1.

If the vectors AB=3hati+4hatk and AC=(5hati-2hatj+4hatk) are the sides of a DeltaABC, then the length of the median through A is

Answer»

`sqrt(18)`
`sqrt(72)`
`sqrt(33)`
`sqrt(45)`

Solution :We know that, the sum of three vectors of a triangle is zero.
`therefore AB+BC=CA=0`
`rArr BC=AC-AB`
`rArr BM=(AC-AB)/(2) `[SINCE, M is mid-point of BC]

ALSO, `AB+BM+MA=0 ` [ property of triangle]
`rArr AB+(AC-AB)/(2) =AM`
`rArr AM=(AB+AC)/(2) = AM`
`rArr AM=(AB+AC)/(2) = (3 hati+4hatk+5hati-2hatj+4hatk)/(2)`
`=4hati - hatj +4 hatk`
`rArr |AM| = sqrt(4^(2)+1^(2)+4^(2)) = sqrt(33)`


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