Ex 11.3, 14 (b) - Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

Last updated at March 22, 2023 by Teachoo

Ex 11.3, 14 (b) - Find distance of (3, -2, 1) from plane 2x-y+2z+3=0

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

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Ex 11.3, 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 (3, −2, 1) So, 𝑥_1 = 3, 𝑦_1 = −2, 𝑧_1 = 1 a And the equation of the plane is 2x − y + 2z + 3 = 0 2x − y + 2z = −3 −(2x − y + 2z) = 3 −2x + y − 2z = 3 Comparing with Ax + By + Cz = D, A = −2, B = 1, C = −2, D = 3 Now, Distance of the point form the plane = |((−2 × 3) + (1 × −2) + (−2 × 1) − 3)/√((−2)^2 + 1^2 + (2)^2 )| = |((−6) + (−2) + (−2) − 3)/√(4 + 1 + 4)| = |(−13)/√9| = |(−13)/3| = 𝟏𝟑/𝟑

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