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

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. `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» Correct Answer - D 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`

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

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