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The position vectors of the vertices A,B...

The position vectors of the vertices A,B and C of a triangle are `hati-hatj-3hatk,2hati+hatj-2hatk` and `-5hati+2hatj-6hatk`, respectively. The length of the bisector AD of the `angleBAC`, where D is on the segment BC, is

A

`(3)/(4)sqrt(3)`

B

`(1)/(4)`

C

`(11)/(2)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A

`|AB|=|(2hati+hatj-2hatk)-(hati-hatj-3hatk)|`
`=|hati+2hatj+hatk|`
`=sqrt(1^(2)+2^(2)+1^(2))=sqrt(6)`
`|AC|=|(-5hati+2hatj-6hatk)-(hati-hatj-3hatk)|`
`=|-6hati+3hatj-3hatk|`
`=sqrt((6-6)^(2)+3^(2)+(-3)^(2))=sqrt(54)=3sqrt(6)`
`BD:DC=AB:AC=(sqrt(6))/(3sqrt(6))=(1)/(3)`
`therefore`Position vector of `D=(1(-5hati+2hatj-6hatk)+3(2hati+hatj-2hatk))/(1+3)`
`=(1)/(4)(hati+5hatj-12hatk)`
`thereforeAD=`Position vector of D-position vector of A
`AD=(1)/(4)(hati+5hatj-12hatk)-(hati-hatj-3hatk)=(1)/(4)(-3hati+9hatj)`
`=(3)/(4)(-hati+3hatj) `
`|AD|=(3)/(4)sqrt((-1)^(2)+3^(2))=(3)/(4)sqrt(10)`
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