To determine the motion of particles A and B, we extract their initial positions and velocities from the slopes and intercepts of the provided $x-t$, $y-t$, and $z-t$ graphs.

For Particle A (Blue):

• $x$-axis: Starts at $x=9$ with slope $v_{xA} = \frac{4-9}{5} = -1$ m/s.
• $y$-axis: Starts at $y=3$ with slope $v_{yA} = \frac{13-3}{5} = 2$ m/s.
• $z$-axis: Starts at $z=7$ with slope $v_{zA} = \frac{1-7}{3} = -2$ m/s.

Position vector: $\vec{r}_A = (9 – t)\hat{i} + (3 + 2t)\hat{j} + (7 – 2t)\hat{k}$

Velocity vector: $\vec{v}_A = -\hat{i} + 2\hat{j} – 2\hat{k}$

For Particle B (Red):

• $x$-axis: Starts at $x=12$ with slope $v_{xB} = \frac{0-12}{3} = -4$ m/s.
• $y$-axis: Starts at $y=7$ with slope $v_{yB} = \frac{0-7}{3.5} = -2$ m/s.
• $z$-axis: Starts at $z=7$ with slope $v_{zB} = \frac{16-7}{3} = 3$ m/s.

Position vector: $\vec{r}_B = (12 – 4t)\hat{i} + (7 – 2t)\hat{j} + (7 + 3t)\hat{k}$

Velocity vector: $\vec{v}_B = -4\hat{i} – 2\hat{j} + 3\hat{k}$

We analyze the motion of particle B with respect to particle A. The relative separation vector $\vec{r}_{rel}$ is:

$$ \vec{r}_{rel} = \vec{r}_B – \vec{r}_A $$ $$ \vec{r}_{rel} = [(12 – 4t) – (9 – t)]\hat{i} + [(7 – 2t) – (3 + 2t)]\hat{j} + [(7 + 3t) – (7 – 2t)]\hat{k} $$ $$ \vec{r}_{rel} = (3 – 3t)\hat{i} + (4 – 4t)\hat{j} + (5t)\hat{k} $$

The relative velocity $\vec{v}_{rel}$ is constant:

$$ \vec{v}_{rel} = \vec{v}_B – \vec{v}_A = (-4 – (-1))\hat{i} + (-2 – 2)\hat{j} + (3 – (-2))\hat{k} $$ $$ \vec{v}_{rel} = -3\hat{i} – 4\hat{j} + 5\hat{k} $$

The particles are closest when the relative position vector is perpendicular to the relative velocity vector ($\vec{r}_{rel} \perp \vec{v}_{rel}$), or by minimizing the square of the distance $|\vec{r}_{rel}|^2$.

Using the condition $\vec{r}_{rel} \cdot \vec{v}_{rel} = 0$:

$$ [(3 – 3t)\hat{i} + (4 – 4t)\hat{j} + (5t)\hat{k}] \cdot [-3\hat{i} – 4\hat{j} + 5\hat{k}] = 0 $$ $$ -3(3 – 3t) – 4(4 – 4t) + 5(5t) = 0 $$ $$ -9 + 9t – 16 + 16t + 25t = 0 $$ $$ 50t – 25 = 0 $$ $$ 50t = 25 \implies t = 0.5 \text{ s} $$

Substitute $t = 0.5$ s back into the relative position vector equation to find the minimum distance:

$$ \vec{r}_{min} = (3 – 3(0.5))\hat{i} + (4 – 4(0.5))\hat{j} + (5(0.5))\hat{k} $$ $$ \vec{r}_{min} = 1.5\hat{i} + 2\hat{j} + 2.5\hat{k} $$

The magnitude is:

$$ |\vec{r}_{min}| = \sqrt{(1.5)^2 + (2)^2 + (2.5)^2} $$ $$ |\vec{r}_{min}| = \sqrt{2.25 + 4 + 6.25} $$ $$ |\vec{r}_{min}| = \sqrt{12.5} = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}} $$ $$ |\vec{r}_{min}| = 2.5\sqrt{2} \text{ m} $$