If  a^ is a unit vector and  i^−3j^+5k^ bisects the angle between  a^ and  −i^+2j^+2k^,  then

We must have λ(i^−3j^+5k^)=a^+2k^+2j^−i^3Therefore, 3a^=3λ(i^−3j^+5k^)−(2k^+2j^−i^)=i^(3λ+1)−j^(2+9λ)+k^(15λ−2)or  3|a^|=(3λ+1)2+(2+9λ)2+(15λ−2)2or  9=(3λ+1)2+(2+9λ)2+(15λ−2)2or  315λ2−18λ=0⇒λ=0,235 If λ=0,a→=i^−2j^−2k^ (not acceptable)  For λ=235,a→=41105i^−88105j^−40105k^