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The position vectors of the vertices A,B,C of DeltaABC are hati-hatj-3hatk,2hati+hatj-2hatk and -5hati+2hatj-6hatk respectively. The length of the bisector AD of the angle /_BAC where D is on the line segment BC is |
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Answer» `15/2` `VEC(AB)=hati+2hatj+hatk, vec(AC)-6hati+3hatj-3hatk` `implies|vec(AB)|=sqrt(6)` and `|vec(AC)|=3sqrt(6)` Clearly, point D divides BC in the ratio `AB:AC` i.e. `1:3`. `:.` Position vector ofDis `((-5hati+2hatj-6hatk)+3(2hati+hatj-2hatk))/(1+3)` `implies` Position vector of D is `=1/4(hati+5hatj-12hatk)` `:.vec(AD)=1/4(hati+5hatj-12hatk)-(hati-hatj-3hatk)` `vec(AD)=3/4(-hati+3hatj)` `implies|vec(AD)|=3/4sqrt(10)` |
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