Question:

**Step 1:** Let $O$ be the point on the ground. Let the higher plane be at point $A$ at a height of $\sqrt{3}$ km, and the lower plane be at point $B$ at a height of $h$ km. Since one plane passes vertically above the other, both planes are on the same vertical line. Let $C$ be the point on the ground directly below both planes, such that $OC$ is the horizontal distance from $O$ to the vertical line containing $A$ and $B$. **Step 2:** For plane $A$, the angle of elevation from $O$ is $60^\circ$. Using the tangent function: $$ \tan(60^\circ) = \frac{\text{height of plane A}}{OC} $$ $$ \sqrt{3} = \frac{\sqrt{3}}{OC} $$ Solving for $OC$: $$ OC = 1 \text{ km} $$ **Step 3:** For plane $B$, the angle of elevation from $O$ is $30^\circ$. Using the tangent function: $$ \tan(30^\circ) = \frac{\text{height of plane B}}{OC} $$ Substituting the value of $OC$ and $\tan(30^\circ)$: $$ \frac{1}{\sqrt{3}} = \frac{h}{1} $$ Solving for $h$: $$ h = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \text{ km} $$ **Step 4:** The distance between the two planes at that instant is the difference in their heights, as they are vertically aligned. $$ \text{Distance} = \text{height of plane A} - \text{height of plane B} $$ $$ \text{Distance} = \sqrt{3} - \frac{\sqrt{3}}{3} $$ $$ \text{Distance} = \frac{3\sqrt{3} - \sqrt{3}}{3} $$ $$ \text{Distance} = \frac{2\sqrt{3}}{3} \text{ km} $$