JEE Main 2020 (Online) 4th September Morning ShiftQuestion 1
Let $x_0$ be the point of local maxima of $f(x)=\vec a\cdot(\vec b\times\vec c)$, where $\vec a=x\hat i-2\hat j+3\hat k$, $\vec b=-2\hat i+x\hat j-\hat k$ 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 :
Vector Algebra
Solution
$f(x)=\vec a\cdot(\vec b\times\vec c)=\det\begin{pmatrix}x&-2&3\\-2&x&-1\\7&-2&x\end{pmatrix}=x^3-27x+26$. $f'=3x^2-27=0\Rightarrow x=\pm3$; $f''=6x<0$ at $x=-3$, so $x_0=-3$. $\vec a=(-3,-2,3)$, $\vec b=(-2,-3,-1)$, $\vec c=(7,-2,-3)$. $\vec a\cdot\vec b=6+6-3=9$, $\vec b\cdot\vec c=-14+6+3=-5$, $\vec c\cdot\vec a=-21+4-9=-26$. Sum $=-22$. Option (B).