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 12 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, โ€“ 4) and which is parallel to the vector 3๐‘–ย ฬ‚ + 2๐‘—ย ฬ‚ โ€“ 8๐‘˜ย ฬ‚ . Vector equation Equation of a line passing through a point with position vector ๐‘Žย โƒ— , and parallel to a vector ๐‘ย โƒ— is ๐‘Ÿย โƒ— = ๐‘Žย โƒ— + ๐œ†๐‘ย โƒ— Since line passes through (5, 2, โˆ’ 4) ๐‘Žย โƒ— = 5๐‘–ย ฬ‚ + 2๐‘—ย ฬ‚ โˆ’ 4๐‘˜ย ฬ‚ Since line is parallel to 3๐‘–ย ฬ‚ + 2๐‘—ย ฬ‚ โˆ’ 8๐‘˜ย ฬ‚ ๐‘ย โƒ— = 3๐‘–ย ฬ‚ + 2๐‘—ย ฬ‚ โˆ’ 8๐‘˜ย ฬ‚ Equation of line ๐‘Ÿย โƒ— = ๐‘Žย โƒ— + ๐œ†๐‘ย โƒ— ๐’“ย โƒ— = (5๐’Šย ฬ‚ + 2๐’‹ย ฬ‚ โˆ’ 4๐’Œย ฬ‚) + ๐œ† (3๐’Šย ฬ‚ + 2๐’‹ย ฬ‚ โˆ’ 8๐’Œย ฬ‚) Therefore, equation of line in vector form is ๐‘Ÿย โƒ— = (5๐‘–ย ฬ‚ + 2๐‘—ย ฬ‚ โˆ’ 4๐‘˜ย ฬ‚) + ๐œ† (3๐‘–ย ฬ‚ + 2๐‘—ย ฬ‚ โˆ’ 8๐‘˜ย ฬ‚) 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 (๐‘ฅ โˆ’ ๐‘ฅ1)/๐‘Ž = (๐‘ฆ โˆ’ ๐‘ฆ1)/๐‘ = (๐‘ง โˆ’ ๐‘ง1)/๐‘ Since line passes through (5, 2, โˆ’4) ๐‘ฅ1 = 5, y1 = 2 , z1 = โˆ’4 Also, line is parallel to 3๐‘–ย ฬ‚ + 2๐‘—ย ฬ‚ โˆ’ 8๐‘˜ย ฬ‚ , ๐‘Ž = 3, b = 2, c = โˆ’ 8 Equation of line in Cartesian form is (๐‘ฅ โˆ’ ๐‘ฅ1)/๐‘Ž = (๐‘ฆ โˆ’ ๐‘ฆ1)/๐‘ = (๐‘ง โˆ’ ๐‘ง1)/๐‘ (๐‘ฅ โˆ’ 5)/3 = (๐‘ฆ โˆ’ 2)/2 = (๐‘ง โˆ’ ( โˆ’ 4))/( โˆ’ 8) (๐’™ โˆ’ ๐Ÿ“)/๐Ÿ‘ = (๐’š โˆ’ ๐Ÿ)/๐Ÿ = (๐’› + ๐Ÿ’)/(โˆ’๐Ÿ–)

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