Find the equation of a plane passing through the points A (2, 1, 2) and B (4,  −2, 1) and perpendicular to plane r . (i − 2k) = 5. Also, find the coordinates of the point, where the line passing through the points (3, 4, 1) and (5, 1, 6) crosses the plane thus obtained.

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  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Transcript

Question 36 Find the equation of a plane passing through the points A (2, 1, 2) and B (4, โˆ’2, 1) and perpendicular to plane ๐‘Ÿ โƒ— . (๐‘– ฬ‚ โˆ’ 2๐‘˜ ฬ‚) = 5. Also, find the coordinates of the point, where the line passing through the points (3, 4, 1) and (5, 1, 6) crosses the plane thus obtained. Given plane passes through A (2, 1, 2) and B (4, โ€“2, 1) Let point P (x, y, z) be any point in the required plane Now, AP lies in the required plane Also, AB lies in the required plane Therefore, Normal of required plane = (๐ด๐‘ƒ) โƒ— ร— (๐ด๐ต) โƒ— It is also given that Required plane is perpendicular to ๐‘Ÿ โƒ— . (๐‘– ฬ‚ โˆ’ 2๐‘˜ ฬ‚) = 5 So, its normal = ๐‘› โƒ— = ๐‘– ฬ‚ โˆ’ 2๐‘˜ ฬ‚ Since required plane and other plane are perpendicular Both their normals will be perpendicular So, their dot product is 0 i.e. Normal of Required Plane . Normal of other plane = 0 ((๐ด๐‘ƒ) โƒ— ร— (๐ด๐ต) โƒ—). ๐‘› โƒ— = 0 ๐‘› โƒ— . ((๐ด๐‘ƒ) โƒ— ร— (๐ด๐ต) โƒ—) = 0 This is scalar triple product ๐‘Ž โƒ—.(๐‘ โƒ—ร—๐‘ โƒ— )= [โ– 8(๐‘Ž โƒ—&๐‘ โƒ—&๐‘ โƒ— )] [โ– 8(๐‘› โƒ—&(๐ด๐‘ƒ) โƒ—&(๐ด๐ต) โƒ— )] = 0 Now, Points are A (2, 1, 2), B (4, โˆ’2, 1) and P(x, y, z) Therefore, (๐ด๐‘ƒ) โƒ— = (๐‘ฅโˆ’2) ๐‘– ฬ‚+(๐‘ฆโˆ’1) ๐‘— ฬ‚+(๐‘งโˆ’2) ๐‘˜ ฬ‚ (๐ด๐ต) โƒ— = (4โˆ’2) ๐‘– ฬ‚+(โˆ’2โˆ’1) ๐‘— ฬ‚+(1โˆ’2) ๐‘˜ ฬ‚ = = 2๐‘– ฬ‚โˆ’3๐‘— ฬ‚โˆ’๐‘˜ ฬ‚ And the other plane is ๐‘Ÿ โƒ— . (๐‘– ฬ‚ โˆ’ 2๐‘˜ ฬ‚) = 5 So, its normal = ๐‘› โƒ— = ๐‘– ฬ‚ โˆ’ 2๐‘˜ ฬ‚ = ๐‘– ฬ‚ + 0๐‘— ฬ‚ โˆ’ 2๐‘˜ ฬ‚ Putting these values in (3) [โ– 8(๐‘› โƒ—&(๐ด๐‘ƒ) โƒ—&(๐ด๐ต) โƒ— )] = 0 |โ– 8(1&0&โˆ’2@(๐‘ฅโˆ’2)&(๐‘ฆโˆ’1)&(๐‘งโˆ’2)@2&โˆ’3&โˆ’1)| = 0 1(โ€“(y โ€“ 1) + 3(z โ€“ 2)) + 0 โ€“ 2 (โ€“3(x โ€“ 2) โ€“ 2 (y โ€“ 1)) = 0 โ€“(y โ€“ 1) + 3(z โ€“ 2) + 6(x โ€“ 2) + 4 (y โ€“ 1) = 0 โ€“y + 1 + 3z โ€“ 6 + 6x โ€“ 12 + 4y โ€“ 4 = 0 6x + 3y + 3z = 21 2x + y + z = 7 So, the equation of plane is 2x + y + z = 7 Now, the next part of the question is Also, find the coordinates of the point, where the line passing through the points (3, 4, 1) and (5, 1, 6) crosses the plane thus obtained. Equation of line passing through (3, 4, 1) and (5, 1, 6) is (๐‘ฅ โˆ’ ๐‘ฅ_1)/(๐‘ฅ_2 โˆ’ ๐‘ฅ_1 ) = (๐‘ฆ โˆ’ ๐‘ฆ_1)/(๐‘ฆ_2 โˆ’ ๐‘ฆ_1 ) = (๐‘ง โˆ’ ๐‘ง_1)/(๐‘ง_2 โˆ’ ๐‘ง_1 ) (๐‘ฅ โˆ’ 3)/(5 โˆ’ 3) = (๐‘ฆ โˆ’ 4)/(1 โˆ’ 4) = (๐‘ง โˆ’ 1)/(6 โˆ’ 1) (๐‘ฅ โˆ’ 3)/2 = (๐‘ฆ โˆ’ 4)/(โˆ’3) = (๐‘ง โˆ’ 1)/5 Now, we need to find intersection point of line and plane Equation of line is (๐‘ฅ โˆ’ 3)/2 = (๐‘ฆ โˆ’ 4)/(โˆ’3) = (๐‘ง โˆ’ 1)/5 = k So, point on the line is x = 2k + 3, y = โ€“3k + 4, z = 5k + 1 Putting these values in equation of plane 2x + y + z = 7 2(2k + 3) + (โ€“3k + 4) + (5k + 1) = 7 4k + 6 โ€“ 3k + 4 + 5k + 1 = 7 6k + 11 = 7 6k = 7 โ€“ 11 6k = โ€“4 k = (โˆ’2)/3 Thus, coordinates of points are x = 2k + 3 = 2 ร— ((โˆ’2)/3) + 3 = (โˆ’4)/3+3 = (โˆ’4 + 9)/3 = 5/3 y = โ€“3k + 4 = โ€“3 ร— ((โˆ’2)/3) + 4 = 6/3+4 = 2 + 4 = 6 z = 5k + 1 = 5 ร— ((โˆ’2)/3) + 1 = (โˆ’10)/3+1 = (โˆ’10 + 3)/3 = (โˆ’7)/3 Thus, the required point is (5/3,6,(โˆ’7)/3)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.