Let `vecc=xhati+yhatj+zhatk`, where `x^(2)+y^(2)+z^(2)=1`.
Unit vector along `3hati+4hatj` is `(3hati+4hatj)/(5)`.
The bisector of these two is `-hati+hatj-hatk` (given). Therefore,
`” “-hati+hatj-hatk=lamda(xhati+yhatj+zhatk+(3hati+4hatj)/(5))`
`” ” -hati+hatj-hatk=(1)/(5) lamda [(5x+3)hati+(5y+4)hatj +5zhatk]`
`” “(lamda)/(5)(5x+3)=-1, (lamda)/(5)(5y+4)=1, (lamda)/(5)5z=-1`
`” “x=-(5+3lamda)/(5lamda), y=(5-4lamda)/(5lamda),z=-(1)/(lamda)`
Putting these values in (i), i.e., `x^(2)+y^(2)+z^(2)=1`, we get
`” “(5+3lamda)^(2)+(5-4lamda)^(2)+25=25lamda^(2)`
`” “25lamda^(2)-10lamda+75=25lamda^(2)`
`” “lamda=(15)/(2)`
`therefore” “vecc=(1)/(15)(-11hatik+10hatj-2hatk)`