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The position vectors of the vertices A,B,C of a triangleABC are hati-hatj-3hatk,2hati+hatj-2hatk and -5hati+2hatj-6hatk respectively . The length of the bisector AD of the angle angleBAC where D is on the line segment BC, is : - |
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Answer» `15/2` `rArr |vec(AB)|=sqrt6` and `|vec(AC)|=3sqrt6` CLEARLY , POINT D divides BC in the RATIO AB : AC i.e. 1:3 `therefore` Position vector of D is `((-5hati+2hatj-6hatk)+3(2hati+hatj-2hatk))/(1+3)` `rArr` Position vector of D is `=1/4(hati + 5hatj-12hatk)` `therefore vec(AD) = 1/4(hati+5hatj-12hatk)-(hati-hatj-3hatk)` `vec(AD)=3/4(-hati+3hatj)` `rArr |vec(AD)|=3/4sqrt10` |
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