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Find the vector and the Cartesian equations of the plane containing the point ๐‘– ฬ‚+2๐‘— ฬ‚−๐‘˜ ฬ‚ and parallel to the lines ๐‘Ÿ โƒ— = (๐‘– ฬ‚+2๐‘— ฬ‚+2๐‘˜ ฬ‚ ) + s (2๐‘– ฬ‚−3๐‘— ฬ‚+2๐‘˜ ฬ‚ ) = 0 and ๐‘Ÿ โƒ— = (3๐‘– ฬ‚+๐‘— ฬ‚−2๐‘˜ ฬ‚ ) + t (๐‘– ฬ‚−3๐‘— ฬ‚+๐‘˜ ฬ‚ ) = 0

This question is similar to Misc 13 Chapter 11 Class 12 - Three Dimensional Geometry

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Transcript

Question 10 (Choice 2) Find the vector and the Cartesian equations of the plane containing the point ๐‘– ฬ‚+2๐‘— ฬ‚โˆ’๐‘˜ ฬ‚ and parallel to the lines ๐‘Ÿ โƒ— = (๐‘– ฬ‚+2๐‘— ฬ‚+2๐‘˜ ฬ‚ ) + s (2๐‘– ฬ‚โˆ’3๐‘— ฬ‚+2๐‘˜ ฬ‚ ) = 0 and ๐‘Ÿ โƒ— = (3๐‘– ฬ‚+๐‘— ฬ‚โˆ’2๐‘˜ ฬ‚ ) + t (๐‘– ฬ‚โˆ’3๐‘— ฬ‚+๐‘˜ ฬ‚ ) = 0 Given that Plane is parallel to lines ๐‘Ÿ โƒ— = (๐‘– ฬ‚+2๐‘— ฬ‚+2๐‘˜ ฬ‚ ) + s (2๐‘– ฬ‚โˆ’3๐‘— ฬ‚+2๐‘˜ ฬ‚ ) and ๐‘Ÿ โƒ— = (3๐‘– ฬ‚+๐‘— ฬ‚โˆ’2๐‘˜ ฬ‚ ) + t (๐‘– ฬ‚โˆ’3๐‘— ฬ‚+๐‘˜ ฬ‚ ) Thus, Normal will be perpendicular to both their parallel vectors โˆด Normal of plane = (๐Ÿ๐’Š ฬ‚โˆ’๐Ÿ‘๐’‹ ฬ‚+๐Ÿ๐’Œ ฬ‚ ) ร— (๐’Š ฬ‚โˆ’๐Ÿ‘๐’‹ ฬ‚+๐’Œ ฬ‚ ) = |โ– 8(๐‘– ฬ‚&๐‘— ฬ‚&๐‘˜ ฬ‚@2&โˆ’3&2@1&โˆ’3&1)| = ๐‘– ฬ‚ (โˆ’3(1) โ€“ (โˆ’3)(2)) โ€“ ๐‘— ฬ‚ (2(1) โ€“ 1(2)) + ๐‘˜ ฬ‚(2(โˆ’3) โ€“ 1(โˆ’3)) = ๐‘– ฬ‚ (โˆ’3 + 6) โ€“ ๐‘— ฬ‚ (2 โˆ’ 2) + ๐‘˜ ฬ‚(โˆ’6 + 3) = 3๐’Š ฬ‚ โ€“ 3๐’Œ ฬ‚ Now, 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 Thus, Equation of plane passing through ๐’‚ โƒ— =๐‘– ฬ‚+2๐‘— ฬ‚โˆ’๐‘˜ ฬ‚ & perpendicular to ๐’ โƒ— = 3๐‘– ฬ‚ โ€“ 3๐‘˜ ฬ‚ is [๐’“ โƒ—โˆ’(๐’Š ฬ‚+๐Ÿ๐’‹ ฬ‚โˆ’๐’Œ ฬ‚)]. (๐Ÿ‘๐’Š ฬ‚โˆ’๐Ÿ‘๐’Œ ฬ‚) = 0 Finding Cartesian form Putting ๐’“ โƒ— = x๐’Š ฬ‚ + y๐’‹ ฬ‚ + z๐’Œ ฬ‚ [(๐‘ฅ๐‘– ฬ‚+๐‘ฆ๐‘— ฬ‚+๐‘ง๐‘˜ ฬ‚ )โˆ’(๐‘– ฬ‚+2๐‘— ฬ‚โˆ’๐‘˜ ฬ‚)]. (3๐‘– ฬ‚ โˆ’ 3๐‘˜ ฬ‚) = 0 [(๐‘ฅโˆ’1) ๐‘– ฬ‚+(๐‘ฆโˆ’2) ๐‘— ฬ‚+ (๐‘งโˆ’(โˆ’ 1))๐‘˜ ฬ‚ ]. (3๐‘– ฬ‚ โˆ’ 3๐‘˜ ฬ‚) = 0 3(x โˆ’ 1) + 0 (y โˆ’ 2) + (โˆ’3)(z + 1) = 0 3x โˆ’ 3 โˆ’ 3z โˆ’ 3 = 0 3x โˆ’ 3z โˆ’ 6 = 0 3(x โˆ’ z โˆ’ 2) = 0 x โˆ’ z โˆ’ 2 = 0

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.