Last updated at May 29, 2018 by Teachoo

Transcript

Example 28 Find the distance between the point P(6, 5, 9) and the plane determined by the points A (3, – 1, 2), B (5, 2, 4) and C(– 1, – 1, 6). The equation of a plane passing through points A( 𝑥1, 𝑦1, 𝑧1) B( 𝑥2, 𝑦2, 𝑧2) and C ( 𝑥3, 𝑦3, 𝑧3) is 𝒙− 𝒙𝟏𝒚− 𝒚𝟏𝒛− 𝒛𝟏 𝒙𝟐− 𝒙𝟏 𝒚𝟐− 𝒚𝟏 𝒛𝟐− 𝒛𝟏 𝒙𝟑− 𝒙𝟏 𝒚𝟑− 𝒚𝟏 𝒛𝟑− 𝒛𝟏 = 0 Given, the three points are Equation of plane is 𝑥−3𝑦−(−1)𝑧−25−32−(−1)4−2−1−3−1−(−1)6−2 = 0 𝑥−3𝑦+1𝑧−2232−404 = 0 (x − 3) 3×4−(0×2) − (y + 1) (2×4)− −4×2 + (z − 2) 2×0−(−4×3) (x − 3) 12−0 − (y + 1) 8+8 +(𝑧−2) 0+12 = 0 12(x − 3) – 16(y + 1) + 12(z − 2) = 0 3(x − 3) – 4(y + 1) + 3(z − 2) = 0 3x − 9 – 4y – 4 + 3z − 6 = 0 3x – 4y + 3z − 19 = 0 Therefore, equation of plane is 3x – 4y + 3z = 19 Now, the distance between a point P( 𝑥1, 𝑦1, 𝑧1) and the plane Ax + By + Cz = D is 𝑨 𝒙𝟏 + 𝑩 𝒚𝟏 + 𝑪 𝒛𝟏− 𝑫 𝑨𝟐 + 𝑩𝟐 + 𝑪𝟐 Given, the point is P(6, 5, 9) So, 𝑥1= 6 , 𝑦1= 5 𝑧1= 9 The equation of plane is 3x – 4y + 3z = 19 Comparing with Ax + By + Cz = D, A = 3, B = –4, C = 3 d = 19 Now, Distance of the point from the plane = 3 × 6 + −4 × 5 + 3 × 9 − 19 32 + 42 + 32 = 18 − 20 + 27 − 19 9 + 16 + 9 = 6 34 = 6 34 = 6 34 × 34 34 = 6 3434 = 𝟑 𝟑𝟒𝟏𝟕

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Example 28 You are here

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Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

About the Author

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.