The position vectores of the vertices A,B, C of a `/_\ ABC` are`hati - hatj - 3 hatk , 2 hati +hatj - 2hatk and - 5 hati + 2 hatj - 6 hatk` respectively. The length of the bisector AD of the angle ` angle BAC,` where D is on the line segments BC, is

Options

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

Answer

D

Explanation

We have ` OA = hati - hatj - 3hatk , OB = 2 hati + hatj - 2 hatk`
`and OC =- 5 hati + 2hatj - 6 hatk`
` AB=OB -OA = hati + 2hatj +hatk`
`AC =OC -OA =- 6 hati +3hatj - 3hatk`
`|AB|= sqrt(1^(2)+2^(2)+1^(2))= sqrt(6)`
` and |AC|=sqrt((-6)^(2) +3^(2) +(-3)^(2))=3sqrt(6)`

Clearly , point D divides BC in the ratio `AB : AC i .e `1 :3`
` therefore Position vector of ` D =((-5 hati +2 hatj - 6 hatk ) +3 ( 2 hati + hatj - 2 hatk ))/( 1+3)`
Position vector of` =(1)/(4) ( hati +5hatj- 12 hatk)`
`Now AD=(1)/(4) (hati +5hatj- 12 hatk)-(hati - hatj - 3 hatk)=(3)/(4) (-hati+3hatj)`
` therefore AD =|AD|=(3)/(4) qrt(10)`