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MHT CET · Maths · Three Dimensional Geometry

The distance of the point \((2,4,0)\) from the point of intersection of the lines \(\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}\) and \(\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}\) is

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

Answer & Solution

Correct Answer

(A) 3 units

Step-by-step Solution

Detailed explanation

Line 1: \(\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}=k_1 \implies x=3k_1-6, y=2k_1, z=k_1-1\) Line 2: \(\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}=k_2 \implies x=4k_2+7, y=3k_2+9, z=2k_2+4\)