Last updated at May 29, 2018 by Teachoo

Transcript

Example 13 Find the vector equation of the plane which is at a distance of 6 29 from the origin and its normal vector from the origin is 2 𝑖 − 3 𝑗 + 4 𝑘. Vector equation of a plane at a distance ‘d’ from the origin and unit vector to normal from origin 𝑛 is 𝒓. 𝒏 = d Unit vector of 𝑛 = 𝑛 = 1 𝑛 ( 𝑛) Now, distance from origin = d = 6 29 𝑛 = 2 𝑖 − 3 𝑗 + 4 𝑘 Magnitude of 𝑛 = 22+ −32+ 42 𝑛 = 4+9+16 = 29 Now, 𝑛 = 1 𝑛 ( 𝑛) = 1 29 (2 𝑖−3 𝑗+4 𝑘) = 2 29 𝑖 − 3 29 𝑗 + 4 29 𝑘 Vector equation is 𝑟. 𝑛 = d 𝑟 . 2 29 𝑖− 3 29 𝑗+ 4 29 𝑘 = 6 29 Therefore, vector equation of the plane is 𝒓 . 𝟐 𝟐𝟗 𝒊− 𝟑 𝟐𝟗 𝒋+ 𝟒 𝟐𝟗 𝒌 = 𝟔 𝟐𝟗 Cartesian equation Equation of a plane in Cartesian form which is at a distance ‘d’ from the origin and has a normal vector 𝑛 = 𝑎 𝑖 + b 𝑗 + c 𝑘 is lx + my + nz = d where l, m, n are direction cosines of 𝑛 l = 𝑎 𝑎2 + 𝑏2 + 𝑐2 , m = 𝑏 𝑎2 + 𝑏2 + 𝑐2 , n = 𝑐 𝑎2 + 𝑏2 + 𝑐2 Distance form origin = d = 6 29 𝑛 = 2 𝑖 − 3 𝑗 + 4 𝑘 Comparing with 𝑛 = 𝑎 𝑖 + 𝑏 𝑗 + 𝑐 𝑘, 𝑎 = 2, b = −3, c = 4 Also, 𝑎2+𝑏2+𝑐2 = 22+(− 3)2+42 = 4+9+16 = 29 So, direction cosines are l = 2 29 , m = − 3 29 , n = 4 29 ∴ Cartesian equations of plane is lx + my + nz = d 2 29 x + − 3 29y + 4 29 z = 6 29 2x − 3y + 4z = 6 Therefore, the equation of plane in Cartesian form is 2x − 3y + 4z = 6

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Chapter 11 Class 12 Three Dimensional Geometry

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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.