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

સમતલ \(a x+b y+c z+8=0\) એ બિંદુ \((-1,0,-2)\) માંથી પસાર થાય છે અને આપેલ સમતલો \(2 x+y-\) \(z=2\) અને  \(x-y-z=3\) ને લંબ હોય તો \(a+b+c\) ની કિમંત મેળવો.

  1. A \(8\)
  2. B \(4\)
  3. C \(3\)
  4. D \(5\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(4\)

Step-by-step Solution

Detailed explanation

Normal of req. plane \((2 \hat{i}+\hat{j}-\hat{k}) \times(\hat{i}-\hat{j}-\hat{k})\) \(=-2 \hat{i}+\hat{j}-3 \hat{k}\) Equation of plane \(-2(x+1)+1(y-0)-3(z+2)=0\) \(-2 x+y-3 z-8=0\) \(2 x-y+3 z+8=0\) \(a+b+c=4\)
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