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:
- A \(3\)
- B \(5\)
- C \(6\)
- D \(8\)
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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