Question

Two point charges $q_{1}$ and $q_{2}$ are placed at $(0,0,0)$ and $(1,2,2) m$ respectively. They repel each other with force of $3 \,N$. The force on $q_{2}$ due to $q_{1}$ is $F_{21}=(x \hat{i}+y \hat{j}+z \hat{k}) N$. The value of $x+y+z$ is ___

Answer 1

$F_{21}=F_{21} \hat{e}=F_{21} \frac{A B}{|A B|}=F_{21} \frac{(\hat{i}+2 \hat{j}+2 \hat{k})}{\sqrt{1+4+4}}$
$=\frac{3}{3}(\hat{i}+2 \hat{j}+2 \hat{k})=\hat{i}+2 \hat{j}+2 \hat{k}$
$\therefore x=1, y=2, z=2$
$\therefore x+y+z=5$