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

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  1. Chapter 11 Class 12 Three Dimensional Geometry
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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‘– ฬ‚ + 2๐‘— ฬ‚ โ€“ 8๐‘˜ ฬ‚ . Vector equation Equation of a line passing through a point with position vector ๐‘Ž โƒ— , and parallel to a vector ๐‘ โƒ— is ๐‘Ÿ โƒ— = ๐‘Ž โƒ— + ๐œ†๐‘ โƒ— Since line passes through (5, 2, โˆ’ 4) ๐‘Ž โƒ— = 5๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚ Since line is parallel to 3๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 8๐‘˜ ฬ‚ ๐‘ โƒ— = 3๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 8๐‘˜ ฬ‚ Equation of line ๐‘Ÿ โƒ— = ๐‘Ž โƒ— + ๐œ†๐‘ โƒ— ๐’“ โƒ— = (5๐’Š ฬ‚ + 2๐’‹ ฬ‚ โˆ’ 4๐’Œ ฬ‚) + ๐œ† (3๐’Š ฬ‚ + 2๐’‹ ฬ‚ โˆ’ 8๐’Œ ฬ‚) Therefore, equation of line in vector form is ๐‘Ÿ โƒ— = (5๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚) + ๐œ† (3๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 8๐‘˜ 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‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 8๐‘˜ 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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