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Ex 11.2, 5 - Find equation of line in vector, cartesian form - Ex 11.2

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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise
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Ex 11.2, 5 Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2 ๐‘–๏ทฏ โˆ’ ๐‘—๏ทฏ + 4 ๐‘˜๏ทฏ and is in the direction ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ โˆ’ ๐‘˜๏ทฏ . Equation of a line passing though a point with position vector ๐‘Ž๏ทฏ and parallel to vector ๐‘๏ทฏ is ๐‘Ÿ๏ทฏ = ๐‘Ž๏ทฏ + ๐œ† ๐‘๏ทฏ Here, ๐‘Ž๏ทฏ = 2 ๐‘–๏ทฏ โˆ’ ๐‘—๏ทฏ + 4 ๐‘˜๏ทฏ & ๐‘๏ทฏ = ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ โˆ’ ๐‘˜๏ทฏ So, ๐‘Ÿ๏ทฏ = (2 ๐’Š๏ทฏ โˆ’ ๐’‹๏ทฏ + 4 ๐’Œ๏ทฏ) + ๐œ† ( ๐’Š๏ทฏ + 2 ๐’‹๏ทฏ โˆ’ ๐’Œ๏ทฏ) โˆด Equation of line in vector form is (2 ๐‘–๏ทฏ โˆ’ ๐‘—๏ทฏ + 4 ๐‘˜๏ทฏ) + ๐œ† ( ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ โˆ’ ๐‘˜๏ทฏ) Equation of a line passing though (x1, y1, z1) and parallel to a line having direction ratios a, b, c is ๐‘ฅ โˆ’ ๐‘ฅ1๏ทฎ๐‘Ž๏ทฏ = ๐‘ฆ โˆ’ ๐‘ฆ1๏ทฎ๐‘๏ทฏ = ๐‘ง โˆ’ ๐‘ง1๏ทฎ๐‘๏ทฏ Since the line passes through a point with position vector 2 ๐‘–๏ทฏ โˆ’ ๐‘—๏ทฏ + 4 ๐‘˜๏ทฏ, โˆด ๐‘ฅ1 = 2, y1 = โˆ’ 1 z1 = 4 Also, line is in the direction of ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ โˆ’ ๐‘˜๏ทฏ, Direction ratios : ๐‘Ž = 1, b = 2 , c = โˆ’1 Equation of line in Cartesian form is ๐‘ฅ โˆ’ 2๏ทฎ1๏ทฏ = ๐‘ฆ โˆ’ ( โˆ’ 1)๏ทฎ2๏ทฏ = ๐‘ง โˆ’ 4๏ทฎ โˆ’ 1๏ทฏ ๐’™ โˆ’ ๐Ÿ๏ทฎ๐Ÿ๏ทฏ = ๐’š + ๐Ÿ๏ทฎ๐Ÿ๏ทฏ = ๐’› โˆ’ ๐Ÿ’๏ทฎ โˆ’ ๐Ÿ๏ทฏ

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