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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 |
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Answer» `(15)/(2) ` `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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