Example 13 - Find vector equation of plane which is at distance

Example 13 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Example 13 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Example 13 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4

Go Ad-free

Transcript

Question 3 Find the vector equation of the plane which is at a distance of 6/āˆš29 from the origin and its normal vector from the origin is 2š‘– Ģ‚ āˆ’ 3š‘— Ģ‚ + 4š‘˜ Ģ‚.Vector equation of a plane at a distance ā€˜dā€™ from the origin and unit vector to normal from origin š‘› Ģ‚ is š’“ āƒ—.š’ Ģ‚ = d Unit vector of š‘› āƒ— = š‘› Ģ‚ = 1/|š‘› āƒ— | (š‘› āƒ—) Now, distance from origin = d = 6/āˆš29 š‘› āƒ— = 2š‘– Ģ‚ āˆ’ 3š‘— Ģ‚ + 4š‘˜ Ģ‚ Magnitude of š‘› Ģ‚ = āˆš(22+(āˆ’3)^2+4^2 ) |š‘› āƒ— | = āˆš(4+9+16) = āˆš29 Now, š‘› Ģ‚ = 1/|š‘› āƒ— | " (" š‘› āƒ—")" = 1/āˆš29 " (2" š‘– Ģ‚āˆ’"3" š‘— Ģ‚+"4" š‘˜ Ģ‚")" = 2/āˆš29 š‘– Ģ‚ āˆ’ 3/āˆš29 š‘— Ģ‚ + 4/āˆš29 š‘˜ Ģ‚ Vector equation of plane is š‘Ÿ āƒ—.š‘› Ģ‚ = d š’“ āƒ— . (šŸ/āˆššŸšŸ— š’Š Ģ‚āˆ’šŸ‘/āˆššŸšŸ— " " š’‹ Ģ‚+ šŸ’/āˆššŸšŸ— " " š’Œ Ģ‚ ) = šŸ”/āˆššŸšŸ— Cartesian equation Equation of a plane in Cartesian form which is at a distance ā€˜dā€™ from the origin and has a normal vector š‘› āƒ— = š‘Žš‘– Ģ‚ + bš‘— Ģ‚ + cš‘˜ Ģ‚ is lx + my + nz = d where l, m, n are direction cosines of š‘› āƒ— l = š‘Ž/āˆš(š‘Ž^2 + š‘^2 + š‘^2 ) , m = š‘/āˆš(š‘Ž^2 + š‘^2 + š‘^2 ) , n = š‘/āˆš(š‘Ž^2 + š‘^2 + š‘^2 ) Distance form origin = d = 6/āˆš29 š‘› āƒ— = 2š‘– Ģ‚ āˆ’ 3š‘— Ģ‚ + 4š‘˜ Ģ‚ Comparing with š‘› āƒ— = š‘Žš‘– Ģ‚ + š‘š‘— Ģ‚ + š‘š‘˜ Ģ‚, š‘Ž = 2, b = āˆ’3, c = 4 Also, āˆš(š‘Ž2+š‘2+š‘2) = āˆš(22+(āˆ’3)2+42) = āˆš(4+9+16) = āˆš29 So, direction cosines are l = 2/āˆš29 , m = ( āˆ’3)/āˆš29 , n = 4/āˆš29 āˆ“ Cartesian equations of plane is lx + my + nz = d 2/āˆš29 x + (( āˆ’3)/āˆš29)y + 4/āˆš29 z = 6/āˆš29 2x āˆ’ 3y + 4z = 6 Therefore, the equation of plane in Cartesian form is 2x āˆ’ 3y + 4z = 6

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