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JEE Mains · Maths · STD 11 - 11. introduction to three dimensional geometry

The square of the distance of the point \((-2, -8, 6)\) from the line \(\dfrac{x-1}{1} = \dfrac{y-1}{2} = \dfrac{z}{-1}\) along the line \(\dfrac{x+5}{1} = \dfrac{y+5}{-1} = \dfrac{z}{2}\) is equal to:

  1. A \(3\)
  2. B \(6\)
  3. C \(8\)
  4. D \(12\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(6\)

Step-by-step Solution

Detailed explanation

Let the given point be \(P(-2, -8, 6)\). The distance is measured along the line \(\dfrac{x+5}{1} = \dfrac{y+5}{-1} = \dfrac{z}{2}\), which has direction ratios \((1, -1, 2)\). The equation of the line passing through \(P\) and parallel to this line is:…
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