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

The square of the distance of the point \(P(5, 6, 7)\) from the line \(\dfrac{x-2}{2} = \dfrac{y-5}{3} = \dfrac{z-2}{4}\) is equal to:

  1. A \(3\)
  2. B \(5\)
  3. C \(6\)
  4. D \(8\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(6\)

Step-by-step Solution

Detailed explanation

Let the foot of the perpendicular from \(P(5, 6, 7)\) to the given line be \(Q\). Any point on the line \(\dfrac{x-2}{2} = \dfrac{y-5}{3} = \dfrac{z-2}{4} = \lambda\) is \(Q(2\lambda + 2, 3\lambda + 5, 4\lambda + 2)\). The direction ratios of \(PQ\) are…
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