the hour of a clock is 6cm long.the magnitude of the displacement of the tip of hour hand between 1:00 pm and 5:00 pm is : options are A. 6cm B. 6sqrt3 C.12 cm D.3 sqrt3cm please explain me

Dear Naba

The angle traced by hour hand in 12 hours =360 (as the hour hand is back to its original position after 12 hours)

thus Angle traced in one hor =360/12=30

Angle traced by hour hand in 4 hours(between 1 and 5 pm) =4* 30=120

So the vector of length 6cm is displaced by 120° in 4 hrs

Since for any two vectors u and v with angle θ between them

|v-u|² = |v|² + |u|²  -2uvcosθ

Here v and u have mod = 6 cm and angle =120°

so |v-u|²=36+36-72*-1/2=36*3

so magnitude of displacement = |v-u| =6√3

thank you very much...............................but i had a query why did we used (v-u)² why not (v+u)² formula........