# Example 28

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 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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Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .