1. Analyze the Condition for Motion

The particle starts moving parallel to the vector $\vec{v} = \hat{i} – \hat{j}$. According to Newton’s Second Law, the initial acceleration—and thus the net force $\vec{F}_{net}$—must be in the direction of motion.

Let the net force be $\vec{R}$. We can express it as a scalar multiple of the direction vector: $$ \vec{R} = \lambda (\hat{i} – \hat{j}) $$ where $\lambda$ is a positive constant.

2. Minimization Logic

The net force is the vector sum of the individual forces: $$ \vec{R} = \vec{F}_1 + \vec{F}_2 $$ Rearranging for the unknown force: $$ \vec{F}_2 = \vec{R} – \vec{F}_1 $$ Geometrically, $\vec{F}_2$ is the vector connecting the tip of $\vec{F}_1$ to a point on the line defined by vector $\vec{v}$. To minimize the magnitude $|\vec{F}_2|$, this vector must be perpendicular to the direction of motion $\vec{v}$.

Therefore, the dot product of $\vec{F}_2$ and $\vec{v}$ must be zero: $$ \vec{F}_2 \cdot (\hat{i} – \hat{j}) = 0 $$

3. Calculation

Let the net force be $\vec{R} = a(\hat{i} – \hat{j})$. Then: $$ \vec{F}_2 = a(\hat{i} – \hat{j}) – \vec{F}_1 $$ Given $\vec{F}_1 = 3\hat{i} – \hat{j} – \hat{k}$, we substitute: $$ \vec{F}_2 = (a\hat{i} – a\hat{j}) – (3\hat{i} – \hat{j} – \hat{k}) $$ $$ \vec{F}_2 = (a-3)\hat{i} + (-a+1)\hat{j} + \hat{k} $$

Apply the perpendicularity condition $\vec{F}_2 \cdot (\hat{i} – \hat{j}) = 0$: $$ [(a-3)\hat{i} + (1-a)\hat{j} + \hat{k}] \cdot [\hat{i} – \hat{j}] = 0 $$ $$ (a-3)(1) + (1-a)(-1) + (1)(0) = 0 $$ $$ a – 3 – 1 + a = 0 $$ $$ 2a – 4 = 0 \implies a = 2 $$

Now, substitute $a=2$ back into the expression for $\vec{F}_2$: $$ \vec{F}_2 = (2-3)\hat{i} + (1-2)\hat{j} + \hat{k} $$ $$ \vec{F}_2 = -\hat{i} – \hat{j} + \hat{k} $$