Question 14 (a) - Plane - Chapter 11 Class 12 Three Dimensional Geometry

Last updated at April 16, 2024 by

Ex 11.3, 14 (a) - Find distance of (0, 0, 0) from plane 3x-4y+12z=3

Ex 11.3, 14 (a) - Chapter 11 Class 12 Three Dimensional Geometry - Part 2

 

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Question 14 In the following cases, find the distance of each of the given points from the corresponding given plane. The distance of the point (x1, y1, z1) from the plane Ax + By + Cz = D is |(𝑨𝒙_𝟏 + 〖𝑩𝒚〗_𝟏 +〖 𝑪𝒛〗_𝟏 − 𝑫)/√(𝑨^𝟐 + 𝑩^𝟐 + 𝑪^𝟐 )| Given, the point is (0, 0, 0) So, 𝑥_1 = 0, 𝑦_1 = 0, 𝑧_1 = 0 and the equation of plane is 3x − 4y + 12z = 3 Comparing with Ax + By + Cz = D, A = 3, B = −4, C = 12, D = 3 Now, Distance of point from the plane is = |((3 × 0) + (−4 × 0) + (12 × 0) − 3)/( √(3^2 + (−4)^2 + 〖12〗^2 ))| = |(0 + 0 + 0 − 3)/( √(9 + 16 + 144))| = |3/( √169)| = 𝟑/𝟏𝟑

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