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

Options

A. `hata = (1)/(150)(41hati+88hatj-40hatk)`
B. `hata=(1)/(105)(41hati + 88hatj + 40hatk)`
C. `hata = (1)/(105) (-41hati + 88hatj-40hatk)`
D. `hata= (1)/(105) (41hati - 88hatj - 40 hatk )`

Answer

D

Explanation

We must have `lamda (hati-3hatj +5hatk) = hata + (2hatk+2hatj-hati)/(3)`
Therefore
`3hata = 3lamda = (hati-3hatj +5hatj) - (2hatk +2hatj -hati)`
`" "= hati (3lamda +1) -hatj( 2+ 9lamda) + hatk ( 15lamda -2)`
or `" "3|hata| = sqrt((3lamda +1)^(2) + (2+ 9lamda)^(2) + (15lamda -)^(2))`
or ` " "9 = (3lamda +1)^(2) + (2 + 9lamda )^(2) + (15lamda -2)^(2)`
or `" "315lamda ^(2) - 18lamda =0 rArr lamda =0, (2)/(35)`
if `lamda =0, veca =hati-2hatj -hatk` ( not acceptable)
For `lamda = (2)/(35) , veca = (41)/(105)hati - (88)/(105) hatj - (40/(105)hatk`