1.

The position vectores of the vertices A,B, C of a delta ABC arehati- hatj- 3 hatk , 2hati+hatj - 2hatkand- 5 hati+ 2 hatj- 6 hatk respectively. The length of the bisector AD of the angleangle BAC, where D is on the line segments BC, is

Answer»

`(15)/(2) `
`(11)/(2)`
`(1)/(4)`
None of these

Solution :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 , pointD dividesBCin theratio `AB : AC i .e`1 :3`
` thereforePositionvectorof ` D =((-5 hati +2 hatj- 6 hatk ) +3 ( 2 hati + hatj- 2 hatk ))/( 1+3)`
Positionvectorof` =(1)/(4) ( hati+5hatj- 12 hatk)`
`NowAD=(1)/(4) (hati +5hatj- 12 hatk)-(hati - hatj - 3 hatk)=(3)/(4) (-hati+3hatj)`
` thereforeAD =|AD|=(3)/(4)qrt(10)`


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