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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) / 2Β  = (y -2) / 3 = (z -3) / 4Β  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 hereΒ  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


Transcript

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

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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.