The vectors `vec(AB)=3hati+4hatk and vec(AC)=5hati-2hatj+4hatk` are the sides of a triangle ABC. The length of the median through A is (A) `sqrt(72)` (B) `sqrt(33)` (C) `sqrt(2880` (D) `sqrt(18)`

Options

A. `sqrt(18)`
B. `sqrt(72)`
C. `sqrt(33)`
D. `sqrt(45)`

Answer

C

Explanation

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)`