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

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Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.