If `hati-3hatj+5hatk` bisects the angle between `hata and -hati+2hatj+2hatk`, where `hata` is a unit vector, then

Options

A. `a=(1)/(105)(41hati+88hatj-40hatk)`
B. `a=(1)/(105)(41hati+88hatj+40hatk)`
C. `a=(1)/(105)(-41hati+88hatj-40hatk)`
D. `a=(1)/(105)(41hati-88hatj-40hatk)`

Answer

D

Explanation

We must have `lamda(hati-3hatj+5hatk)=a+(2hatk+2hatj-hati)/(3)`
therefore, `3a=3lamda(hati-3hatj+5hatk)-(2hatk+2hatj-hati)`
`=hati(3lamda+1)-hatj(2+9lamda)+hatk(15lamda-2)`
or `3|a|=sqrt((3lamda+1)^(2)+(2+9lamda)^(2)+(15lamda-2)^(2))`
or `9=(3lamda+1)^(2)+(2+9lamda)^(2)+(15lamda-2)^(2)`
or `315lamda^(2)-18lamda=0 implies lamda=0, (2)/(35)`.
if `lamda=0,a=hati-2hatj-2hatk` (not acceptable).
for `lamda=(2)/(35),a=(41)/(105)hati-(88)/(105)hatj-(40)/(105)hatk`