Let $\vec{a} = x \hat{i} - 2 \hat{j} + 3 \hat{k}$, $\vec{b} = -2 \hat{i} + x \hat{j} - \hat{k}$, and…

Let $\vec{a} = x \hat{i} - 2 \hat{j} + 3 \hat{k}$, $\vec{b} = -2 \hat{i} + x \hat{j} - \hat{k}$, 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 $x = x_0$, where $x_0$ is the point of local maxima of $f(x) = \vec{a} \cdot (\vec{b} \times \vec{c})$, is:
  1. -4
  2. -30
  3. 14
  4. -22

Solution

fx=a→  b→  c→=x      −2      3−2      x     −17       −2      x

=xx2−2+2−2x+7+34−7x

=x3−2x−4x+14+12−21x

fx=x3−27x+26

f'x=3x2−27=3x−3x+3

           

so local maxima point is x0=-3

Now a→.b→+b→.c→+c→.a→=−2x−2x−3-14−2x−x+7x+4+3x=3x−13

at x=x0=−3

a→.b→+b→.c→+c→.a→=−9−13=−22

Asked in: JEE Main 2020 (04 Sep Shift 1)

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