JEE Mains · Maths · STD 12 - 11. three dimension geometry
An angle between the plane, \(x + y + z = 5\) and the line of intersection of the planes, \(3x + 4y + z- 1 = 0\) and \(5x + 8y + 2z+ 14 = 0\) , is
- A \({\cos ^{ - 1}}\left( {\frac{3}{{\sqrt {17} }}} \right)\)
- B \({\cos ^{ - 1}}\left( {\sqrt {\frac{3}{{17}}} } \right)\)
- C \({\sin ^{ - 1}}\left( {\frac{3}{{\sqrt {17} }}} \right)\)
- D \({\sin ^{ - 1}}\left( {\sqrt {\frac{3}{{17}}} } \right)\)
Answer & Solution
Correct Answer
(D) \({\sin ^{ - 1}}\left( {\sqrt {\frac{3}{{17}}} } \right)\)
Step-by-step Solution
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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