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

यदि बिंन्दु \((1,1,2)\) से होकर जाने वाला तथा रेखा \(x-3 y+2 z-1=04 x-y+z\) के लंबवत समतल का समीकरण \(\mathrm{Ax}+\mathrm{By}+\mathrm{Cz}=1\) है तो \(140(\mathrm{C}-\mathrm{B}+\mathrm{A})\) बराबर है __________________. 

  1. A \(14\)
  2. B \(13\)
  3. C \(12\)
  4. D \(15\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(15\)

Step-by-step Solution

Detailed explanation

\(x-3 y+2 z-1=0\) \(4 x-y+z=0\) \(\overrightarrow{ n }_1 \times \overrightarrow{ n }_2=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & -3 & 2 \\ 4 & -1 & 1\end{array}\right|\) \(=-\hat{ i }+7 \hat{ j }+11 \hat{ k }\) \(\operatorname{Dr}^5\) of normal to the…
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