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Find the vector and the Cartesian equations of the plane containing the point ๐‘–ย ฬ‚+2๐‘—ย ฬ‚โˆ’๐‘˜ย ฬ‚ and parallel to the lines ๐‘Ÿย โƒ— = (๐‘–ย ฬ‚+2๐‘—ย ฬ‚+2๐‘˜ย ฬ‚ ) + s (2๐‘–ย ฬ‚โˆ’3๐‘—ย ฬ‚+2๐‘˜ย ฬ‚ ) = 0 and ๐‘Ÿย โƒ— = (3๐‘–ย ฬ‚+๐‘—ย ฬ‚โˆ’2๐‘˜ย ฬ‚ ) + t (๐‘–ย ฬ‚โˆ’3๐‘—ย ฬ‚+๐‘˜ย 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.