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Example 6 - Line through (5, 2, -4), parallel to 3i + 2j - 8k - Equation of line  - given point and //vector

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  1. Chapter 11 Class 12th Three Dimensional Geometry
  2. Serial order wise
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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 ̂ + 2𝑗 ̂ – 8π‘˜Β Μ‚ . Vector equation Equation of a line passing through a point with position vector π‘ŽΒ βƒ— , and parallel to a vector 𝑏 ⃗ is π‘ŸΒ βƒ— = π‘ŽΒ βƒ— + πœ†π‘Β βƒ— Since line passes through (5, 2, βˆ’ 4) π‘ŽΒ βƒ— = 5𝑖 ̂ + 2𝑗 ̂ βˆ’ 4π‘˜Β Μ‚ Since line is parallel to 3𝑖 ̂ + 2𝑗 ̂ βˆ’ 8π‘˜Β Μ‚ 𝑏 ⃗ = 3𝑖 ̂ + 2𝑗 ̂ βˆ’ 8π‘˜Β Μ‚ Equation of line π‘ŸΒ βƒ— = π‘ŽΒ βƒ— + πœ†π‘Β βƒ— 𝒓 ⃗ = (5π’ŠΒ Μ‚ + 2𝒋 ̂ βˆ’ 4π’ŒΒ Μ‚) + πœ† (3π’ŠΒ Μ‚ + 2𝒋 ̂ βˆ’ 8π’ŒΒ Μ‚) Therefore, equation of line in vector form is π‘ŸΒ βƒ— = (5𝑖 ̂ + 2𝑗 ̂ βˆ’ 4π‘˜Β Μ‚) + πœ† (3𝑖 ̂ + 2𝑗 ̂ βˆ’ 8π‘˜Β 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 Also, line is parallel to 3𝑖 ̂ + 2𝑗 ̂ βˆ’ 8π‘˜Β Μ‚ , π‘Ž = 3, b = 2, c = βˆ’ 8 Equation of line in Cartesian form is (π‘₯ βˆ’ π‘₯1)/π‘Ž = (𝑦 βˆ’ 𝑦1)/𝑏 = (𝑧 βˆ’ 𝑧1)/𝑐 (π‘₯ βˆ’ 5)/3 = (𝑦 βˆ’ 2)/2 = (𝑧 βˆ’ ( βˆ’ 4))/( βˆ’ 8) (𝒙 βˆ’ πŸ“)/πŸ‘ = (π’š βˆ’ 𝟐)/𝟐 = (𝒛 + πŸ’)/(βˆ’πŸ–)

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