Find the least positive integral value of `x` for which the angle between vectors ` vec a=x hat i-3 hat j- hat k` and `vec b=2x hat i+x hat j- hat k` is acute.

Answer

2

Explanation

Let `veca = x hati - 3hatj -hatk and vecb = 2x hati+ xhatj -hatk` be the adjacent sides of the parallelogram.
Now angle between `veca and vecb` is acute, i.e.,
`" "|veca = vec b| gt |veca - vecb|`
`rArr |3xhati + (x-3)hatj- 2hatk|^(2) gt |-xhati -(x+3)hatj|^(2)`
or `" "9x^(2) + (x-3)^(2)+ 4 gt x^(2) + (x+3)^(2)`

or ` " " 8x^(2) - 12x +4 gt 0`
or `" " 2x^(2) - 3x + 1 gt 0`
or `" "(2x-1)(x-1) gt 0`
`rArr x lt 1//2 or x gt 1 `
Hence, the least positive integral value is 2.