The problem consists of a pyramid $P-ABCD$, where the base $ABCD$ is a square with side length 3, and point $P$ is located such that $PD \perp \text{plane } ABCD$. You are tasked with solving two parts:

1. Prove that $BC \perp \text{plane } CDP$.
2. Given that the angle between line $AD$ and line $BP$ is $60^\circ$, find the length of $DP$.

Part 1: Proof that $BC \perp \text{plane } CDP$

Since $PD$ is perpendicular to the plane $ABCD$, and since $BC$ is a side of the square base, we know that:

• $BC \parallel \text{plane } ABCD$,
• $PD \perp ABCD$.

Because $BC$ lies in the plane $ABCD$ and $PD \perp ABCD$, line $BC$ must be perpendicular to any line passing through $P$ and extending outside the plane, like $DP$. Therefore, $BC \perp \text{plane } CDP$, as desired.

Part 2: Finding the length of $DP$

To find the length of $DP$, we'll use the fact that the angle between lines $AD$ and $BP$ is $60^\circ$. Here are the steps:

1. Coordinates setup: Since the base $ABCD$ is a square with side length 3, we can place the points on a coordinate plane:

   • $A(0, 0, 0)$,
   • $B(3, 0, 0)$,
   • $C(3, 3, 0)$,
   • $D(0, 3, 0)$.

   Let $P(0, 3, h)$, where $h$ is the height we need to determine for $P$, and hence for $DP$.

2. Using the angle information: We know that the angle $\angle(AD, BP) = 60^\circ$. This allows us to use the dot product formula between vectors $\overrightarrow{AD}$ and $\overrightarrow{BP}$:

   $\cos 60^\circ = \frac{\overrightarrow{AD} \cdot \overrightarrow{BP}}{|\overrightarrow{AD}| |\overrightarrow{BP}|}$

   First, compute the vectors:

   • $\overrightarrow{AD} = (0 - 0, 3 - 0, 0 - 0) = (0, 3, 0)$,
   • $\overrightarrow{BP} = (0 - 3, 3 - 0, h - 0) = (-3, 3, h)$.

   Next, apply the dot product: $\overrightarrow{AD} \cdot \overrightarrow{BP} = (0, 3, 0) \cdot (-3, 3, h) = 0 \cdot (-3) + 3 \cdot 3 + 0 \cdot h = 9$

   Using the magnitudes:

   • $|\overrightarrow{AD}| = \sqrt{0^2 + 3^2 + 0^2} = 3$,
   • $|\overrightarrow{BP}| = \sqrt{(-3)^2 + 3^2 + h^2} = \sqrt{9 + 9 + h^2} = \sqrt{18 + h^2}$.

   Now, substitute into the dot product formula: $\frac{9}{3 \times \sqrt{18 + h^2}} = \cos 60^\circ = \frac{1}{2}$

   Simplifying: $\frac{9}{3\sqrt{18 + h^2}} = \frac{1}{2} \quad \Rightarrow \quad 18 + h^2 = 36 \quad \Rightarrow \quad h^2 = 18 \quad \Rightarrow \quad h = 3\sqrt{2}$

3. Finding $DP$: Now that we have $h = 3\sqrt{2}$, we can find the length of $DP$: $DP = \sqrt{(0 - 0)^2 + (3 - 3)^2 + (3\sqrt{2} - 0)^2} = \sqrt{0 + 0 + (3\sqrt{2})^2} = \sqrt{18} = 3\sqrt{2}.$

Thus, the length of $DP$ is $3\sqrt{2}$.

Would you like further details or have any questions?

Here are 5 related questions for deeper understanding:

1. What would change in the proof if the base $ABCD$ were a rectangle instead of a square?
2. How do you find the perpendicular distance from a point to a plane in 3D geometry?
3. What is the significance of using vector dot products to find angles between lines?
4. How can you verify if two planes are perpendicular in space?
5. What are the conditions for a line to be perpendicular to a plane?

Tip: In geometry problems involving 3D shapes, always start by placing points on a coordinate system and use vector analysis for clarity and accuracy.