The position vectors of the vectices A, B, C of a `triangle ABC " are " hati - hat j - 3 hat k , 2 hati + hat j - 2 hat k and - 5 hati + 2 hat j - 6 hat k `
respectively. The length of the bisector AD of the angle `angle BAC ` where D is on the line segment BC, is

Options

A. `(15)/2`
B. `(11)/(2)`
C. `(1)/(4)`
D. none of these

Answer

D

Explanation

We have,
`vec(AB)= hati + 2 hat j + hat k , vec(AC) = -6 hati + 3 hat j -3 hat k`
`rArr |vec(AB)| = sqrt(6) and | vec(AC)| = 3 sqrt(6)`
Clearly, point D divides BC in the ratio AB : AC i.e. `1 : 3`.
`therefore " Position vector of " D= ((-5 hati+2 hat j-6hat k)+3(2hat i + hat j - 2hat k))/(1+3)`
`rArr " Position vector of " D=(1)/(4) (hat i+5 hat j -12 hat k)`
`therefore vec(AD) = (1)/(4) (hat i + 5 hat j-12 hat k)`
`therefore vec(AD) = (1)/(4) (hat i + 5 hat j-12 hat k) - ( hat i - hat j - 3hat k )= (3)/(4) (- hat i + 3 hat j)`
`rArr AD = | vec(AD)| = (3)/(4) sqrt(10)`