Two particles of masses 1 kg and 3 kg have position vectors `2hati+3hatj+4hatk and-2hati+3hatj-4hatk` respectively. The centre of mass has a position vector

Options

A. `hati+3hatj-2hatk`
B. `-hati-3hatj-2hatk`
C. `-hati+3hatj+2hatk`
D. `-hati+3hatj-2hatk`

Answer

D

Explanation

Here, `m_(1)=1kg,` `m_(2)=3kg`
`vecr_(CM)= (m_(1)vecr_(1)+m_(2)vecr_(2))/(m_(1)+m_(2))`
`=((1)(2hati+3hatj+4hatk)+(3)(-2hati+3hatj-4hatk))/(1+3)`
`=(2hati+3hatj+4hatk-6hati+9hatj-12hatk)/(4)`
`=(-4hati+12hatj-8hatk)/(4)=-hati+3hatj-2hatk`