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

मान लीजिए कि बिंदु A, बिंदु \(P(a, b, 0)\) से रेखा \(\dfrac{x-1}{2} = \dfrac{y-2}{1} = \dfrac{z-\alpha}{3}\) पर डाले गए लंब का पाद है। यदि रेखाखंड PA का मध्य-बिंदु \(\left(0, \dfrac{3}{4}, \dfrac{-1}{4}\right)\) है, तो \(a^2 + b^2 + \alpha^2\) का मान बराबर है:

  1. A \(1\)
  2. B \(2\)
  3. C \(6\)
  4. D \(9\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(1\)

Step-by-step Solution

Detailed explanation

Let \(P = (a, b, 0)\) and the midpoint of \(PA\) be \(M = \left(0, \dfrac{3}{4}, -\dfrac{1}{4}\right)\). Using the midpoint formula, the coordinates of \(A\) are given by \(2M - P\): \(A = \left(-a, \dfrac{3}{2} - b, -\dfrac{1}{2}\right)\) Since \(A\) lies on the given line…
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