Question

Let the point \( A \) be the foot of perpendicular drawn from the point \( P(a, b, 0) \) on the line} \[ \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - \alpha}{3}. \] If the midpoint of the line segment \( PA \) is \( \left( \frac{3}{4}, \frac{4}{3}, -\frac{1}{4} \right) \), then the value of \( a^2 + b^2 + \alpha^2 \) is:

• A. 21
• B. 76
• C. 62
• D. 9

Answer 1

The Correct Option is A

The point \( A \) lies on the given line. The parametric equations of the line are: \[ x = 2t + 1, \quad y = t + 2, \quad z = 3t + \alpha. \] Let the coordinates of point \( A \) be \( (2t + 1, t + 2, 3t + \alpha) \). The midpoint of the line segment \( PA \) is the average of the coordinates of \( P(a, b, 0) \) and \( A(2t + 1, t + 2, 3t + \alpha) \), which is given by: \[ \left( \frac{a + (2t + 1)}{2}, \frac{b + (t + 2)}{2}, \frac{0 + (3t + \alpha)}{2} \right) = \left( \frac{3}{4}, \frac{4}{3}, -\frac{1}{4} \right). \] Equating the components: \[ \frac{a + (2t + 1)}{2} = \frac{3}{4}, \quad \frac{b + (t + 2)}{2} = \frac{4}{3}, \quad \frac{3t + \alpha}{2} = -\frac{1}{4}. \] Solving these equations for \( a, b, \alpha \), we get: \[ a = 1, \quad b = 2, \quad \alpha = -3. \] Now, calculate \( a^2 + b^2 + \alpha^2 \): \[ a^2 + b^2 + \alpha^2 = 1^2 + 2^2 + (-3)^2 = 1 + 4 + 9 = 21. \]
Final Answer: 21