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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).

This is a question of CBSE Sample Paper - Class 12 - 2017/18.

You can download the question paper here  https://www.teachoo.com/cbse/sample-papers/


Find equation of line which intersects lines x+2/1 = y-3/2 = z+1/4

Question 21 - CBSE Class 12 Sample Paper for 2018 Boards - Part 2
Question 21 - CBSE Class 12 Sample Paper for 2018 Boards - Part 3 Question 21 - CBSE Class 12 Sample Paper for 2018 Boards - Part 4 Question 21 - CBSE Class 12 Sample Paper for 2018 Boards - Part 5 Question 21 - CBSE Class 12 Sample Paper for 2018 Boards - Part 6 Question 21 - CBSE Class 12 Sample Paper for 2018 Boards - Part 7 Question 21 - CBSE Class 12 Sample Paper for 2018 Boards - Part 8

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Question 21 Find the equation of the line which intersects the lines (š‘„ + 2)/1 = (š‘¦ āˆ’ 3)/2 = (š‘§ +1)/4 and (š‘„ āˆ’1)/2 = (š‘¦ āˆ’2)/3 = (š‘§ āˆ’ 3)/4 and passes through the point (1, 1, 1). We need to find equation of line which intersects the lines (š‘„ + 2)/1 = (š‘¦ āˆ’ 3)/2 = (š‘§ + 1)/4 (š‘„ āˆ’ 1)/2 = (š‘¦ āˆ’ 2)/3 = (š‘§ āˆ’ 3)/4 and passes through (1, 1, 1) Now, Equation of a line passing through two points (x1, y1, z1) and (x1, y1, z1) (š‘„ āˆ’ š‘„1)/(š‘„2 āˆ’ š‘„1) = (š‘¦ āˆ’ š‘¦1)/(š‘¦2 āˆ’ š‘¦1) = (š‘§ āˆ’ š‘§1)/(š‘§2 āˆ’ š‘§1) We use (x1, y1, z1 ) = (1, 1, 1) If point is in Line (1) (š‘„ + 2)/1 = (š‘¦ āˆ’ 3)/2 = (š‘§ + 1)/4 General point is (š‘„ + 2)/1 = (š‘¦ āˆ’ 3)/2 = (š‘§ +1)/4 = p So, (š‘„ + 2)/1 = p ā‡’ x = p ā€“ 2 So, (š‘¦ āˆ’ 3)/2 = p ā‡’ y = 2p + 3 So, (š‘§ +1)/4 = p z = 4p ā€“ 1 If point is in Line (1) (š‘„ āˆ’ 1)/2 = (š‘¦ āˆ’ 2)/3 = (š‘§ āˆ’ 3)/4 General point is (š‘„ āˆ’ 1)/2 = (š‘¦ āˆ’ 2)/3 = (š‘§ āˆ’ 3)/4 = q So, (š‘„ āˆ’ 1)/2 = q ā‡’ x = 2q + 1 So, (š‘¦ āˆ’ 2)/3 = q ā‡’ y = 3q + 2 So, (š‘§ āˆ’ 3)/4 = q z = 4q + 3 So, point is (p ā€“ 2, 2p + 3, 4p ā€“ 1) i.e. (x2, y2, z2) = (pā€“2, 2p+3, 4pā€“1) Thus, equation of line will be (š‘„ āˆ’ š‘„1)/(š‘„2 āˆ’ š‘„1) = (š‘¦ āˆ’ š‘¦1)/(š‘¦ āˆ’ š‘¦1) = (š‘§ āˆ’ š‘§1)/(š‘§2 āˆ’ š‘§1) Putting (x1, y1, z1 ) = (1, 1, 1) & (x2, y2, z2) = (pā€“2, 2p+3, 4pā€“1) (š‘„ āˆ’ 1)/((š‘āˆ’2)āˆ’ 1) = (š‘¦ āˆ’ 1)/((2š‘+3) āˆ’ 1) = (š‘§ āˆ’ 1)/((4š‘āˆ’1) āˆ’ 1) (š‘„ āˆ’ 1)/(š‘ āˆ’ 3) = (š‘¦ āˆ’ 1)/(2š‘ + 2) = (š‘§ āˆ’ 1)/(4š‘ āˆ’ 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 (š‘„ āˆ’ š‘„1)/(š‘„2 āˆ’ š‘„1) = (š‘¦ āˆ’ š‘¦1)/(š‘¦ āˆ’ š‘¦1) = (š‘§ āˆ’ š‘§1)/(š‘§2 āˆ’ š‘§1) Putting (x1, y1, z1 ) = (1, 1, 1) & (x2, y2, z2) = (2q+1, 3q+2, 4q+3) (š‘„ āˆ’ 1)/((2š‘ž+1)āˆ’ 1) = (š‘¦ āˆ’ 1)/((3š‘ž+2) āˆ’ 1) = (š‘§ āˆ’ 1)/((4š‘ž+3) āˆ’ 1) (š‘„ āˆ’ 1)/2š‘ž = (š‘¦ āˆ’ 1)/(3š‘ž + 1) = (š‘§ āˆ’ 1)/(4š‘ž + 2) Since (3) & (4) are the same lines The denominator i.e. the direction cosines of line are proportional (š‘ āˆ’ 3)/2š‘ž = (2š‘ + 2)/(3š‘ž + 1) = (4š‘ āˆ’ 2)/(4š‘ž + 2) Let (š‘ āˆ’ 3)/2š‘ž = (2š‘ + 2)/(3š‘ž + 1) = (4š‘ āˆ’ 2)/(4š‘ž + 2) = k (š‘ āˆ’ 3)/2š‘ž = k p ā€“ 3 = 2qk p = 2qk + 3 (2š‘ + 2)/(3š‘ž + 1) = k 2p + 2 = k(3q + 1) 2p + 2 = 3qk + k 2p = 3qk + k ā€“ 2 p = (3š‘žš‘˜ + š‘˜ āˆ’ 2)/2 (4š‘ āˆ’ 2)/(4š‘ž + 2) = k 4p ā€“ 2 = k (4q + 2) 4p ā€“ 2 = 4qk + 2k 4p = 4qk + 2k + 2 p = (4š‘žš‘˜ + 2š‘˜ + 2)/4 Comparing (5) & (6) 2qk + 3 = (3š‘žš‘˜ + š‘˜ āˆ’ 2)/2 4qk + 6 = 3qk + k ā€“ 2 4qk ā€“ 3qk ā€“ k = ā€“2 ā€“ 6 qk ā€“ k = ā€“8 k(q ā€“ 1) = ā€“8 k = (āˆ’8)/(š‘ž āˆ’1) Comparing (6) & (7) (3š‘žš‘˜ + š‘˜ āˆ’ 2)/2 = (4š‘žš‘˜ + 2š‘˜ + 2)/4 4 Ɨ (3š‘žš‘˜ + š‘˜ āˆ’ 2)/2 = 4qk + 2k + 2 2(3qk + k ā€“ 2) = 4qk + 2k + 2 6qk + 2k ā€“ 4 = 4qk + 2k + 2 6qk ā€“ 4qk + 2k ā€“ 2k = 2 + 4 2qk = 6 qk = 3 Putting k = (āˆ’8)/(š‘ž āˆ’ 1) q Ɨ ((āˆ’8)/(š‘ž āˆ’ 1)) = 3 ā€“8q = 3(q ā€“ 1) ā€“8q = 3q ā€“ 3 ā€“8q ā€“ 3q = ā€“3 ā€“11q = ā€“3 q = (āˆ’3)/(āˆ’11) q = 3/11 Now, putting value of q in equation (4) (š‘„ āˆ’ 1)/2š‘ž = (š‘¦ āˆ’ 1)/(3š‘ž + 1) = (š‘§ āˆ’ 1)/(4š‘ž + 2) (š‘„ āˆ’ 1)/2(3/11) = (š‘¦ āˆ’ 1)/(3(3/11) + 1) = (š‘§ āˆ’ 1)/(4(3/11) + 2) (š‘„ āˆ’ 1)/(6/11) = (š‘¦ āˆ’ 1)/(9/11 + 1) = (š‘§ āˆ’ 1)/(12/11 + 2) (š‘„ āˆ’ 1)/(6/11) = (š‘¦ āˆ’ 1)/((9 + 11)/11) = (š‘§ āˆ’ 1)/((12 + 2(11))/11) (š‘„ āˆ’ 1)/(6/11) = (š‘¦ āˆ’ 1)/(20/11) = (š‘§ āˆ’ 1)/(34/11) Here, in the denominator, 1/11 is common, so we remove it (š‘„ āˆ’ 1)/6 = (š‘¦ āˆ’ 1)/20 = (š‘§ āˆ’ 1)/34 (š‘„ āˆ’ 1)/(2 Ɨ 3) = (š‘¦ āˆ’ 1)/(2 Ɨ 10) = (š‘§ āˆ’ 1)/(2 Ɨ 17) Here, in the denominator, 2 is common, so we remove it (š’™ āˆ’ šŸ)/šŸ‘ = (š’š āˆ’ šŸ)/šŸšŸŽ = (š’› āˆ’ šŸ)/šŸšŸ•

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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.