1. **Problem Statement:** We have a rectangular prism with vertices D, A, B, C, E, F, G, H, and a point J on edge EF dividing it in ratio 2:1. Given dimensions: DH = 55 mm, HG = 90 mm, FG = 40 mm. Find: (a) Length HJ (b) Length DJ (c) Angle HDJ (d) Angle between DJ and plane EFGH 2. **Set up coordinate system:** Place point H at origin $(0,0,0)$. Since HG = 90 mm along x-axis, G is at $(90,0,0)$. Since FG = 40 mm along z-axis, F is at $(90,0,40)$. Since DH = 55 mm along y-axis, D is at $(0,55,0)$. E is vertically above F along y-axis, so E is at $(90,55,40)$. 3. **Find coordinates of J:** J divides EF in ratio 2:1, so $J = E + \frac{2}{3}(F - E)$. Calculate: $$F - E = (90,0,40) - (90,55,40) = (0,-55,0)$$ $$J = (90,55,40) + \frac{2}{3}(0,-55,0) = (90,55 - \frac{110}{3},40) = (90,\frac{55}{3},40)$$ 4. **Calculate length HJ:** $$HJ = \sqrt{(90-0)^2 + \left(\frac{55}{3} - 0\right)^2 + (40-0)^2}$$ $$= \sqrt{90^2 + \left(\frac{55}{3}\right)^2 + 40^2} = \sqrt{8100 + \frac{3025}{9} + 1600}$$ $$= \sqrt{8100 + 336.11 + 1600} = \sqrt{10036.11} \approx 100.18\text{ mm}$$ 5. **Calculate length DJ:** Coordinates of D: $(0,55,0)$ $$DJ = \sqrt{(90-0)^2 + \left(\frac{55}{3} - 55\right)^2 + (40-0)^2}$$ $$= \sqrt{90^2 + \left(-\frac{110}{3}\right)^2 + 40^2} = \sqrt{8100 + \frac{12100}{9} + 1600}$$ $$= \sqrt{8100 + 1344.44 + 1600} = \sqrt{11044.44} \approx 105.08\text{ mm}$$ 6. **Calculate angle HDJ:** Vectors: $$\overrightarrow{DH} = H - D = (0-0,0-55,0-0) = (0,-55,0)$$ $$\overrightarrow{DJ} = J - D = (90-0, \frac{55}{3} - 55, 40-0) = (90, -\frac{110}{3}, 40)$$ Dot product: $$\overrightarrow{DH} \cdot \overrightarrow{DJ} = 0 \times 90 + (-55) \times \left(-\frac{110}{3}\right) + 0 \times 40 = \frac{6050}{3}$$ Magnitudes: $$|\overrightarrow{DH}| = 55$$ $$|\overrightarrow{DJ}| = 105.08$$ Angle: $$\cos \theta = \frac{\overrightarrow{DH} \cdot \overrightarrow{DJ}}{|\overrightarrow{DH}||\overrightarrow{DJ}|} = \frac{\frac{6050}{3}}{55 \times 105.08} = \frac{2016.67}{5779.4} \approx 0.349$$ $$\theta = \cos^{-1}(0.349) \approx 69.56^\circ$$ 7. **Calculate angle between DJ and plane EFGH:** Plane EFGH is parallel to xz-plane at y=constant. Normal vector to plane EFGH is along y-axis: $\vec{n} = (0,1,0)$. Vector $\overrightarrow{DJ} = (90, -\frac{110}{3}, 40)$. Angle $\phi$ between $\overrightarrow{DJ}$ and plane is complementary to angle between $\overrightarrow{DJ}$ and normal: $$\cos \alpha = \frac{|\overrightarrow{DJ} \cdot \vec{n}|}{|\overrightarrow{DJ}||\vec{n}|} = \frac{|-\frac{110}{3}|}{105.08 \times 1} = \frac{36.67}{105.08} \approx 0.349$$ $$\alpha = \cos^{-1}(0.349) = 69.56^\circ$$ Angle between $\overrightarrow{DJ}$ and plane: $$\phi = 90^\circ - 69.56^\circ = 20.44^\circ$$ **Final answers:** (a) $HJ \approx 100.18$ mm (b) $DJ \approx 105.08$ mm (c) $\angle HDJ \approx 69.56^\circ$ (d) Angle between $DJ$ and plane $EFGH \approx 20.44^\circ$