1.

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 : -

Answer»

`15/2`
`3/4sqrt10`
`4/3sqrt10`
None of these

Solution :We have , `vec(AB)=HATI+2hatj+hatk, vec(AC)-6hati+3hatj-3hatk`
`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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