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Recent questions tagged telegram-57400

(chatquestion)...EXAMPLE 18 Find the equation of the plane passing through the points (-1, 1, 1) and (1,-1, 1) and ls perpendicular to the plane x + 2y +2z 5 O Solution The equation of a plane passing through the point (-1, 1, 1) is a(x+1) + b(y-1) + c(z-1) = 0 ...(1) [where a, b, c are constants] Now, the plane (1) passes through the point (1, -1, 1) a(1 + 1) + b(-1 - 1) + c(1-1) = 0 2a - 2b = 0 or, a - b+0 c= 0 or, ...(2) Since the plane (1) is perpendicular to the plane x + 2y + 2z = 5 (given), so, normals to the planes (1) and x + 2y + 2z = 5 are are perpendicular to each other. a + 2b + 2c = 0 ..*(3) Hence, from (2) and (3) [by cross-multiplication] we get, 9. -k (+0, say) %3D 3. a = -2k, b =-2k, c= 3k in (1) we get, - 2k(x + 1) - 2k(y-1)+3k(z-1) = 0 - 2x- 2-2y + 2 + 3z -3 = 0 [: k#0] or, 2x + 2y-3z+3 = 0 Therefore, the equation of the required plane is or, 2x + 2y-3z +3 = 0

asked Feb 17, 2020 by balaji.thirumalai

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