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.

 

Find foot of perpendicular drawn from point (-1, 3, -6) to plane 2x+y

Question 37 (Choice 2) - CBSE Class 12 Sample Paper for 2021 Boards - Part 2
Question 37 (Choice 2) - CBSE Class 12 Sample Paper for 2021 Boards - Part 3
Question 37 (Choice 2) - CBSE Class 12 Sample Paper for 2021 Boards - Part 4
Question 37 (Choice 2) - CBSE Class 12 Sample Paper for 2021 Boards - Part 5 Question 37 (Choice 2) - CBSE Class 12 Sample Paper for 2021 Boards - Part 6

 

 

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/

Go Ad-free

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

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.