Misc 19 - Find the vector equation of the line passing through (1, 2, 3) and parallel to planes r.(i-j+2k) = 5 & r.(3i+j+k) = 6 - Equation of line under planes condition

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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise
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Misc 19 (Method 1) Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes ๐‘Ÿ๏ทฏ . ( ๐‘–๏ทฏ โˆ’ ๐‘—๏ทฏ + 2 ๐‘˜๏ทฏ) = 5 and ๐‘Ÿ๏ทฏ . (3 ๐‘–๏ทฏ + ๐‘—๏ทฏ + ๐‘˜๏ทฏ) = 6 . The vector equation of a line passing through a point with position vector ๐‘Ž๏ทฏ and parallel to a vector ๐‘๏ทฏ is ๐’“๏ทฏ = ๐’‚๏ทฏ + ๐œ† ๐’ƒ๏ทฏ Given, the line passes through (1, 2, 3) So, ๐‘Ž๏ทฏ = 1 ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ + 3 ๐‘˜๏ทฏ Given, line is parallel to both planes โˆด Line is perpendicular to normal of both planes. i.e. ๐‘๏ทฏ is perpendicular to normal of both planes. We know that ๐‘Ž๏ทฏ ร— ๐‘๏ทฏ is perpendicular to both ๐‘Ž๏ทฏ & ๐‘๏ทฏ So, ๐‘๏ทฏ is cross product of normals of planes ๐‘Ÿ๏ทฏ . ( ๐‘–๏ทฏ โˆ’ ๐‘—๏ทฏ + 2 ๐‘˜๏ทฏ) = 5 and ๐‘Ÿ๏ทฏ . (3 ๐‘–๏ทฏ + ๐‘—๏ทฏ + ๐‘˜๏ทฏ) = 6 Required normal = ๐‘–๏ทฏ๏ทฎ ๐‘—๏ทฏ๏ทฎ ๐‘˜๏ทฏ๏ทฎ1๏ทฎโˆ’1๏ทฎ2๏ทฎ3๏ทฎ1๏ทฎ1๏ทฏ๏ทฏ = ๐‘–๏ทฏ (โ€“1(1) โ€“ 1(2)) โ€“ ๐‘—๏ทฏ (1(1) โ€“ 3(2)) + ๐‘˜๏ทฏ(1(1) โ€“ 3(โ€“1)) = ๐‘–๏ทฏ (โ€“1 โ€“ 2) โ€“ ๐‘—๏ทฏ (1 โ€“ 6) + ๐‘˜๏ทฏ(1 + 3) = โ€“3 ๐‘–๏ทฏ + 5 ๐‘—๏ทฏ + 4 ๐‘˜๏ทฏ Thus, ๐‘๏ทฏ = โˆ’3 ๐‘–๏ทฏ + 5 ๐‘—๏ทฏ + 4 ๐‘˜๏ทฏ Now, Putting value of ๐‘Ž๏ทฏ & ๐‘๏ทฏ in formula ๐‘Ÿ๏ทฏ = ๐‘Ž๏ทฏ + ๐œ† ๐‘๏ทฏ = ( ๐’Š๏ทฏ + 2 ๐’‹๏ทฏ + 3 ๐’Œ๏ทฏ) + ๐œ† (โˆ’3 ๐’Š๏ทฏ + 5 ๐’‹๏ทฏ + 4 ๐’Œ๏ทฏ) Therefore, the equation of the line is ( ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ + 3 ๐‘˜๏ทฏ) + ๐œ† (โˆ’3 ๐‘–๏ทฏ + 5 ๐‘—๏ทฏ + 4 ๐‘˜๏ทฏ). Misc 19 (Method 2) Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes ๐‘Ÿ๏ทฏ . ( ๐‘–๏ทฏ โˆ’ ๐‘—๏ทฏ + 2 ๐‘˜๏ทฏ) = 5 and ๐‘Ÿ๏ทฏ . (3 ๐‘–๏ทฏ + ๐‘—๏ทฏ + ๐‘˜๏ทฏ) = 6 . The vector equation of a line passing through a point with position vector ๐‘Ž๏ทฏ and parallel to a vector ๐‘๏ทฏ is ๐’“๏ทฏ = ๐’‚๏ทฏ + ๐œ† ๐’ƒ๏ทฏ Given, the line passes through (1, 2, 3) So, ๐‘Ž๏ทฏ = 1 ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ + 3 ๐‘˜๏ทฏ Let ๐‘๏ทฏ = ๐‘๏ทฎ1๏ทฏ ๐‘–๏ทฏ + ๐‘๏ทฎ2๏ทฏ ๐‘—๏ทฏ + ๐‘๏ทฎ3๏ทฏ ๐‘˜๏ทฏ A line parallel to a plane is perpendicular to the normal of the plane. And two lines ๐‘๏ทฏ and ๐‘ž๏ทฏ are perpendicular if ๐‘๏ทฏ. ๐‘ž๏ทฏ = 0 Given, the line is parallel to planes So, our equations are ๐‘1 โˆ’ ๐‘2 + 2๐‘3 = 0 3๐‘1 + ๐‘2 + ๐‘3 = 0 Thus, ๐‘๏ทฏ = ๐‘๏ทฎ1๏ทฏ ๐‘–๏ทฏ + ๐‘๏ทฎ2๏ทฏ ๐‘—๏ทฏ + ๐‘๏ทฎ3๏ทฏ ๐‘˜๏ทฏ = โˆ’3k ๐‘–๏ทฏ + 5k ๐‘—๏ทฏ + 4k ๐‘˜๏ทฏ Now, Putting value of ๐‘Ž๏ทฏ & ๐‘๏ทฏ in formula ๐‘Ÿ๏ทฏ = ๐‘Ž๏ทฏ + ๐œ† ๐‘๏ทฏ โˆด ๐‘Ÿ๏ทฏ = (1 ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ + 3 ๐‘˜๏ทฏ) + ๐œ† (โˆ’3k ๐‘–๏ทฏ + 5k ๐‘—๏ทฏ + 4k ๐‘˜๏ทฏ) = ( ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ + 3 ๐‘˜๏ทฏ) + ๐œ†k (โˆ’3 ๐‘–๏ทฏ + 5 ๐‘—๏ทฏ + 4 ๐‘˜๏ทฏ) = ( ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ + 3 ๐‘˜๏ทฏ) + ๐œ† (โˆ’3 ๐‘–๏ทฏ + 5 ๐‘—๏ทฏ + 4 ๐‘˜๏ทฏ) Therefore, the equation of the line is ( ๐’Š๏ทฏ + 2 ๐’‹๏ทฏ + 3 ๐’Œ๏ทฏ) + ๐œ† (โˆ’3 ๐’Š๏ทฏ + 5 ๐’‹๏ทฏ + 4 ๐’Œ๏ทฏ).

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.