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

ધારો કે \((\alpha, \beta, \gamma)\) એ બિંદૂ \((8,5,7)\) નું રેખા \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{5}\) પરનું પ્રતિબિંબ છે. તો \(\alpha+\beta+\gamma\) \(=\) ...........

  1. A \(16\)
  2. B \(18\)
  3. C \(14\)
  4. D \(20\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(14\)

Step-by-step Solution

Detailed explanation

\( \overrightarrow{\mathrm{AM}} \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})=0 \) \( (2 \lambda-7)(2)+(3 \lambda-6)(3)+(5 \lambda-5)(5)=0 \) \( 38 \lambda=57 \) \( \lambda=\frac{3}{2} \) \( \mathrm{M}\left(4, \frac{7}{2}, \frac{19}{2}\right) \)…
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