}-27=0 \Rightarrow x=-3,3$

The local maxima of $f(x)$ is, $x_{0}=-3$.

Then $\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}$

$=-2 x-2 x-3-14-2 x-x+7 x+4+3 x=3 x-13$

So, value at $x=x_{0},=\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}=3 x-13$

$=3 \times(-3)-13=-22$