3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
If $10y - 8x - (x^2 + y^2 + z^2) = 40, P_1 = \max\left\{\sqrt{(x+2)^2 + (y-3)^2 + z^2}\right\}, P_2 = \min\left\{\sqrt{(x+2)^2 + (y-3)^2 + z^2}\right\}$, then $P_1 - P_2$ is __________.
Step-by-Step Solution
Key Concept: Rearrange the given condition into the standard form of a sphere equation to identify its centre and radius.
Given $\vec{r} \cdot \vec{i} + \vec{j} + \vec{k} = \vec{r}$, the condition $\vec{r} \cdot (10\vec{i} - 8\vec{j} - \vec{r}) = 41$ becomes $-\vec{r} \cdot (10\vec{i} - 8\vec{j}) + r^2 + 40 = 0$, which represents a sphere. The equation is a sphere with centre at $\frac{10\vec{i} - 8\vec{j}}{2}$ and radius 1, since $P_1 - P_2 = 2 \times 1 = 2$.
Correct Answer: 10