1. **Problem statement:** We have a rectangular box with dimensions 9 cm (length), 6 cm (width), and 6 cm (height). The lid is open at an angle of 30°. We need to calculate: (a) The angle between line BD' and the plane BCC'B'. (b) The angle between lines AC' and BD'. (c) The angle between line BP and the plane ABCD. 2. **Understanding the box and notation:** - Let the base rectangle be ABCD with AB = 9 cm and BC = 6 cm. - The top rectangle (lid) is A'B'C'D' parallel to ABCD, with height 6 cm. - The lid is open at 30°, so the plane A'B'C'D' is tilted by 30° from the base plane ABCD. 3. **(a) Angle between BD' and plane BCC'B':** - Plane BCC'B' is vertical along the width side BC. - Vector BD' connects B (base) to D' (top corner of the lid). - To find the angle between BD' and the plane BCC'B', we find the angle between BD' and the normal vector to the plane. 4. **Coordinates assignment:** - Let B = (0,0,0), C = (0,6,0), D = (9,6,0), A = (9,0,0) - Since the lid is open at 30°, point D' is rotated about edge BC by 30°. - The rotation axis is BC from (0,6,0) to (0,6,6). - Coordinates of D' after rotation: - Original D' if lid closed: (9,6,6) - After rotation by 30° about BC axis: - Translate so BC axis at origin: D'' = D' - C = (9,0,6) - Rotate D'' by 30° around y-axis (since BC is along y): $$x' = x\cos30° + z\sin30° = 9 \times \frac{\sqrt{3}}{2} + 6 \times \frac{1}{2} = 7.794 + 3 = 10.794$$ $$y' = y = 0$$ $$z' = -x\sin30° + z\cos30° = -9 \times \frac{1}{2} + 6 \times \frac{\sqrt{3}}{2} = -4.5 + 5.196 = 0.696$$ - Translate back: D' = (10.794, 6, 0.696) 5. **Vector BD':** $$\vec{BD'} = D' - B = (10.794, 6, 0.696) - (0,0,0) = (10.794, 6, 0.696)$$ 6. **Plane BCC'B':** - Points: B(0,0,0), C(0,6,0), C'(0,6,6), B'(0,0,6) - This plane is the plane x=0 (yz-plane). - Normal vector to plane BCC'B' is along x-axis: $$\vec{n} = (1,0,0)$$ 7. **Angle between BD' and plane BCC'B':** - Angle between vector and plane is $$\theta = 90° - \alpha$$ where $$\alpha$$ is angle between vector and normal. - Compute $$\cos \alpha = \frac{\vec{BD'} \cdot \vec{n}}{|\vec{BD'}||\vec{n}|} = \frac{10.794}{\sqrt{10.794^2 + 6^2 + 0.696^2} \times 1}$$ - Calculate magnitude: $$|\vec{BD'}| = \sqrt{10.794^2 + 6^2 + 0.696^2} = \sqrt{116.53 + 36 + 0.484} = \sqrt{153.014} = 12.37$$ - So, $$\cos \alpha = \frac{10.794}{12.37} = 0.8727$$ - $$\alpha = \cos^{-1}(0.8727) = 29.1°$$ - Therefore, $$\theta = 90° - 29.1° = 60.9°$$ 8. **(b) Angle between AC' and BD':** - Coordinates: - A = (9,0,0) - C' = (0,6,6) - Vector AC': $$\vec{AC'} = C' - A = (0-9, 6-0, 6-0) = (-9, 6, 6)$$ - Vector BD' from step 5: (10.794, 6, 0.696) 9. **Calculate angle between AC' and BD':** $$\cos \phi = \frac{\vec{AC'} \cdot \vec{BD'}}{|\vec{AC'}||\vec{BD'}|}$$ - Dot product: $$(-9)(10.794) + 6(6) + 6(0.696) = -97.146 + 36 + 4.176 = -56.97$$ - Magnitude of AC': $$\sqrt{(-9)^2 + 6^2 + 6^2} = \sqrt{81 + 36 + 36} = \sqrt{153} = 12.37$$ - Magnitude of BD' from step 7: 12.37 - So, $$\cos \phi = \frac{-56.97}{12.37 \times 12.37} = \frac{-56.97}{153} = -0.372$$ - $$\phi = \cos^{-1}(-0.372) = 111.9°$$ - Angle between lines is the smaller angle, so $$180° - 111.9° = 68.1°$$ 10. **(c) Angle between BP and plane ABCD:** - Point P is not defined in the problem, so assuming P is the midpoint of the lid edge or a point on the lid. - Without P's coordinates, cannot calculate. **Final answers:** - (a) Angle between BD' and plane BCC'B' is approximately **60.9°**. - (b) Angle between AC' and BD' is approximately **68.1°**. - (c) Insufficient data to calculate angle between BP and plane ABCD.