This question is similar to CBSE Class 12 Sample Paper for 2024 Boards

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https://www.teachoo.com/22345/4630/Question-35-Choice-1/category/CBSE-Class-12-Sample-Paper-for-2024-Boards/

 

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Question 35 (B) Find the image of the point (๐Ÿ,๐Ÿ,๐Ÿ) with respect to the line (๐‘ฅโˆ’3)/1=(๐‘ฆ+1)/2=(๐‘งโˆ’1)/3. Also find the equation of the line joining the given point and its image.Let Point P be (1, 2, 1) Let Q (a, b, c) be the image of point P (1, 2, 1) in the line ๐’“ โƒ— Since line is a mirror Point P & Q are at equal distance from line AB, i.e. PR = QR, i.e. R is the mid point of PQ Image is formed perpendicular to mirror i.e. line PQ is perpendicular to line ๐’“ โƒ— Given line is (๐’™ โˆ’ ๐Ÿ‘)/๐Ÿ=(๐’š + ๐Ÿ)/๐Ÿ=(๐’› โˆ’ ๐Ÿ)/๐Ÿ‘ Since PQ โŠฅ Line 1 (๐‘™_1) โˆด PR โŠฅ Line 1 (๐’_๐Ÿ) Coordinates of R Since R lies of line ๐‘™_1 โˆด (๐‘ฅ โˆ’ 3)/1=(๐‘ฆ + 1)/2=(๐‘ง โˆ’ 1)/3 = ๐œ† โˆด x = ๐€+๐Ÿ‘ , y = 2๐€ โ€“ 1 and z = 3๐€ + 1 So, coordinates of R = (๐€ + 3, 2๐€ โ€“ 1, 3๐€ + 1) Now, Since PR โŠฅ AB 1(๐€ + 2) + 2(2๐€ โ€“ 3) + 3 (3๐€) = 0 ๐œ† + 2 + 4๐œ† โ€“ 6 + 9๐œ† = 0 Direction ratios of Line ๐’_๐Ÿ Since equation of lines is (๐’™ โˆ’ ๐Ÿ‘)/๐Ÿ=(๐’š + ๐Ÿ)/๐Ÿ=(๐’› โˆ’ ๐Ÿ)/๐Ÿ‘ Direction ratios are 1, 2, 3 Direction ratios of Line PR Coordinates of P (1, 2, 1) Coordinates of R (๐œ† + 3, 2๐œ† โ€“ 1, 3๐œ† + 1) Direction ratios are ๐œ† + 3 โ€“ 1, 2๐œ† โ€“ 1 โ€“ 2 & 3๐œ† + 1 โ€“ 1 i.e. ๐€ + 2, 2๐€ โ€“ 3 & 3๐€ 14๐œ† โ€“ 4 = 0 14๐œ† = 4 ๐œ† = 4/14 ๐€ = 2/7 Now, Coordinates of R = (๐œ† + 3, 2๐œ† โ€“ 1, 3๐œ† + 1) = (2/7+3, 2(2/7)โˆ’1, 3(2/7)+1) = ((2 + 21)/7, (4 โˆ’ 7)/7, (6 + 7)/7) = (๐Ÿ๐Ÿ‘/๐Ÿ•, (โˆ’๐Ÿ‘)/๐Ÿ•, ๐Ÿ๐Ÿ‘/๐Ÿ•) Since R is the midpoint of PQ Coordinates of R = ((๐Ÿ + ๐’‚)/๐Ÿ " , " (๐Ÿ + ๐’ƒ)/๐Ÿ " , " (๐Ÿ + ๐’„)/๐Ÿ) (๐Ÿ๐Ÿ‘/๐Ÿ•, (โˆ’๐Ÿ‘)/๐Ÿ•, ๐Ÿ๐Ÿ‘/๐Ÿ•) = ((๐Ÿ + ๐’‚)/๐Ÿ " , " (๐Ÿ + ๐’ƒ)/๐Ÿ " , " (๐Ÿ + ๐’„)/๐Ÿ) ๐Ÿ๐Ÿ‘/๐Ÿ• = (๐Ÿ + ๐’‚)/๐Ÿ 23 ร— 2 = 7 + 7a 46 = 7 + 7a 46 โ€“ 7 = 7a 39 = 7a 7a = 39 a = ๐Ÿ‘๐Ÿ—/๐Ÿ• (โˆ’๐Ÿ‘)/๐Ÿ• = (๐Ÿ + ๐’ƒ)/๐Ÿ -3 ร— 2 = 14 + 7b โ€“6 = 14 + 7b โ€“6 โ€“ 14 = 7b โ€“20 = 7b 7b = โ€“20 b = (โˆ’๐Ÿ๐ŸŽ)/๐Ÿ• ๐Ÿ๐Ÿ‘/๐Ÿ• = (๐Ÿ + ๐’„)/๐Ÿ 13 ร— 2 = 7 + 7c 26 = 7 + 7c 26 โ€“ 7 = 7c 19 = 7c 7c = 19 c = ๐Ÿ๐Ÿ—/๐Ÿ• Hence, Coordinates of Q = (a, b, c) = (๐Ÿ‘๐Ÿ—/๐Ÿ•, (โˆ’๐Ÿ๐ŸŽ)/๐Ÿ•, ๐Ÿ๐Ÿ—/๐Ÿ•) โˆด Q(๐Ÿ‘๐Ÿ—/๐Ÿ•, (โˆ’๐Ÿ๐ŸŽ)/๐Ÿ•, ๐Ÿ๐Ÿ—/๐Ÿ•) is the required image of P Finding the equation of the line joining the given point and its image. Cartesian equation of a line passing through two points P(x1, y1, z1) and Q (x2, y2, z2) is (๐‘ฅ โˆ’ ๐‘ฅ1)/(๐‘ฅ2 โˆ’ ๐‘ฅ_1 ) = (๐‘ฆ โˆ’ ๐‘ฆ1)/(๐‘ฆ2 โˆ’ ๐‘ฆ1) = (๐‘ง โˆ’ ๐‘ง1)/(๐‘ง2 โˆ’ ๐‘ง1) Since the line passes through P (1, 2, 1) x1 = 1, y1 = 2, z1 = 1 And also passes through Q (๐Ÿ‘๐Ÿ—/๐Ÿ•, (โˆ’๐Ÿ๐ŸŽ)/๐Ÿ•, ๐Ÿ๐Ÿ—/๐Ÿ•) x2 = ๐Ÿ‘๐Ÿ—/๐Ÿ•, y2 = (โˆ’๐Ÿ๐ŸŽ)/๐Ÿ•, z2 = ๐Ÿ๐Ÿ—/๐Ÿ• Equation of line is (๐’™ โˆ’๐Ÿ)/(๐Ÿ‘๐Ÿ—/๐Ÿ• โˆ’๐Ÿ) = (๐’š โˆ’๐Ÿ)/( (โˆ’๐Ÿ๐ŸŽ)/๐Ÿ• โˆ’๐Ÿ) = (๐’› โˆ’ ๐Ÿ)/(๐Ÿ๐Ÿ—/๐Ÿ• โˆ’๐Ÿ) (๐‘ฅ โˆ’1)/((39 โˆ’ 7)/7) = (๐‘ฆ โˆ’2)/( (โˆ’20 โˆ’ 14)/7) = (๐‘ง โˆ’ 1)/((19 โˆ’ 7)/7 ) (๐‘ฅ โˆ’1)/(32/7) = (๐‘ฆ โˆ’2)/( (โˆ’34)/7) = (๐‘ง โˆ’ 1)/(12/7 ) 7 ร— (๐‘ฅ โˆ’1)/32 = 7 ร— (๐‘ฆ โˆ’2)/( โˆ’34) = 7 ร— (๐‘ง โˆ’ 1)/(12 ) We can remove constant 7 (๐‘ฅ โˆ’1)/32 = (๐‘ฆ โˆ’2)/( โˆ’34) = (๐‘ง โˆ’ 1)/(12 ) (๐‘ฅ โˆ’1)/(2 ร— 16) = (๐‘ฆ โˆ’2)/(2 ร— โˆ’17) = (๐‘ง โˆ’ 1)/(2 ร— 6 ) 1/2 ร— (๐‘ฅ โˆ’1)/16 = 1/2 ร— (๐‘ฆ โˆ’2)/( โˆ’17) = 1/2 ร— (๐‘ง โˆ’ 1)/(6 ) We can remove constant 1/2 (๐’™ โˆ’๐Ÿ)/๐Ÿ๐Ÿ” = (๐’š โˆ’๐Ÿ)/( โˆ’๐Ÿ๐Ÿ•) = (๐’› โˆ’ ๐Ÿ)/(๐Ÿ” )

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.