Last updated at Dec. 8, 2016 by Teachoo

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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๏ทฏ . ( ๐๏ทฏ โ ๐๏ทฏ + 2 ๐๏ทฏ) = 5 and ๐๏ทฏ . (3 ๐๏ทฏ + ๐๏ทฏ + ๐๏ทฏ) = 6 . The vector equation of a line passing through a point with position vector ๐๏ทฏ and parallel to a vector ๐๏ทฏ is ๐๏ทฏ = ๐๏ทฏ + ๐ ๐๏ทฏ Given, the line passes through (1, 2, 3) So, ๐๏ทฏ = 1 ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ Given, line is parallel to both planes โด Line is perpendicular to normal of both planes. i.e. ๐๏ทฏ is perpendicular to normal of both planes. We know that ๐๏ทฏ ร ๐๏ทฏ is perpendicular to both ๐๏ทฏ & ๐๏ทฏ So, ๐๏ทฏ is cross product of normals of planes ๐๏ทฏ . ( ๐๏ทฏ โ ๐๏ทฏ + 2 ๐๏ทฏ) = 5 and ๐๏ทฏ . (3 ๐๏ทฏ + ๐๏ทฏ + ๐๏ทฏ) = 6 Required normal = ๐๏ทฏ๏ทฎ ๐๏ทฏ๏ทฎ ๐๏ทฏ๏ทฎ1๏ทฎโ1๏ทฎ2๏ทฎ3๏ทฎ1๏ทฎ1๏ทฏ๏ทฏ = ๐๏ทฏ (โ1(1) โ 1(2)) โ ๐๏ทฏ (1(1) โ 3(2)) + ๐๏ทฏ(1(1) โ 3(โ1)) = ๐๏ทฏ (โ1 โ 2) โ ๐๏ทฏ (1 โ 6) + ๐๏ทฏ(1 + 3) = โ3 ๐๏ทฏ + 5 ๐๏ทฏ + 4 ๐๏ทฏ Thus, ๐๏ทฏ = โ3 ๐๏ทฏ + 5 ๐๏ทฏ + 4 ๐๏ทฏ Now, Putting value of ๐๏ทฏ & ๐๏ทฏ in formula ๐๏ทฏ = ๐๏ทฏ + ๐ ๐๏ทฏ = ( ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ) + ๐ (โ3 ๐๏ทฏ + 5 ๐๏ทฏ + 4 ๐๏ทฏ) Therefore, the equation of the line is ( ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ) + ๐ (โ3 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๏ทฏ . ( ๐๏ทฏ โ ๐๏ทฏ + 2 ๐๏ทฏ) = 5 and ๐๏ทฏ . (3 ๐๏ทฏ + ๐๏ทฏ + ๐๏ทฏ) = 6 . The vector equation of a line passing through a point with position vector ๐๏ทฏ and parallel to a vector ๐๏ทฏ is ๐๏ทฏ = ๐๏ทฏ + ๐ ๐๏ทฏ Given, the line passes through (1, 2, 3) So, ๐๏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๏ทฏ are perpendicular if ๐๏ทฏ. ๐๏ทฏ = 0 Given, the line is parallel to planes So, our equations are ๐1 โ ๐2 + 2๐3 = 0 3๐1 + ๐2 + ๐3 = 0 Thus, ๐๏ทฏ = ๐๏ทฎ1๏ทฏ ๐๏ทฏ + ๐๏ทฎ2๏ทฏ ๐๏ทฏ + ๐๏ทฎ3๏ทฏ ๐๏ทฏ = โ3k ๐๏ทฏ + 5k ๐๏ทฏ + 4k ๐๏ทฏ Now, Putting value of ๐๏ทฏ & ๐๏ทฏ in formula ๐๏ทฏ = ๐๏ทฏ + ๐ ๐๏ทฏ โด ๐๏ทฏ = (1 ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ) + ๐ (โ3k ๐๏ทฏ + 5k ๐๏ทฏ + 4k ๐๏ทฏ) = ( ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ) + ๐k (โ3 ๐๏ทฏ + 5 ๐๏ทฏ + 4 ๐๏ทฏ) = ( ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ) + ๐ (โ3 ๐๏ทฏ + 5 ๐๏ทฏ + 4 ๐๏ทฏ) Therefore, the equation of the line is ( ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ) + ๐ (โ3 ๐๏ทฏ + 5 ๐๏ทฏ + 4 ๐๏ทฏ).

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .