Question

Three forces $\overrightarrow{F_1}=(3\hat{i}+2\hat{j}-\hat{k})N,\overrightarrow{F_2}=(3\hat{i}+4\hat{j}-5\hat{k})N$ and $\overrightarrow{F_3}=A(\hat{i}+\hat{j}-\hat{k})N$ act simultaneously on a particle. In order that the particle remains in equilibrium, the value of A should be

• A. -6
• B. 6
• C. 9
• D. -9

Answer 1

The correct option is A -6

For particle to remain in equilibrium,
$\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}=0$
$\Rightarrow 3\hat{i}+2\hat{j}-\hat{k}+3\hat{i}+4\hat{j}-5\hat{k}+A\hat{i}+A\hat{j}-A\hat{k}=\overrightarrow{0}$
$\Rightarrow \hat{i}(6+A)+\hat{j}(6+A)+\hat{k}(-6-A)=0\hat{i}+0\hat{j}+0\hat{k}$
Comparing the coefficient of $\hat{i}$ in L.H.S and R.H.S, we get
$\Rightarrow 6+A=0$
$\Rightarrow A=-6$