Find the equation of the line which intersects the lines (x + 2) / 1 = (y - 3) / 2 = (z +1) / 4 and (x -1) / 2  = (y -2) / 3 = (z -3) / 4  and passes through the point (1, 1, 1).

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Question 21 Find the equation of the line which intersects the lines (๐‘ฅ + 2)/1 = (๐‘ฆ โˆ’ 3)/2 = (๐‘ง +1)/4 and (๐‘ฅ โˆ’1)/2 = (๐‘ฆ โˆ’2)/3 = (๐‘ง โˆ’ 3)/4 and passes through the point (1, 1, 1). We need to find equation of line which intersects the lines (๐‘ฅ + 2)/1 = (๐‘ฆ โˆ’ 3)/2 = (๐‘ง + 1)/4 (๐‘ฅ โˆ’ 1)/2 = (๐‘ฆ โˆ’ 2)/3 = (๐‘ง โˆ’ 3)/4 and passes through (1, 1, 1) Now, Equation of a line passing through two points (x1, y1, z1) and (x1, y1, z1) (๐‘ฅ โˆ’ ๐‘ฅ1)/(๐‘ฅ2 โˆ’ ๐‘ฅ1) = (๐‘ฆ โˆ’ ๐‘ฆ1)/(๐‘ฆ2 โˆ’ ๐‘ฆ1) = (๐‘ง โˆ’ ๐‘ง1)/(๐‘ง2 โˆ’ ๐‘ง1) We use (x1, y1, z1 ) = (1, 1, 1) If point is in Line (1) (๐‘ฅ + 2)/1 = (๐‘ฆ โˆ’ 3)/2 = (๐‘ง + 1)/4 General point is (๐‘ฅ + 2)/1 = (๐‘ฆ โˆ’ 3)/2 = (๐‘ง +1)/4 = p So, (๐‘ฅ + 2)/1 = p โ‡’ x = p โ€“ 2 So, (๐‘ฆ โˆ’ 3)/2 = p โ‡’ y = 2p + 3 So, (๐‘ง +1)/4 = p z = 4p โ€“ 1 If point is in Line (1) (๐‘ฅ โˆ’ 1)/2 = (๐‘ฆ โˆ’ 2)/3 = (๐‘ง โˆ’ 3)/4 General point is (๐‘ฅ โˆ’ 1)/2 = (๐‘ฆ โˆ’ 2)/3 = (๐‘ง โˆ’ 3)/4 = q So, (๐‘ฅ โˆ’ 1)/2 = q โ‡’ x = 2q + 1 So, (๐‘ฆ โˆ’ 2)/3 = q โ‡’ y = 3q + 2 So, (๐‘ง โˆ’ 3)/4 = q z = 4q + 3 So, point is (p โ€“ 2, 2p + 3, 4p โ€“ 1) i.e. (x2, y2, z2) = (pโ€“2, 2p+3, 4pโ€“1) Thus, equation of line will be (๐‘ฅ โˆ’ ๐‘ฅ1)/(๐‘ฅ2 โˆ’ ๐‘ฅ1) = (๐‘ฆ โˆ’ ๐‘ฆ1)/(๐‘ฆ โˆ’ ๐‘ฆ1) = (๐‘ง โˆ’ ๐‘ง1)/(๐‘ง2 โˆ’ ๐‘ง1) Putting (x1, y1, z1 ) = (1, 1, 1) & (x2, y2, z2) = (pโ€“2, 2p+3, 4pโ€“1) (๐‘ฅ โˆ’ 1)/((๐‘โˆ’2)โˆ’ 1) = (๐‘ฆ โˆ’ 1)/((2๐‘+3) โˆ’ 1) = (๐‘ง โˆ’ 1)/((4๐‘โˆ’1) โˆ’ 1) (๐‘ฅ โˆ’ 1)/(๐‘ โˆ’ 3) = (๐‘ฆ โˆ’ 1)/(2๐‘ + 2) = (๐‘ง โˆ’ 1)/(4๐‘ โˆ’ 2) So, point is (2q + 1, 3q + 2, 4q + 3) i.e. (x2, y2, z2) = (2q+1, 3q+2, 4q+3) Thus, equation of line will be (๐‘ฅ โˆ’ ๐‘ฅ1)/(๐‘ฅ2 โˆ’ ๐‘ฅ1) = (๐‘ฆ โˆ’ ๐‘ฆ1)/(๐‘ฆ โˆ’ ๐‘ฆ1) = (๐‘ง โˆ’ ๐‘ง1)/(๐‘ง2 โˆ’ ๐‘ง1) Putting (x1, y1, z1 ) = (1, 1, 1) & (x2, y2, z2) = (2q+1, 3q+2, 4q+3) (๐‘ฅ โˆ’ 1)/((2๐‘ž+1)โˆ’ 1) = (๐‘ฆ โˆ’ 1)/((3๐‘ž+2) โˆ’ 1) = (๐‘ง โˆ’ 1)/((4๐‘ž+3) โˆ’ 1) (๐‘ฅ โˆ’ 1)/2๐‘ž = (๐‘ฆ โˆ’ 1)/(3๐‘ž + 1) = (๐‘ง โˆ’ 1)/(4๐‘ž + 2) Since (3) & (4) are the same lines The denominator i.e. the direction cosines of line are proportional (๐‘ โˆ’ 3)/2๐‘ž = (2๐‘ + 2)/(3๐‘ž + 1) = (4๐‘ โˆ’ 2)/(4๐‘ž + 2) Let (๐‘ โˆ’ 3)/2๐‘ž = (2๐‘ + 2)/(3๐‘ž + 1) = (4๐‘ โˆ’ 2)/(4๐‘ž + 2) = k (๐‘ โˆ’ 3)/2๐‘ž = k p โ€“ 3 = 2qk p = 2qk + 3 (2๐‘ + 2)/(3๐‘ž + 1) = k 2p + 2 = k(3q + 1) 2p + 2 = 3qk + k 2p = 3qk + k โ€“ 2 p = (3๐‘ž๐‘˜ + ๐‘˜ โˆ’ 2)/2 (4๐‘ โˆ’ 2)/(4๐‘ž + 2) = k 4p โ€“ 2 = k (4q + 2) 4p โ€“ 2 = 4qk + 2k 4p = 4qk + 2k + 2 p = (4๐‘ž๐‘˜ + 2๐‘˜ + 2)/4 Comparing (5) & (6) 2qk + 3 = (3๐‘ž๐‘˜ + ๐‘˜ โˆ’ 2)/2 4qk + 6 = 3qk + k โ€“ 2 4qk โ€“ 3qk โ€“ k = โ€“2 โ€“ 6 qk โ€“ k = โ€“8 k(q โ€“ 1) = โ€“8 k = (โˆ’8)/(๐‘ž โˆ’1) Comparing (6) & (7) (3๐‘ž๐‘˜ + ๐‘˜ โˆ’ 2)/2 = (4๐‘ž๐‘˜ + 2๐‘˜ + 2)/4 4 ร— (3๐‘ž๐‘˜ + ๐‘˜ โˆ’ 2)/2 = 4qk + 2k + 2 2(3qk + k โ€“ 2) = 4qk + 2k + 2 6qk + 2k โ€“ 4 = 4qk + 2k + 2 6qk โ€“ 4qk + 2k โ€“ 2k = 2 + 4 2qk = 6 qk = 3 Putting k = (โˆ’8)/(๐‘ž โˆ’ 1) q ร— ((โˆ’8)/(๐‘ž โˆ’ 1)) = 3 โ€“8q = 3(q โ€“ 1) โ€“8q = 3q โ€“ 3 โ€“8q โ€“ 3q = โ€“3 โ€“11q = โ€“3 q = (โˆ’3)/(โˆ’11) q = 3/11 Now, putting value of q in equation (4) (๐‘ฅ โˆ’ 1)/2๐‘ž = (๐‘ฆ โˆ’ 1)/(3๐‘ž + 1) = (๐‘ง โˆ’ 1)/(4๐‘ž + 2) (๐‘ฅ โˆ’ 1)/2(3/11) = (๐‘ฆ โˆ’ 1)/(3(3/11) + 1) = (๐‘ง โˆ’ 1)/(4(3/11) + 2) (๐‘ฅ โˆ’ 1)/(6/11) = (๐‘ฆ โˆ’ 1)/(9/11 + 1) = (๐‘ง โˆ’ 1)/(12/11 + 2) (๐‘ฅ โˆ’ 1)/(6/11) = (๐‘ฆ โˆ’ 1)/((9 + 11)/11) = (๐‘ง โˆ’ 1)/((12 + 2(11))/11) (๐‘ฅ โˆ’ 1)/(6/11) = (๐‘ฆ โˆ’ 1)/(20/11) = (๐‘ง โˆ’ 1)/(34/11) Here, in the denominator, 1/11 is common, so we remove it (๐‘ฅ โˆ’ 1)/6 = (๐‘ฆ โˆ’ 1)/20 = (๐‘ง โˆ’ 1)/34 (๐‘ฅ โˆ’ 1)/(2 ร— 3) = (๐‘ฆ โˆ’ 1)/(2 ร— 10) = (๐‘ง โˆ’ 1)/(2 ร— 17) Here, in the denominator, 2 is common, so we remove it (๐’™ โˆ’ ๐Ÿ)/๐Ÿ‘ = (๐’š โˆ’ ๐Ÿ)/๐Ÿ๐ŸŽ = (๐’› โˆ’ ๐Ÿ)/๐Ÿ๐Ÿ•

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.