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Example 6 - Line through (5, 2, -4), parallel to 3i + 2j - 8k

Example, 6 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Example, 6 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3

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Transcript

Example 6 Find the vector and the Cartesian equations of the line through the point (5, 2, – 4) and which is parallel to the vector 3𝑖 Μ‚ + 2𝑗 Μ‚ – 8π‘˜ Μ‚ . Vector equation Equation of a line passing through a point with position vector π‘Ž βƒ— , and parallel to a vector 𝑏 βƒ— is π‘Ÿ βƒ— = π‘Ž βƒ— + πœ†π‘ βƒ— Since line passes through (5, 2, βˆ’4) π‘Ž βƒ— = 5𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚ Since line is parallel to 3𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 8π‘˜ Μ‚ 𝑏 βƒ— = 3𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 8π‘˜ Μ‚ Equation of line π‘Ÿ βƒ— = π‘Ž βƒ— + πœ†π‘ βƒ— 𝒓 βƒ— = (5π’Š Μ‚ + 2𝒋 Μ‚ βˆ’ 4π’Œ Μ‚) + πœ† (3π’Š Μ‚ + 2𝒋 Μ‚ βˆ’ 8π’Œ Μ‚) Therefore, equation of line in vector form is π‘Ÿ βƒ— = (5𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚) + πœ† (3𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 8π‘˜ Μ‚) Cartesian equation Equation of a line passing through a point (x, y, z) and parallel to a line with direction ratios a, b, c is (π‘₯ βˆ’ π‘₯1)/π‘Ž = (𝑦 βˆ’ 𝑦1)/𝑏 = (𝑧 βˆ’ 𝑧1)/𝑐 Since line passes through (5, 2, βˆ’4) π‘₯1 = 5, y1 = 2 , z1 = βˆ’4 .....................................

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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.