Question

If the image of the point $ P(1, 0, 3) $ in the line joining the points $ A(4, 7, 1) $ and $ B(3, 5, 3) $ is $ Q(\alpha, \beta, \gamma) $, then $ \alpha + \beta + \gamma $ is equal to:

• A. 47/3
• B. 46/3
• C. 18
• D. 13

Hint:

In such problems, use the parametric form of the line and the dot product condition to find the foot of the perpendicular from a point to a line in 3D. Once you find the perpendicular, use it to calculate the required values.

Answer 1

The Correct Option is B

Given: \( P(1, 0, 3) \), \( A(4, 7, 1) \), \( B(3, 5, 3) \) Line \( AB \) is represented as: \[ \text{Line AB} \Rightarrow \frac{x - 3}{1} = \frac{y - 5}{2} = \frac{z - 3}{-2} = \lambda \] Let the foot of the perpendicular from \( P \) on \( AB \) be \( R \). \[ R = (\lambda + 3, 2\lambda + 5, -2\lambda + 3) \] Equating the components, we get: \[ (\lambda + 3 - 1)(1) + (2\lambda + 5 - 0)(2) + (-2\lambda + 3 - 3)(2) = 0 \] \[ \Rightarrow \lambda + 2 + 4\lambda + 10 + 4\lambda = 0 \] \[ \Rightarrow \lambda = -\frac{4}{3} \] Substitute \( \lambda = -\frac{4}{3} \) into the equations for the coordinates of \( R \): \[ R = \left( \frac{5}{3}, \frac{7}{3}, \frac{17}{3} \right) \] Now calculate the coordinates of \( Q \): \[ Q = \left( \frac{10}{3}, \frac{14}{3}, \frac{34}{3} \right) \] Now calculate \( \alpha + \beta + \gamma \): \[ \alpha + \beta + \gamma = \frac{7 + 14 + 25}{3} = \frac{46}{3} \] Hence, the answer is \( \frac{46}{3} \).