Example 17 - Find vector and cartesian equation of plane which passes

 

 

Example 17 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2

Example 17 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Example 17 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4
Example 17 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5 Example 17 - Chapter 11 Class 12 Three Dimensional Geometry - Part 6

 


Transcript

Question 7 (introduction) Find the vector and cartesian equations of the plane which passes through the point (5, 2, โ€“ 4) and perpendicular to the line with direction ratios 2, 3, โ€“ 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 [๐‘Ÿ โƒ— โˆ’(๐‘ฅ1๐‘– ฬ‚ + ๐‘ฆ1๐‘— ฬ‚ + ๐‘ง1๐‘˜ ฬ‚)]. (A๐‘– ฬ‚ + B๐‘— ฬ‚ + C๐‘˜ ฬ‚) = 0 or (๐‘Ÿ โƒ— โˆ’ ๐‘Ž โƒ—).๐‘› โƒ— = 0 ("A" ๐‘ƒ) โƒ— is perpendicular to "n" โƒ— So, ("A" P) โƒ— . "n" โƒ— = 0 ("r" โƒ— โˆ’ "a" โƒ—)."n" โƒ— = 0 Question 7 Find the vector and Cartesian equations of the plane which passes through the point (5, 2, โ€“ 4) and perpendicular to the line with direction ratios 2, 3, โ€“ 1.Vector form Equation of plane passing through point A whose position vector is ๐’‚ โƒ— & perpendicular to ๐’ โƒ— is (๐’“ โƒ— โˆ’ ๐’‚ โƒ—) . ๐’ โƒ— = 0 Given Plane passes through (5, 2, โˆ’4) So ๐’‚ โƒ— = 5๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚ Direction ratios of line perpendicular to plane = 2, 3, โˆ’1 So, "n" โƒ— = 2๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ 1๐‘˜ ฬ‚ Equation of plane in vector form is (๐‘Ÿ โƒ— โˆ’ ๐‘Ž โƒ—) . ๐‘› โƒ— = 0 [๐’“ โƒ—โˆ’(๐Ÿ“๐’Š ฬ‚+๐Ÿ๐’‹ ฬ‚โˆ’๐Ÿ’๐’Œ ฬ‚)]. (๐Ÿ๐’Š ฬ‚+๐Ÿ‘๐’‹ ฬ‚โˆ’๐’Œ ฬ‚) = 0 Now, Finding Cartesian form using two methods Cartesian form (Method 1) Vector equation is [๐‘Ÿ โƒ—โˆ’(5๐‘– ฬ‚+2๐‘— ฬ‚โˆ’4๐‘˜ ฬ‚)]. (2๐‘– ฬ‚+3๐‘— ฬ‚โˆ’๐‘˜ ฬ‚) = 0 Put ๐’“ โƒ— = x๐’Š ฬ‚ + y๐’‹ ฬ‚ + z๐’Œ ฬ‚ [(๐‘ฅ๐‘– ฬ‚+๐‘ฆ๐‘— ฬ‚+๐‘ง๐‘˜ ฬ‚ )โˆ’(5๐‘– ฬ‚+2๐‘— ฬ‚โˆ’4๐‘˜ ฬ‚)].(2๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ ๐‘˜ ฬ‚) = 0 [(๐‘ฅโˆ’5) ๐‘– ฬ‚+(๐‘ฆโˆ’2) ๐‘— ฬ‚+ (๐‘งโˆ’(โˆ’ 4))๐‘˜ ฬ‚ ].(2๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ ๐‘˜ ฬ‚) = 0 2(x โˆ’ 5) + 3 (y โˆ’ 2) + (โˆ’ 1)(z + 4) = 0 2x โˆ’ 10 + 3y โˆ’ 6 โˆ’ z โˆ’ 4 = 0 2x + 3y โˆ’ z โˆ’ 20 = 0 2x + 3y โˆ’ z = 20 Therefore equation of plane in Cartesian form is 2x + 3y โˆ’ 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 โˆ’ x1) + B(y โˆ’ y1) + c (z โˆ’ z1) = 0 Since the plane passes through (5, 2, โˆ’4) x1 = 5, y1 = 2, z1 = โˆ’4 Direction ratios of line perpendicular to plane = 2, 3, โˆ’1 โˆด A = 2, B = 3, C = โˆ’1 Therefore, equation of line in Cartesian form is 2(x โˆ’ 5) + 3 (y โˆ’ 2) + (โˆ’1) (x โˆ’ (โˆ’4)) = 0 2 (x โˆ’ 5) + 3(y โˆ’ 2) โˆ’ 1 (z + 4) = 0 2x โˆ’ 10 + 3y โˆ’ 6 โˆ’ z โˆ’ 4 = 0 2x + 3y โˆ’ z โˆ’ 20 = 0 2x + 3y โˆ’ z = 20 Therefore equation of plane in Cartesian form is 2x + 3y โˆ’ z = 20

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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.