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Last updated at Feb. 1, 2020 by Teachoo

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Example 13 Find the vector equation of the plane which is at a distance of 6/โ29 from the origin and its normal vector from the origin is 2๐ ฬ โ 3๐ ฬ + 4๐ ฬ.Vector equation of a plane at a distance โdโ from the origin and unit vector to normal from origin ๐ ฬ is ๐ โ.๐ ฬ = d Unit vector of ๐ โ = ๐ ฬ = 1/|๐ โ | (๐ โ) Now, distance from origin = d = 6/โ29 ๐ โ = 2๐ ฬ โ 3๐ ฬ + 4๐ ฬ Magnitude of ๐ ฬ = โ(22+(โ3)^2+4^2 ) |๐ โ | = โ(4+9+16) = โ29 Now, ๐ ฬ = 1/|๐ โ | " (" ๐ โ")" = 1/โ29 " (2" ๐ ฬโ"3" ๐ ฬ+"4" ๐ ฬ")" = 2/โ29 ๐ ฬ โ 3/โ29 ๐ ฬ + 4/โ29 ๐ ฬ Vector equation of plane is ๐ โ.๐ ฬ = d ๐ โ . (๐/โ๐๐ ๐ ฬโ๐/โ๐๐ " " ๐ ฬ+ ๐/โ๐๐ " " ๐ ฬ ) = ๐/โ๐๐ Cartesian equation Equation of a plane in Cartesian form which is at a distance โdโ from the origin and has a normal vector ๐ โ = ๐๐ ฬ + b๐ ฬ + c๐ ฬ is lx + my + nz = d where l, m, n are direction cosines of ๐ โ l = ๐/โ(๐^2 + ๐^2 + ๐^2 ) , m = ๐/โ(๐^2 + ๐^2 + ๐^2 ) , n = ๐/โ(๐^2 + ๐^2 + ๐^2 ) Distance form origin = d = 6/โ29 ๐ โ = 2๐ ฬ โ 3๐ ฬ + 4๐ ฬ Comparing with ๐ โ = ๐๐ ฬ + ๐๐ ฬ + ๐๐ ฬ, ๐ = 2, b = โ3, c = 4 Also, โ(๐2+๐2+๐2) = โ(22+(โ3)2+42) = โ(4+9+16) = โ29 So, direction cosines are l = 2/โ29 , m = ( โ3)/โ29 , n = 4/โ29 โด Cartesian equations of plane is lx + my + nz = d 2/โ29 x + (( โ3)/โ29)y + 4/โ29 z = 6/โ29 2x โ 3y + 4z = 6 Therefore, the equation of plane in Cartesian form is 2x โ 3y + 4z = 6

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Example, 9 Not in Syllabus - CBSE Exams 2021

Example 10 Not in Syllabus - CBSE Exams 2021

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Example 13 Important You are here

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Example 22 Not in Syllabus - CBSE Exams 2021

Example 23 Important Not in Syllabus - CBSE Exams 2021

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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 9 years. He provides courses for Maths and Science at Teachoo.