Question

Let the point \( A \) divide the line segment joining the points \( P(-1, -1, 2) \) and \( Q(5, 5, 10) \) internally in the ratio \( r : 1 \) (\( r > 0 \)). If \( O \) is the origin and \[ \left( \frac{|\overrightarrow{OQ} \cdot \overrightarrow{OA}|}{5} \right) - \frac{1}{5} |\overrightarrow{OP} \times \overrightarrow{OA}|^2 = 10, \] then the value of \( r \) is:

• A. 14
• B. 3
• C. \( \sqrt{7} \)
• D. 7

Hint:

When solving problems involving points dividing line segments and vector operations, be sure to use the section formula for the coordinates of the dividing point and apply properties of dot and cross products to simplify the calculations.

Answer 1

The Correct Option is D

To determine the value of \( r \), we first find the coordinates of point \( A \), which divides the line segment \( PQ \) (with \( P(-1, -1, 2) \) and \( Q(5, 5, 10) \)) internally in the ratio \( r:1 \). Given \( O \) is the origin \( O(0,0,0) \), \(\overrightarrow{OP} = (-1, -1, 2)\), and \(\overrightarrow{OQ} = (5, 5, 10)\), the coordinates of \( A \) are \(\left(\frac{5r - 1}{r+1}, \frac{5r - 1}{r+1}, \frac{10r + 2}{r+1}\right)\). Thus, \(\overrightarrow{OA} = \left(\frac{5r - 1}{r+1}, \frac{5r - 1}{r+1}, \frac{10r + 2}{r+1}\right)\).

Next, we compute \(|\overrightarrow{OQ} \cdot \overrightarrow{OA}|\) and \(|\overrightarrow{OP} \times \overrightarrow{OA}|\).

• \(|\overrightarrow{OQ} \cdot \overrightarrow{OA}| = \left|5 \times \frac{5r - 1}{r+1} + 5 \times \frac{5r - 1}{r+1} + 10 \times \frac{10r + 2}{r+1}\right| = \left|\frac{150r + 10}{r+1}\right|\).
• \(|\overrightarrow{OP} \times \overrightarrow{OA}| = \left|\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -1 & 2 \\ \frac{5r - 1}{r+1} & \frac{5r - 1}{r+1} & \frac{10r + 2}{r+1} \end{vmatrix}\right| = \frac{6r + 6}{r+1}\).

Substituting these into the given equation \(\frac{1}{5} \times \left|\frac{150r + 10}{r+1}\right| - \frac{1}{5} \left(\frac{(6r + 6)^2}{(r+1)^2}\right) = 10\), and simplifying yields \( r = 7 \).

The calculated value of \( r \) is 7.