Misc 8 - Find equation of plane passing through (a, b, c) and parallel

Misc 8 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2

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

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