Ex 11.3, 14 (d) - Find distance of (-6, 0, 0) from plane 2x-3y+6z-2=0

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

Go Ad-free

Transcript

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 (−6, 0, 0) So, 𝑥_1 = −6, 𝑦_1 = 0, 𝑧_1 = 0 and the equation of plane is 2x − 3y + 6z − 2 = 0 2x − 3y + 6z = 2 Comparing with Ax + By + Cz = D, A = 2, B = −3, C = 6 D = 3 Now, Distance of the point from the plane = |((2 × −6) + (−3 × 0) + (6 × 0)− 2 )/√(2^2+(−3)^2+6^2 )| = |(−12 + 0 + 0 − 2)/√(4 + 9 + 36)| = |(−14)/√49| = |(−14)/7| = |−2| = 2

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo