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JEE Mains · Maths · STD 12 - 11. three dimension geometry

Let \(P(\alpha,\beta,\gamma)\) be the point on the line \(\frac{x-1}{2}=\frac{y+1}{-3}=z\) at a distance \(4\sqrt{14}\) from the point (1, -1, 0) and nearer to the origin. Then the shortest distance, between the lines \(\frac{x-\alpha}{1}=\frac{y-\beta}{2}=\frac{z-\gamma}{3}\) and \(\frac{x+5}{2}=\frac{y-10}{1}=\frac{z-3}{1}\), is equal to

  1. A \(7\sqrt{\frac{5}{4}}\)
  2. B \(4\sqrt{\frac{7}{5}}\)
  3. C \(4\sqrt{\frac{5}{7}}\)
  4. D \(2\sqrt{\frac{7}{4}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(4\sqrt{\frac{7}{5}}\)

Step-by-step Solution

Detailed explanation

Let \(P (2 \lambda+1,-3 \lambda-1, \lambda)\) Then \(4 \lambda^2+9 \lambda^2+\lambda^2=16 \cdot 14 \Rightarrow \lambda= \pm 4 \Rightarrow-4\) (nearer to origin) \(\therefore P (-7,11,-4)\) ∴ Shortest distance…