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JEE Mains · Maths · STD 12 - 11. three dimension geometry

ધારો કે \((-1,2,3)\) માંથી પસાર થતી એક રેખા, રેખાઓ \(L_1: \frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{-2}\) ને \(M(\alpha, \beta, \gamma)\) આગળ અને \(L_2: \frac{x+2}{-3}=\frac{y-2}{-2}=\frac{z-1}{4}\) ને \(N(a, b, c)\) આગળ છેદ છે. તો \(\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}\) નું મૂલ્ય ........... છે.

  1. A \(100\)
  2. B \(196\)
  3. C \(150\)
  4. D \(190\)
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Answer & Solution

Correct Answer

(B) \(196\)

Step-by-step Solution

Detailed explanation

\(\mathrm{M}(3 \lambda+1,2 \lambda+2,-2 \lambda-1) \quad \therefore \alpha+\beta+\gamma=3 \lambda+2 \) \(\mathrm{~N}(-3 \mu-2,-2 \mu+2,4 \mu+1) \quad \therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=-\mu+1\)…
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