Find the foot of the perpendicular drawn from the point (-1, 3, -6) to the plane 2π‘₯ + 𝑦 − 2𝑧 + 5 = 0. Also find the equation and length of the perpendicular.

 

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Note : This is similar to Example 16 of NCERT – Chapter 11 Class 12 Three Dimensional Geometry

Check the answer here

https://www.teachoo.com/3572/756/Example-16---Find-coordinates-of-foot-of-perpendicular-from/category/Examples/

  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Transcript

Question 37 (Choice 2) Find the foot of the perpendicular drawn from the point (βˆ’1, 3, βˆ’6) to the plane 2π‘₯ + 𝑦 βˆ’ 2𝑧 + 5 = 0. Also find the equation and length of the perpendicular. Let point P(x1, y1, z1) be foot of perpendicular from point X (βˆ’1, 3, βˆ’6) Since perpendicular to plane is parallel to normal vector Vector (𝑿𝑷) βƒ— is parallel to normal vector 𝒏 βƒ— Given equation of the plane is 2x + y βˆ’ 2z + 5 = 0 2x + y βˆ’ 2z = βˆ’5 So, Normal vector = 𝒏 βƒ— = 2π’Š Μ‚ + 𝒋 Μ‚ βˆ’ 2π’Œ Μ‚ Since, (𝑿𝑷) βƒ— and 𝒏 βƒ— are parallel their direction ratios are proportional. Finding direction ratios (𝑿𝑷) βƒ— = (x1 + 1)π’Š Μ‚ + (y1 βˆ’ 3)𝒋 Μ‚ + (z1 + 6)π’Œ Μ‚ Direction ratios = x1 + 1, y1 βˆ’ 3, z1 + 6 ∴ a1 = x1 + 1 , b1 = y1 βˆ’ 3, c1 = z1 + 6 𝒏 βƒ— = 2π’Š Μ‚ + 𝒋 Μ‚ βˆ’ 2π’Œ Μ‚ Direction ratios = 2, 1, βˆ’2 ∴ a2 = 2 , b2 = 1, c2 = βˆ’2 Direction ratios are proportional π‘Ž_1/π‘Ž_2 = 𝑏_1/𝑏_2 = 𝑐_1/𝑐_2 = k (π‘₯_1 + 1)/2 = (𝑦_1 βˆ’ 3)/( 1) = (𝑧_1 + 6)/(βˆ’2) = k Thus, x1 = 2k βˆ’ 1, y1 = k + 3, z1 = βˆ’2k βˆ’ 6 Also, point P(x1, y1, z1) lies in the plane. Putting P (2k βˆ’ 1, k + 3, βˆ’2k βˆ’ 6) in equation of plane 2x + y βˆ’ 2z = βˆ’5 2(2k βˆ’ 1) + (k + 3) βˆ’ 2(βˆ’2k βˆ’ 6) = βˆ’5 4k βˆ’ 2 + k + 3 + 4k + 12 = βˆ’5 4k + k + 4k βˆ’ 2 + 3 + 12 = βˆ’5 9k + 13 = βˆ’5 9k = βˆ’5 βˆ’ 13 9k = βˆ’18 ∴ k = βˆ’2 Thus, x1 = 2k βˆ’ 1 = 2(βˆ’2) βˆ’ 1 = βˆ’5 y1 = k + 3 = (βˆ’2) + 3 = 1 z1 = βˆ’2k βˆ’ 6 = βˆ’2(βˆ’2) βˆ’ 6 = βˆ’2 Therefore, coordinate of foot of perpendicular are P (βˆ’5, 1, βˆ’2) Equation of perpendicular Equation of perpendicular would be equation of line joining X (βˆ’1, 3, βˆ’6) and P (βˆ’5, 1, βˆ’2) (π‘₯ βˆ’ (βˆ’1))/(βˆ’5 βˆ’ (βˆ’1))=(𝑦 βˆ’ 3)/(1 βˆ’ 3)=(𝑧 βˆ’ (βˆ’6))/(βˆ’2 βˆ’ (βˆ’6)) (π‘₯ + 1)/(βˆ’4)=(𝑦 βˆ’ 3)/(βˆ’2)=(𝑧 + 6)/4 (𝒙 + 𝟏)/(βˆ’πŸ)=(π’š βˆ’ πŸ‘)/(βˆ’πŸ)=(𝒛 + πŸ”)/𝟐 Length of perpendicular X (βˆ’1, 3, βˆ’6) and P (βˆ’5, 1, βˆ’2) Let of Perpendicular is length of PX PX = √((βˆ’5βˆ’(βˆ’1))^2+(1βˆ’3)^2+(βˆ’2βˆ’(βˆ’6))^2 ) PX = √((βˆ’5+1)^2+(βˆ’2)^2+(βˆ’2+6)^2 ) PX = √((βˆ’4)^2+(βˆ’2)^2+(4)^2 ) PX = √(16+4+16) PX = √36 PX = 6 units

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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.