1. **Problem Statement:** You are given a roof with rectangular base OABC where OA = 14 m along unit vector i, OC = 8 m along unit vector j, and the top edge DE is 6 m long and 5 m above the base (vertical direction k). The sloping edges OD, CD, AE, and BE are equal in length. (i) Find the vector \(\overrightarrow{OD}\) in terms of \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) and its magnitude. (ii) Find the angle DOB using the scalar product. 2. **Step (i): Express \(\overrightarrow{OD}\) and find magnitude** - The base coordinates start at \(O = (0,0,0)\). - Since DE is 6 m length horizontally and the sloping edges are equal, point D must lie vertically above a point on the base such that DE is parallel and length 6 m. - Given the rectangle with O at origin, coordinates of A: \(14\mathbf{i}\), C: \(8\mathbf{j}\). - We place \(D = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\), with \(z = 5\) m vertical height. - The length of the sloping edge OD is equal to AE. Since OD and AE are equal, and considering symmetry, the horizontal position of D is \(x = 6\mathbf{i}\) (half of OA) and \(y = 0\) (on the j-axis), but we verify using more precise constraints. - Since DE = 6 m and DE is parallel to the horizontal projection, and D is 5 m above base, find \(\overrightarrow{OD} = 6\mathbf{i} + 0\mathbf{j} + 5\mathbf{k}\). - The magnitude: $$|\overrightarrow{OD}| = \sqrt{6^2 + 0^2 + 5^2} = \sqrt{36 + 25} = \sqrt{61}$$ 3. **Step (ii): Find angle DOB** - Vector \(\overrightarrow{OB} = 14\mathbf{i} + 8\mathbf{j} + 0\mathbf{k}\) because B is at the corner from O with lengths \(OA=14\) and \(OC=8\). - Vector \(\overrightarrow{OD} = 6\mathbf{i} + 0\mathbf{j} + 5\mathbf{k}\) as determined. - Scalar product: $$\overrightarrow{OD} \cdot \overrightarrow{OB} = (6)(14) + (0)(8) + (5)(0) = 84$$ - Magnitudes: $$|\overrightarrow{OD}| = \sqrt{61},\quad |\overrightarrow{OB}| = \sqrt{14^2 + 8^2} = \sqrt{196 + 64} = \sqrt{260}$$ - Use formula for angle \(\theta\) between vectors: $$\cos \theta = \frac{\overrightarrow{OD} \cdot \overrightarrow{OB}}{|\overrightarrow{OD}||\overrightarrow{OB}|} = \frac{84}{\sqrt{61} \times \sqrt{260}} = \frac{84}{\sqrt{15860}}$$ - Numerically: $$\sqrt{15860} \approx 125.95$$ $$\cos \theta \approx \frac{84}{125.95} = 0.6667$$ - Thus: $$\theta = \cos^{-1}(0.6667) \approx 48.19^\circ$$ **Final answers:** (i) \(\overrightarrow{OD} = 6\mathbf{i} + 0\mathbf{j} + 5\mathbf{k}\), magnitude \(\sqrt{61}\). (ii) Angle DOB is approximately \(48.2^\circ\).