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

If `hati-3hatj+5hatk` bisects the angle between `hata and -hati+2hatj+

1.	If `hati-3hatj+5hatk` bisects the angle between `hata and -hati+2hatj+2hatk`, where `hata` is a unit vector, thenA. `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» Correct Answer - D 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`

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