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

बिन्दु \((4,-1,2)\) से होकर जाने वाला समतल जो रेखाओं \(\frac{x+2}{3}=\frac{y-2}{-1}=\frac{z+1}{2}\) तथा \(\frac{x-2}{1}=\frac{y-3}{2}=\frac{z-4}{3}\) के समांतर है, निम्न में से जिस बिन्दु से भी होकर जाता है, वह है

  1. A \((1, 1, -1)\)
  2. B \((1, 1, 1)\)
  3. C \((-1,-1,-1)\)
  4. D \((-1,-1,1)\)
Verified Solution

Answer & Solution

Correct Answer

(B) \((1, 1, 1)\)

Step-by-step Solution

Detailed explanation

\(\vec n = {\vec n_1} \times {\vec n_2} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ 3&{ - 1}&2\\ 1&2&3 \end{array}} \right| = 7\hat i - 7\hat j + 7\hat k\) Equation of plane is \(-7(x-4)-7(y+1)+7(z-2)=0\) \(\Rightarrow-7 x-7 y+7 z+7=0 \Rightarrow x+y-z=1\)
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