Angle between planes MCQ Quiz in मराठी - Objective Question with Answer for Angle between planes - मोफत PDF डाउनलोड करा

Last updated on Mar 29, 2025

पाईये Angle between planes उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). हे मोफत डाउनलोड करा Angle between planes एमसीक्यू क्विझ पीडीएफ आणि बँकिंग, एसएससी, रेल्वे, यूपीएससी, स्टेट पीएससी यासारख्या तुमच्या आगामी परीक्षांची तयारी करा.

Latest Angle between planes MCQ Objective Questions

Angle between planes Question 1:

रेषा \(\frac{{x - 2}}{1} = \frac{{y + 3}}{{ - \;2}} = \frac{{z + 4}}{- \ 3}\) आणि समतल 2x - 3y + z - 5 = 0 मधील कोन शोधा ?

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

Angle between planes Question 1 Detailed Solution

संकल्पना:

जर θ हा रेषा  \(\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\) आणि प्रतलातील a2 x + b2 y + c2 z + d = 0 असेल तर

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

गणना:

दिले आहे: रेषीय समीकरण आहे \(\frac{{x - 2}}{1} = \frac{{y + 3}}{{ - \;2}} = \frac{{z + 4}}{- \ 3}\) आणि प्रतलाचे समीकरण आहे 2x - 3y + z - 5 = 0.

आपल्याला माहित आहे की, रेषा \(\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\) आणि प्रतल a2 x + b2 y + c2 z + d = 0 द्वारे दिला आहे:

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

येथे, a1 = 1, b1 = - 2, c1 = - 3, a2 = 2, b2 = - 3 आणिc2 = 1.

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

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

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

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

Top Angle between planes MCQ Objective Questions

रेषा \(\frac{{x - 2}}{1} = \frac{{y + 3}}{{ - \;2}} = \frac{{z + 4}}{- \ 3}\) आणि समतल 2x - 3y + z - 5 = 0 मधील कोन शोधा ?

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

Angle between planes Question 2 Detailed Solution

Download Solution PDF

संकल्पना:

जर θ हा रेषा  \(\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\) आणि प्रतलातील a2 x + b2 y + c2 z + d = 0 असेल तर

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

गणना:

दिले आहे: रेषीय समीकरण आहे \(\frac{{x - 2}}{1} = \frac{{y + 3}}{{ - \;2}} = \frac{{z + 4}}{- \ 3}\) आणि प्रतलाचे समीकरण आहे 2x - 3y + z - 5 = 0.

आपल्याला माहित आहे की, रेषा \(\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\) आणि प्रतल a2 x + b2 y + c2 z + d = 0 द्वारे दिला आहे:

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

येथे, a1 = 1, b1 = - 2, c1 = - 3, a2 = 2, b2 = - 3 आणिc2 = 1.

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

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

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

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

Angle between planes Question 3:

रेषा \(\frac{{x - 2}}{1} = \frac{{y + 3}}{{ - \;2}} = \frac{{z + 4}}{- \ 3}\) आणि समतल 2x - 3y + z - 5 = 0 मधील कोन शोधा ?

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

Angle between planes Question 3 Detailed Solution

संकल्पना:

जर θ हा रेषा  \(\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\) आणि प्रतलातील a2 x + b2 y + c2 z + d = 0 असेल तर

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

गणना:

दिले आहे: रेषीय समीकरण आहे \(\frac{{x - 2}}{1} = \frac{{y + 3}}{{ - \;2}} = \frac{{z + 4}}{- \ 3}\) आणि प्रतलाचे समीकरण आहे 2x - 3y + z - 5 = 0.

आपल्याला माहित आहे की, रेषा \(\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\) आणि प्रतल a2 x + b2 y + c2 z + d = 0 द्वारे दिला आहे:

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

येथे, a1 = 1, b1 = - 2, c1 = - 3, a2 = 2, b2 = - 3 आणिc2 = 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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