Let x0 be the point of local maxima of f(x)=a→·(b→×c→) where a→=xi-2j+3k and b→=-2i+xj-k and c→=7i-2j+xk. Then the value of a→·b→+b→·c→+c→·a→ atx=x0 is:
-22
-4
-30
14
Explanation for the correct answer:
a→·(b→×c→)=x-23-2x-17-2x=x(x2-2)+2(-2x+7)+3(4-7x)=x3-2x-4x+14+12-21x
f(x)=x3–27x+26f'(x)=3x2-27f''(x)=6x
Now, x=3,f''(3)=6×3=18>0x=-3,f''(-3)=6×3=-18<0
Maximum at, x0=-3
a→=(-3,-2,3)b→=(-2,-3,-1)c→=(7,-2,-3)
So
a→·b→+b→·c→+c→·a→=6+6-3-14+6+3-21+4-9=25-47=-22
Hence, the correct option is (A)
Show that the points whose position vectors are -2i+3j, i+2j+3k and 7i-k are collinear