Find the angle between the planes x + 2y + z = 7 and 2x – y + z = 13.

  1. \(\theta = {\cos ^{ - 1}}\left( {\frac{1}{6}} \right)\)
  2. \(\theta = {\cos ^{ - 1}}\left( {\frac{1}{3}} \right)\)
  3. \(\theta = {\cos ^{ - 1}}\left( {\frac{2}{3}} \right)\)
  4. \(\theta = {\cos ^{ - 1}}\left( {\frac{3}{4}} \right)\)

Answer (Detailed Solution Below)

Option 1 : \(\theta = {\cos ^{ - 1}}\left( {\frac{1}{6}} \right)\)
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NDA 01/2025: English Subject Test
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Detailed Solution

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CONCEPT:

  • Let A1x + B1y + C1z + D1 = 0 and A2x + B2y + C2 z + D2 = 0 are the equations of two planes aligned at an angle θ where A1, B1, Cand A2, B2, C2 are the direction ratios of the normal to the planes, then the cosine of the angle between the two planes is given by:

\(cos\theta = \left| {\frac{{{A_1}{A_2} + {B_1}{B_2} + {C_1}{C_2}}}{{\sqrt {A_1^2 + B_1^2 + C_1^2} \sqrt {A_2^2 + B_2^2 + C_2^2} }}} \right|\)

 

CALCULATION:

Given planes are x + 2y + z = 7 and 2x – y + z = 13.

\(cos\theta = \left| {\frac{{1.2 - 2.1 + 1.1}}{{\sqrt {{1^2} + {2^2} + {1^2}} \sqrt {{2^2} + {{\left( { - 1} \right)}^2} + {1^2}} }}} \right| = \frac{1}{6}\)

∴ \(\theta = {\cos ^{ - 1}}\left( {\frac{1}{6}} \right)\)
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