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

ધારોકે \(a, b \in R.\) જો  રેખા \(\frac{x-3}{7}=\frac{y-2}{5}=\frac{z-1}{-9}\) ની સાપેક્ષે બિંદુ \(P( a, 6,9)\)નું પ્રતિબિંબ \((20, b,-a-9)\) હોય તો \(|a+b| = \, .......\) 

  1. A \(88\)
  2. B \(86\)
  3. C \(84\)
  4. D \(90\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(88\)

Step-by-step Solution

Detailed explanation

\(P (9,6,9)\) \(\frac{ x -3}{7}=\frac{ y -2}{5}=\frac{ z -1}{-9}\) \(Q =(20, b ,- a -9)\) \(\frac{\frac{20+a}{2}-3}{7}=\frac{\frac{b+6}{2}-2}{5}=\frac{-\frac{9}{2}-1}{-9}\) \(\frac{14+9}{14}=\frac{b+2}{10}=\frac{a+2}{18}\) \(\Rightarrow a=-56\) and \(b=-32\)…
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