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Question 5 Find the equation of the plane passing through (a, b, c) and parallel to the plane ๐‘Ÿ โƒ— . (๐‘– ฬ‚ + ๐‘— ฬ‚ + ๐‘˜ ฬ‚) = 2.The equation of plane passing through (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is A(x โˆ’ x1) + B (y โˆ’ y1) + C(z โˆ’ z1) = 0 The plane passes through (a, b, c) So, x1 = ๐‘Ž, y1 = ๐‘, z1 = ๐‘ Since both planes are parallel to each other, their normal will be parallel โˆด Direction ratios of normal = Direction ratios of normal of ๐‘Ÿ โƒ—.(๐‘– ฬ‚ + ๐‘— ฬ‚ + ๐‘˜ ฬ‚) = 2 Direction ratios of normal = 1, 1, 1 โˆด A = 1, B = 1, C = 1 Thus, Equation of plane in Cartesian form is A(x โˆ’ x1) + B (y โˆ’ y1) + C(z โˆ’ z1) = 0 1(x โˆ’ ๐‘Ž) + 1(y โˆ’ b) + 1(z โˆ’ c) = 0 x โˆ’ a + y โˆ’ b + z โˆ’ c = 0 x + y + z โˆ’ (a + b + c) = 0 x + y + z = a + b + c โˆด Direction ratios of normal = Direction ratios of normal of ๐‘Ÿ โƒ—.(๐‘– ฬ‚ + ๐‘— ฬ‚ + ๐‘˜ ฬ‚) = 2 Direction ratios of normal = 1, 1, 1 โˆด A = 1, B = 1, C = 1 Thus, Equation of plane in Cartesian form is A(x โˆ’ x1) + B (y โˆ’ y1) + C(z โˆ’ z1) = 0 1(x โˆ’ ๐‘Ž) + 1(y โˆ’ b) + 1(z โˆ’ c) = 0 x โˆ’ a + y โˆ’ b + z โˆ’ c = 0 x + y + z โˆ’ (a + b + c) = 0 x + y + z = a + b + c

  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo