Example 23 - Find angle between 3x - 6y + 2z = 7 and 2x + 2y

Example 23 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Example 23 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3

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Transcript

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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.