# Example 13

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 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

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 .