We are given that the points \( P \) and \( Q \) trisect the line segment \( AB \), and the position vectors of \( A \) and \( B \) are: \[ \mathbf{A} = 2\hat{i} - 5\hat{j} + 3\hat{k}, \quad \mathbf{B} = 4\hat{i} + \hat{j} - 6\hat{k}. \] We are asked to find the position vector of the point dividing \( PQ \) in the ratio 2:3.
Step 1: Use the section formula to find the position vector of \( P \). Since \( P \) divides \( AB \) in the ratio 1:2, the position vector of \( P \) is given by: \[ \mathbf{P} = \frac{2\mathbf{A} + 1\mathbf{B}}{3}. \] Substitute the values of \( \mathbf{A} \) and \( \mathbf{B} \): \[ \mathbf{P} = \frac{2(2\hat{i} - 5\hat{j} + 3\hat{k}) + 1(4\hat{i} + \hat{j} - 6\hat{k})}{3}. \] Simplifying: \[ \mathbf{P} = \frac{(4\hat{i} - 10\hat{j} + 6\hat{k}) + (4\hat{i} + \hat{j} - 6\hat{k})}{3} = \frac{8\hat{i} - 9\hat{j}}{3}. \] Thus, the position vector of \( P \) is: \[ \mathbf{P} = \frac{8}{3}\hat{i} - 3\hat{j}. \] Step 2: Use the section formula to find the position vector of \( Q \). Since \( Q \) divides \( AB \) in the ratio 2:1, the position vector of \( Q \) is given by: \[ \mathbf{Q} = \frac{1\mathbf{A} + 2\mathbf{B}}{3}. \] Substitute the values of \( \mathbf{A} \) and \( \mathbf{B} \): \[ \mathbf{Q} = \frac{1(2\hat{i} - 5\hat{j} + 3\hat{k}) + 2(4\hat{i} + \hat{j} - 6\hat{k})}{3}. \] Simplifying: \[ \mathbf{Q} = \frac{(2\hat{i} - 5\hat{j} + 3\hat{k}) + (8\hat{i} + 2\hat{j} - 12\hat{k})}{3} = \frac{10\hat{i} - 3\hat{j} - 9\hat{k}}{3}. \] Thus, the position vector of \( Q \) is: \[ \mathbf{Q} = \frac{10}{3}\hat{i} - \hat{j} - 3\hat{k}. \]
Step 3: Use the section formula to find the position vector of the point dividing \( PQ \) in the ratio 2:3. Let the position vector of the point dividing \( PQ \) in the ratio 2:3 be \( \mathbf{R} \). Using the section formula, the position vector of \( R \) is given by: \[ \mathbf{R} = \frac{3\mathbf{P} + 2\mathbf{Q}}{5}. \] Substitute the values of \( \mathbf{P} \) and \( \mathbf{Q} \): \[ \mathbf{R} = \frac{3\left( \frac{8}{3}\hat{i} - 3\hat{j} \right) + 2\left( \frac{10}{3}\hat{i} - \hat{j} - 3\hat{k} \right)}{5}. \] Simplifying: \[ \mathbf{R} = \frac{(8\hat{i} - 9\hat{j}) + \left( \frac{20}{3}\hat{i} - 2\hat{j} - 6\hat{k} \right)}{5}. \] \[ \mathbf{R} = \frac{8\hat{i} - 9\hat{j} + \frac{20}{3}\hat{i} - 2\hat{j} - 6\hat{k}}{5}. \] Combining the terms: \[ \mathbf{R} = \frac{\left( 8 + \frac{20}{3} \right)\hat{i} - 11\hat{j} - 6\hat{k}}{5} = \frac{\frac{24}{3} + \frac{20}{3}}{5}\hat{i} - \frac{11}{5}\hat{j} - \frac{6}{5}\hat{k}. \] \[ \mathbf{R} = \frac{44}{3}\hat{i} - \frac{33}{3}\hat{j} - 18\hat{k}. \] Thus, the position vector of \( R \) is: \[ \mathbf{R} = \frac{1}{15} \left( 44\hat{i} - 33\hat{j} - 18\hat{k} \right). \] Thus, the final answer is: \[ \boxed{\frac{1}{15} \left( 44\hat{i} - 33\hat{j} - 18\hat{k} \right)}. \]
If \( \sqrt{5} - i\sqrt{15} = r(\cos\theta + i\sin\theta), -\pi < \theta < \pi, \) then
\[ r^2(\sec\theta + 3\csc^2\theta) = \]
For \( n \in \mathbb{N} \), the largest positive integer that divides \( 81^n + 20n - 1 \) is \( k \). If \( S \) is the sum of all positive divisors of \( k \), then find \( S - k \).