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MHT CET · Maths · Three Dimensional Geometry

If the point (\(1, \alpha, \beta\)) lies on the line of the shortest distance between the lines \(\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}\) and \(\frac{x+2}{-1}=\frac{y+6}{2}, z=1\), then \(\alpha+\beta=\)

  1. A \(1\)
  2. B \(-3\)
  3. C \(7\)
  4. D \(-7\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(1\)

Step-by-step Solution

Detailed explanation

Let the lines be \(L_1: \frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2} = \lambda\) and \(L_2: \frac{x+2}{-1}=\frac{y+6}{2}, z=1 = \mu\). General points on \(L_1\) and \(L_2\) are \(P_1(-3\lambda-2, 4\lambda+2, 2\lambda+5)\) and \(P_2(-\mu-2, 2\mu-6, 1)\).