Topic: Engineering Mechanics
$P=2\hat{i}-3\hat{j}$, $Q=-3\hat{i}+4\hat{j}-2\hat{k}$, and $\vec{R}$ are in equilibrium, if $\vec{R}$ is
- A. -\hat{i}-\hat{j}+2\hat{k}
- B. \hat{i}-\hat{j}+2\hat{k}
- C. \hat{i}+\hat{j}+2\hat{k}
- D. \hat{i}-\hat{j}-2\hat{k}
Answer: B. \hat{i}-\hat{j}+2\hat{k}
Explanation: For equilibrium of three vectors, $\vec{P}+\vec{Q}+\vec{R}=0$. Given $\vec{P}=2\hat{i}-3\hat{j}$ and $\vec{Q}=-3\hat{i}+4\hat{j}-2\hat{k}$. Let $\vec{R}=x\hat{i}+y\hat{j}+z\hat{k}$. Then $(2-3+x)\hat{i}+(-3+4+y)\hat{j}+(0-2+z)\hat{k}=0 \implies (x-1)\hat{i}+(y+1)\hat{j}+(z-2)\hat{k}=0$. So $x=1, y=-1, z=2$. Therefore, $\vec{R}=\hat{i}-\hat{j}+2\hat{k}$.