1. **Problem Statement:** We have points A and B with position vectors \(\overrightarrow{OA} = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\) and \(\overrightarrow{OB} = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}\). Point P lies on segment AB such that \(AP : PB = \lambda : 1 - \lambda\). We are given \(\overrightarrow{OP} = (1 + \lambda) \mathbf{i} + (2 - 5\lambda) \mathbf{j} + (-2 + 8\lambda) \mathbf{k}\). We need to find: (i) The value of \(\lambda\) for which \(\overrightarrow{OP}\) is perpendicular to \(\overrightarrow{AB}\). (ii) The value of \(\lambda\) for which angles \(AOP\) and \(POB\) are equal. 2. **Step 1: Express \(\overrightarrow{AB}\)** \[ \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (2 - 1)\mathbf{i} + (-3 - 2)\mathbf{j} + (6 + 2)\mathbf{k} = \mathbf{i} - 5\mathbf{j} + 8\mathbf{k} \] 3. **Step 2: Confirm \(\overrightarrow{OP}\) formula** Since \(P\) divides \(AB\) in ratio \(\lambda : 1 - \lambda\), the position vector of \(P\) is \[ \overrightarrow{OP} = \overrightarrow{OA} + \lambda \overrightarrow{AB} = (\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) + \lambda (\mathbf{i} - 5\mathbf{j} + 8\mathbf{k}) = (1 + \lambda)\mathbf{i} + (2 - 5\lambda)\mathbf{j} + (-2 + 8\lambda)\mathbf{k} \] This matches the given expression. 4. **Part (i): Find \(\lambda\) such that \(\overrightarrow{OP} \perp \overrightarrow{AB}\)** Two vectors are perpendicular if their dot product is zero: \[ \overrightarrow{OP} \cdot \overrightarrow{AB} = 0 \] Calculate the dot product: \[ ((1 + \lambda), (2 - 5\lambda), (-2 + 8\lambda)) \cdot (1, -5, 8) = (1 + \lambda)(1) + (2 - 5\lambda)(-5) + (-2 + 8\lambda)(8) \] Simplify: \[ (1 + \lambda) - 5(2 - 5\lambda) + 8(-2 + 8\lambda) = 1 + \lambda - 10 + 25\lambda - 16 + 64\lambda = (1 - 10 - 16) + (\lambda + 25\lambda + 64\lambda) = -25 + 90\lambda \] Set equal to zero: \[ -25 + 90\lambda = 0 \implies 90\lambda = 25 \implies \lambda = \frac{25}{90} = \frac{5}{18} \] 5. **Part (ii): Find \(\lambda\) such that angles \(AOP\) and \(POB\) are equal** The angles \(AOP\) and \(POB\) are angles between vectors: - \(AOP\): between \(\overrightarrow{OA}\) and \(\overrightarrow{OP}\) - \(POB\): between \(\overrightarrow{OP}\) and \(\overrightarrow{OB}\) The cosine of the angle between vectors \(\mathbf{u}\) and \(\mathbf{v}\) is: \[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} \] Set \(\theta_1 = \angle AOP\) and \(\theta_2 = \angle POB\). We want \(\theta_1 = \theta_2\), so: \[ \cos \theta_1 = \cos \theta_2 \] Calculate: \[ \cos \theta_1 = \frac{\overrightarrow{OA} \cdot \overrightarrow{OP}}{|\overrightarrow{OA}| |\overrightarrow{OP}|}, \quad \cos \theta_2 = \frac{\overrightarrow{OP} \cdot \overrightarrow{OB}}{|\overrightarrow{OP}| |\overrightarrow{OB}|} \] Equate and multiply both sides by \(|\overrightarrow{OP}||\overrightarrow{OA}||\overrightarrow{OB}| \[ (\overrightarrow{OA} \cdot \overrightarrow{OP}) |\overrightarrow{OB}| = (\overrightarrow{OP} \cdot \overrightarrow{OB}) |\overrightarrow{OA}| \] Calculate magnitudes: \[ |\overrightarrow{OA}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = 3 \] \[ |\overrightarrow{OB}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = 7 \] Calculate dot products: \[ \overrightarrow{OA} \cdot \overrightarrow{OP} = (1)(1+\lambda) + (2)(2 - 5\lambda) + (-2)(-2 + 8\lambda) = (1 + \lambda) + (4 - 10\lambda) + (4 - 16\lambda) = 9 - 25\lambda \] \[ \overrightarrow{OP} \cdot \overrightarrow{OB} = (1 + \lambda)(2) + (2 - 5\lambda)(-3) + (-2 + 8\lambda)(6) = 2 + 2\lambda - 6 + 15\lambda - 12 + 48\lambda = -16 + 65\lambda \] Set equation: \[ (9 - 25\lambda) \times 7 = (-16 + 65\lambda) \times 3 \] Simplify: \[ 63 - 175\lambda = -48 + 195\lambda \] Bring terms together: \[ 63 + 48 = 195\lambda + 175\lambda \implies 111 = 370\lambda \implies \lambda = \frac{111}{370} \] **Final answers:** (i) \(\lambda = \frac{5}{18}\) (ii) \(\lambda = \frac{111}{370}\)