Example, 25 - Chapter 11 Class 12 Three Dimensional Geometry
Last updated at March 11, 2017 by Teachoo
Transcript
Example, 25 Find the angle between the line 𝑥 + 12 = 𝑦3 = 𝑧 − 36 And the plane 10x + 2y – 11 z = 3. The angle between a line 𝑥 − 𝑥1𝑎 = 𝑦 − 𝑦1𝑏 = 𝑧 − 𝑧1𝑐 and the normal to the plane Ax + By + Cz = D is given by cos θ = 𝐴𝑎 + 𝐵𝑏 + 𝐶𝑐 𝑎2 + 𝑏2 + 𝑐2 𝐴2 + 𝐵2 + 𝐶2 So, angle between line and plane is given by sin 𝝓 = 𝐴𝑎 + 𝐵𝑏 + 𝐶𝑐 𝑎2 + 𝑏2+ 𝑐2+ 𝐴2+ 𝐵2 + 𝐶2 Given, the line is 𝑥 + 12 = 𝑦3 = 𝑧 − 36 𝑥 − (−1)2 = 𝑦 − 03 = 𝑧 − 36 Comparing with 𝑥 − 𝑥1𝑎 = 𝑦 − 𝑦1𝑏 = 𝑧 − 𝑧1𝑐 , 𝑎 = 2, b = 3, c = 6 The plane is 10x + 2y − 11z = 3 Comparing with Ax + By + Cz = D, A = 10, B = 2, C = −11 So, sin ϕ = 10 × 2 + 2 × 3 + (−11 × 6) 22 + 32 + 62 102 + 22 + (−11)2 = 20 + 6 − 66 4 + 9 + 36 100 + 4 + 121 = −407 × 15 = 821 So, sin ϕ = 821 ∴ 𝝓 = 𝒔𝒊𝒏−𝟏 𝟖𝟐𝟏 Therefore, the angle between the given line and plane is sin−1 821.
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Chapter 11 Class 12 Three Dimensional Geometry
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.