Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 17 (introduction) Find the vector and cartesian equations of the plane which passes through the point (5, 2, – 4) and perpendicular to the line with direction ratios 2, 3, – 1. Vector equation of a plane passing through a point (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is 𝑟 −(𝑥1 𝑖 + 𝑦1 𝑗 + 𝑧1 𝑘). (A 𝑖 + B 𝑗 + C 𝑘) = 0 or ( 𝑟 − 𝑎). 𝑛 = 0 Example 17 Find the vector and Cartesian equations of the plane which passes through the point (5, 2, – 4) and perpendicular to the line with direction ratios 2, 3, – 1. Vector form Equation of plane passing through point A whose position vector is 𝒂 & perpendicular to 𝒏 is ( 𝒓 − 𝒂) . 𝒏 = 0 Given Plane passes through (5, 2, −4) So 𝒂 = 5 𝑖 + 2 𝑗 − 4 𝑘 Direction ratios of line perpendicular to plane = 2, 3, −1 So, n = 2 𝑖 + 3 𝑗 − 1 𝑘 Equation of plane in vector form is ( 𝑟 − 𝑎) . 𝑛 = 0 𝒓−(𝟓 𝒊+𝟐 𝒋−𝟒 𝒌). (𝟐 𝒊+𝟑 𝒋− 𝒌) = 0 Cartesian form (Method 1): Vector equation is 𝑟−(5 𝑖+2 𝑗−4 𝑘). (2 𝑖+3 𝑗− 𝑘) = 0 Put 𝒓 = x 𝒊 + y 𝒋 + z 𝒌 𝑥 𝑖+𝑦 𝑗−𝑧 𝑘−(5 𝑖+2 𝑗−4 𝑘).(2 𝑖 + 3 𝑗 − 𝑘) = 0 𝑥−5 𝑖+ 𝑦−2 𝑗+ (𝑧− − 4) 𝑘.(2 𝑖 + 3 𝑗 − 𝑘) = 0 2(x − 5) + 3 (y − 2) + (− 1)(z + 4) = 0 2x − 10 + 3y − 6 − z − 4 = 0 2x − 10 + 3(y − 2) + (− 1) (z + 4) = 0 2x − 10 + 3y − 6 − z − 4 = 0 2x + 3y − z − 20 = 0 2x + 3y − z = 20 Therefore equation of plane in Cartesian form is 2x + 3y − z = 20 Cartesian form (Method 2): Equation of plane passing through (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is A(x − x1) + B(y − y1) + c (z − z1) = 0 Since the plane passes through (5, 2, −4) x1 = 5 , y1 = 2 , z1 = −4 Direction ratios of line perpendicular to plane = 2, 3, −1 ∴ A = 2, B = 3, C = −1 Therefore, equations of line in Cartesian form is 2(x − 5) + 3 (y − 2) + (− 1) (x − (−4)) = 0 2 (x − 5) + 3(y − 2) − 1 (z + 4) = 0 2x − 10 + 3y − 6 − z − 4 = 0 2x + 3y − z − 20 = 0 2x + 3y − z = 20 Therefore equation of plane in Cartesian form is 2x + 3y − z = 20

Example 1

Example, 2

Example, 3 Important

Example, 4

Example, 5

Example, 6 Important

Example, 7

Example 8

Example, 9 Important

Example 10

Example 11

Example 12 Important

Example 13

Example 14

Example 15

Example 16

Example 17 You are here

Example 18

Example 19

Example 20 Important

Example 21 Important

Example 22

Example 23 Important

Example 24 Important

Example, 25 Important

Example 26

Example 27 Important

Example 28

Example 29 Important

Example 30 Important

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 .