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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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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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