JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1}\) lie on the plane \(p x-q y+\) \(z=5\), for some \(p, q \in R\). The shortest distance of the plane from the origin is
- A \(\sqrt{\frac{3}{109}}\)
- B \(\sqrt{\frac{5}{142}}\)
- C \(\sqrt{\frac{5}{71}}\)
- D \(\sqrt{\frac{1}{142}}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\frac{5}{142}}\)
Step-by-step Solution
Detailed explanation
\((2,-1,-3)\) satisfy the given plane. So \(2 p+q=8\) \(\dots(i)\) Also given line is perpendicular to normal plane so \(3 p +2 q -1=0\) \(\dots(ii)\) \(\Rightarrow p =15, q =-22\) Eq. of plane \(15 x -22 y + z -5=0\) its distance from origin…
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