Question 27 (OR 1 st question)

Find the equation of the plane through the line (x - 1)/3 = (y - 4)/2 = (z - 4)/(-2) and parallel to the line (x + 1)/2 = (1 - y)/4 = (z + 2)/1 . Hence, find the shortest distance between the lines

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Question 27 (OR 1st question) Find the equation of the plane through the line (π‘₯ βˆ’ 1)/3 = (𝑦 βˆ’ 4)/2 = (𝑧 βˆ’ 4)/(βˆ’2) and parallel to the line (π‘₯ + 1)/2 = (1 βˆ’ 𝑦)/4 = (𝑧 + 2)/1 . Hence, find the shortest distance between the lines Given lines (π‘₯ βˆ’ 1)/3 = (𝑦 βˆ’ 4)/2 = (𝑧 βˆ’ 4)/(βˆ’2) (π‘₯ + 1)/2 = (1 βˆ’ 𝑦)/4 = (𝑧 + 2)/1 Since the plane passes through line (1) All points of line (1) will be in the plane Now, for line (1) (1, 4, 4) lies in line (1) So, plane passes through (1, 4, 4). Thus, the equation of plane is a(x – 1) + b(y – 4) + c(z – 4) = 0 Where a, b, c are the direction ratios of normal to the plane Now, given that plane is parallel to line (2) ∴ Plane is parallel to line (1) and (2) Thus, Normal of plane is perpendicular to both line (1) and (2) ∴ Normal will be cross product of parallel vectors of line (1) and (2) Now, our lines are (π‘₯ βˆ’ 1)/3 = (𝑦 βˆ’ 4)/2 = (𝑧 βˆ’ 4)/(βˆ’2) (π‘₯ + 1)/2 = (1 βˆ’ 𝑦)/4 = (𝑧 + 2)/1 For line (2), we write y – 1 (π‘₯ + 1)/2 = (𝑦 βˆ’ 1)/(βˆ’4) = (𝑧 + 2)/1 So, Normal of plane = |β– 8(𝑖 Μ‚&𝑗 Μ‚&π‘˜ Μ‚@3&2&βˆ’2@2&βˆ’4&1)| = 𝑖 Μ‚ (2 Γ— 1 – (-4) Γ— (-2)) βˆ’ 𝑗 Μ‚ (3 Γ— 1 βˆ’ 2 Γ— (-2)) + π‘˜ Μ‚ (3 Γ— (-4) βˆ’ 2 Γ— 2) = 𝑖 Μ‚ (2 – 8) βˆ’ 𝑗 Μ‚ (3 + 4) + π‘˜ Μ‚ (–12 – 4) = –6𝑖 Μ‚ βˆ’ 7𝑗 Μ‚ – 16π‘˜ Μ‚ So, direction ratios of normal are –6, –7, –16 ∴ a = –6, b = –7, c = –16 And, equation of plane is a(x – 1) + b(y – 4) + c(z – 4) = 0 (–6)(x – 1) + (–7)(y – 4) + (–16)(z – 4) = 0 6(x – 1) + 7(y – 4) + 16(z – 4) = 0 6x – 6 + 7y – 28 + 16z – 64 = 0 6x + 7y + 16z – 98 = 0 Also, we need to find shortest distance between lines Now, equation of line (2) is (π‘₯ + 1)/2 = (𝑦 βˆ’ 1)/(βˆ’4) = (𝑧 + 2)/1 Shortest distance between lines = Perpendicular distance of point (–1, 1, –2) from plane 6x + 7y + 16z – 98 = 0 = |(𝐴π‘₯_1 + 〖𝐡𝑦〗_1 +γ€– 𝐢𝑧〗_1 βˆ’ 𝐷)/√(𝐴^2 + 𝐡^2 + 𝐢^2 )| Putting π‘₯_1 = βˆ’1, 𝑦_1 = 1, 𝑧_1 = –2 and A = 6, B = 7, C = 16, D = 98 = |((6 Γ— βˆ’1) + (7 Γ— 1) + (16 Γ— βˆ’2) βˆ’ 98 )/√(6^2 + 7^2 +16^2 )| = |(βˆ’6 + 7 βˆ’ 32 βˆ’ 98)/√(36 + 49 + 256)| = |(βˆ’129)/√341| = 129/√341 Thus, shortest distance πŸπŸπŸ—/βˆšπŸ‘πŸ’πŸ units

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Davneet Singh
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.