கோடு \(\frac{{x - 2}}{1} = \frac{{y + 3}}{{ - \;2}} = \frac{{z + 4}}{- \ 3}\) க்கும் விமானம் 2x - 3y + z - 5 = 0 க்கும் இடையே உள்ள கோணத்தைக் கண்டறியவும்

  1. \({\sin ^{ - 1}}\left( {\frac{11}{{14}}} \right)\)
  2. \({\sin ^{ - 1}}\left( {\frac{13}{{14}}} \right)\)
  3. \({\sin ^{ - 1}}\left( {\frac{3}{{14}}} \right)\)
  4. \({\sin ^{ - 1}}\left( {\frac{5}{{14}}} \right)\)

Answer (Detailed Solution Below)

Option 4 : \({\sin ^{ - 1}}\left( {\frac{5}{{14}}} \right)\)
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கருத்து:

θ என்பது கோடு \(\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\) க்கும் விமானம் a2 x + b2 y + c2 z + d = 0 க்கும் இடையே உள்ள கோணம் என்றால்

\(\sin \theta = \frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\left( {\sqrt {a_1^2 + b_1^2 + c_1^2} } \right)\left( {\sqrt {a_2^2 + b_2^2 + c_2^2} } \right)}}\)

கணக்கீடு:

கொடுக்கப்பட்டவை: கோட்டின் சமன்பாடு \(\frac{{x - 2}}{1} = \frac{{y + 3}}{{ - \;2}} = \frac{{z + 4}}{- \ 3}\) மற்றும் விமானத்தின் சமன்பாடு 2x - 3y + z - 5 = 0.

கோடு \(\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\) மற்றும் விமானம் a2 x + b2 y + c2 z + d = 0 ஆகியவற்றுக்கு இடையே உள்ள கோணம் என்பது நமக்குத் தெரியும்: \(\sin \theta = \frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\left( {\sqrt {a_1^2 + b_1^2 + c_1^2} } \right)\left( {\sqrt {a_2^2 + b_2^2 + c_2^2} } \right)}}\)

இங்கே, a1 = 1, b1 = - 2, c1 = - 3, a2 = 2, b2 = - 3 மற்றும் c2 = 1.

⇒ a1 ⋅ a2 + b1 ⋅ b2 + c1 ⋅ c2 = 2 + 6 - 3 = 5

\(⇒ \sqrt {a_1^2 + b_1^2 + c_1^2} = \sqrt {14} \;and\;\sqrt {a_2^2 + b_2^2 + c_2^2} = \sqrt {14} \)

\(\Rightarrow \sin \theta = \frac{5}{{14}}\)

\(\Rightarrow \theta = {\sin ^{ - 1}}\left( {\frac{5}{{14}}} \right)\)

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