Physics - Center of Mass Question with Solution | TestHub

Position of COM of a mass - system is given as, ${\vec{r}}_{\mathrm{com}}=\frac{m_1{\vec{r}}_1+m_2{\vec{r}}_2}{m_1+m_2}=\frac{1(\hat{i}+2\hat{j}+\hat{k})+3(-3\hat{i}-2\hat{j}+\hat{k}}{1+3}$ $=-2\hat{i}-\hat{j}+\hat{k}$ $\left|{\vec{r}}_{cm}\right|=|-2\hat{i}-\hat{j}+\hat{k}|=\sqrt{(2{)}^2+(1{)}^2+(1{)}^2}=\sqrt{6}$ Only option (1) magnitude is $\sqrt{1^2+(-2{)}^2+1^2}=\sqrt{6}$ So option (1) is correct.