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

समतल \(x + y + z =5\) तथा \(\quad\) समतलों \(3 x +4 x + z -1=0\) और \(5 x +8 y +2 z +14=0\) की प्रतिच्छेदन रेखा के बीच का एक कोण है

  1. A \({\cos ^{ - 1}}\left( {\frac{3}{{\sqrt {17} }}} \right)\)
  2. B \({\cos ^{ - 1}}\left( {\sqrt {\frac{3}{{17}}} } \right)\)
  3. C \({\sin ^{ - 1}}\left( {\frac{3}{{\sqrt {17} }}} \right)\)
  4. D \({\sin ^{ - 1}}\left( {\sqrt {\frac{3}{{17}}} } \right)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \({\sin ^{ - 1}}\left( {\sqrt {\frac{3}{{17}}} } \right)\)

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Detailed explanation

\(3 x+4 y+z-1=0\) and \(5 x+8 y+2 z+14=0\) The direction of line of intersection of plans is qiven by \(\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 1 \\ 5 & 8 & 2\end{array}\right|\) \(=\hat{i}(8-8)-\hat{j}(6-5)+\hat{k}(24-20)\) \(=-\hat{j}+4 \hat{k}\)…
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