Chapter 11 Class 12 Three Dimensional Geometry
Ex 11.1, 2
Example, 6 Important
Example, 7
Example 10 Important
Ex 11.2, 5 Important
Ex 11.2, 9 (i) Important
Ex 11.2, 10 Important
Ex 11.2, 12 Important
Ex 11.2, 13 Important
Ex 11.2, 15 Important
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Question 11 Important Deleted for CBSE Board 2024 Exams
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Question 15 Important Deleted for CBSE Board 2024 Exams
Question 4 (a) Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 12 Important Deleted for CBSE Board 2024 Exams
Question 14 (a) Important Deleted for CBSE Board 2024 Exams
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Question 20 Important Deleted for CBSE Board 2024 Exams
Misc 3 Important
Misc 4 Important
Question 10 Important Deleted for CBSE Board 2024 Exams
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Misc 5 Important
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Chapter 11 Class 12 Three Dimensional Geometry
Last updated at April 16, 2024 by Teachoo
Question 13 Find the angle between the two planes 3x – 6y + 2z = 7 and 2x + 2y – 2z =5.Angle between two planes A1x + B1y + C1z = d1 and A2x + B2y + C2z = d2 is given by cos θ = |(𝑨_𝟏 𝑨_𝟐 + 𝑩_𝟏 𝑩_𝟐 + 𝑪_𝟏 𝑪_𝟐)/(√(〖𝑨_𝟏〗^𝟐 + 〖𝑩_𝟏〗^𝟐 + 〖𝑪_𝟏〗^𝟐 ) √(〖𝑨_𝟐〗^𝟐 + 〖𝑩_𝟐〗^𝟐 + 〖𝑪_𝟐〗^𝟐 ))| Given the two planes are 3x − 6y + 2z = 7 Comparing with A1x + B1y + C1z = d1 A1 = 3 , B1 = –6 , C1 = 2 , 𝑑_1= 7 2x + 2y − 2z = 5 Comparing with A2x + B2y + C2z = d2 A2 = 2 , B2 = 2 , C2 = –2 , 𝑑_2= 5 So, cos θ = |((3 × 2) + (−6 × 2) + (2 × −2))/(√(3^2 + 〖(−6)〗^2 + 2^2 ) √(2^2 + 2^2 + 〖(−2)〗^2 ))| = |(6 + (−12) + (−4))/(√(9 + 36 + 4) ×√(4 + 4 + 4))| = |(−10)/(√(49 ) ×√12)| = |(−10)/(7 ×√(4×3))| = 10/(7 × 2 × √3) = 5/(7√3) = 5/(7√3) × √3/√3 = (5√3)/21 So, cos θ = (5√3)/21 ∴ θ = 〖𝒄𝒐𝒔〗^(−𝟏) ((𝟓√𝟑)/𝟐𝟏) Therefore, the angle between the two planes is 〖𝑐𝑜𝑠〗^(−1) ((5√3)/21) E