JEE Mains · Maths · STD 12 - 11. three dimension geometry
A plane \(P\) contains the line \(x+2 y+3 z+1=0=x-y-z-6\) and is perpendicular to the plane \(-2 x+y+z+8=0\). Then which of the following points lies on \(\mathrm{P}\) ?
- A \((-1,1,2)\)
- B \((0,1,1)\)
- C \((1,0,1)\)
- D \((2,-1,1)\)
Answer & Solution
Correct Answer
(B) \((0,1,1)\)
Step-by-step Solution
Detailed explanation
Equation of plane \(P\) can be assumed as \(P: x+2 y+3 z+1+\lambda(x-y-z-6)=0\) \(\Rightarrow P:(1+\lambda) x+(2-\lambda) y+(3-\lambda) z+1-6 \lambda=0\) \(\Rightarrow \vec{n}_{1}=(1+\lambda) \hat{i}+(2-\lambda) \hat{j}+(3-\lambda) \hat{k}\)…
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