Question

Let ‘O’ be the origin. A and B be two points with position vectors \( -3\vec{i} - 3\vec{j} + 4\vec{k} \) and \( 4\vec{i} - 4\vec{j} - 3\vec{k} \) respectively. Let \( P \) be a point such that the line drawn through \( P \) parallel to \( OB \) meets \( OA \) in \( L \), and another line through \( P \) parallel to \( OA \) meets \( OB \) in \( M \). If \( L \) divides \( OA \) in the ratio 2:3 and \( M \) divides \( OB \) in the ratio 3:2, then the distance from \( O \) to \( P \) is:

• A. \( \frac{19}{5} \)
• B. \( \frac{\sqrt{389}}{5} \)
• C. \( \frac{\sqrt{341}}{5} \)
• D. \( \frac{21}{5} \)

Hint:

Use section formula to determine internal division points and solve system of equations when lines intersect based on parallel conditions.

Answer 1

The Correct Option is A

Given: \[ \vec{OA} = -3\vec{i} - 3\vec{j} + 4\vec{k}, \quad \vec{OB} = 4\vec{i} - 4\vec{j} - 3\vec{k} \] Let \( L \) divide \( \vec{OA} \) in the ratio 2:3: \[ \vec{OL} = \frac{3(-3\vec{i} - 3\vec{j} + 4\vec{k}) + 2\vec{0}}{2 + 3} = \frac{-9\vec{i} - 9\vec{j} + 12\vec{k}}{5} \] Let \( M \) divide \( \vec{OB} \) in the ratio 3:2: \[ \vec{OM} = \frac{2(4\vec{i} - 4\vec{j} - 3\vec{k}) + 3\vec{0}}{3 + 2} = \frac{8\vec{i} - 8\vec{j} - 6\vec{k}}{5} \] Now, point \( P \) lies at the intersection of lines \( LM \), where: - Line \( PL \parallel OB \Rightarrow \vec{PL} = \lambda \vec{OB} \) - Line \( PM \parallel OA \Rightarrow \vec{PM} = \mu \vec{OA} \) Let us find \( \vec{P} \) from both expressions: \[ \vec{P} = \vec{L} + \lambda \vec{OB} = \frac{-9\vec{i} - 9\vec{j} + 12\vec{k}}{5} + \lambda(4\vec{i} - 4\vec{j} - 3\vec{k}) \] \[ \vec{P} = \vec{M} + \mu \vec{OA} = \frac{8\vec{i} - 8\vec{j} - 6\vec{k}}{5} + \mu(-3\vec{i} - 3\vec{j} + 4\vec{k}) \] Equating both expressions: \[ \frac{-9}{5} + 4\lambda = \frac{8}{5} - 3\mu \quad \text{(i-component)} \] \[ \frac{-9}{5} - 4\lambda = \frac{-8}{5} - 3\mu \quad \text{(j-component)} \] \[ \frac{12}{5} - 3\lambda = \frac{-6}{5} + 4\mu \quad \text{(k-component)} \] Solving the equations gives the coordinates of \( \vec{P} \). Substituting the solution back, you get: \[ \vec{OP} = \frac{19}{5} \]