Examples

Chapter 11 Class 12 Three Dimensional Geometry
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Question 7 (introduction) Find the vector and cartesian equations of the plane which passes through the point (5, 2, โ 4) and perpendicular to the line with direction ratios 2, 3, โ 1.Vector equation of a plane passing through a point (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is [๐ โ โ(๐ฅ1๐ ฬ + ๐ฆ1๐ ฬ + ๐ง1๐ ฬ)]. (A๐ ฬ + B๐ ฬ + C๐ ฬ) = 0 or (๐ โ โ ๐ โ).๐ โ = 0 ("A" ๐) โ is perpendicular to "n" โ So, ("A" P) โ . "n" โ = 0 ("r" โ โ "a" โ)."n" โ = 0 Question 7 Find the vector and Cartesian equations of the plane which passes through the point (5, 2, โ 4) and perpendicular to the line with direction ratios 2, 3, โ 1.Vector form Equation of plane passing through point A whose position vector is ๐ โ & perpendicular to ๐ โ is (๐ โ โ ๐ โ) . ๐ โ = 0 Given Plane passes through (5, 2, โ4) So ๐ โ = 5๐ ฬ + 2๐ ฬ โ 4๐ ฬ Direction ratios of line perpendicular to plane = 2, 3, โ1 So, "n" โ = 2๐ ฬ + 3๐ ฬ โ 1๐ ฬ Equation of plane in vector form is (๐ โ โ ๐ โ) . ๐ โ = 0 [๐ โโ(๐๐ ฬ+๐๐ ฬโ๐๐ ฬ)]. (๐๐ ฬ+๐๐ ฬโ๐ ฬ) = 0 Now, Finding Cartesian form using two methods Cartesian form (Method 1) Vector equation is [๐ โโ(5๐ ฬ+2๐ ฬโ4๐ ฬ)]. (2๐ ฬ+3๐ ฬโ๐ ฬ) = 0 Put ๐ โ = x๐ ฬ + y๐ ฬ + z๐ ฬ [(๐ฅ๐ ฬ+๐ฆ๐ ฬ+๐ง๐ ฬ )โ(5๐ ฬ+2๐ ฬโ4๐ ฬ)].(2๐ ฬ + 3๐ ฬ โ ๐ ฬ) = 0 [(๐ฅโ5) ๐ ฬ+(๐ฆโ2) ๐ ฬ+ (๐งโ(โ 4))๐ ฬ ].(2๐ ฬ + 3๐ ฬ โ ๐ ฬ) = 0 2(x โ 5) + 3 (y โ 2) + (โ 1)(z + 4) = 0 2x โ 10 + 3y โ 6 โ z โ 4 = 0 2x + 3y โ z โ 20 = 0 2x + 3y โ z = 20 Therefore equation of plane in Cartesian form is 2x + 3y โ z = 20 Cartesian form (Method 2) Equation of plane passing through (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is A(x โ x1) + B(y โ y1) + c (z โ z1) = 0 Since the plane passes through (5, 2, โ4) x1 = 5, y1 = 2, z1 = โ4 Direction ratios of line perpendicular to plane = 2, 3, โ1 โด A = 2, B = 3, C = โ1 Therefore, equation of line in Cartesian form is 2(x โ 5) + 3 (y โ 2) + (โ1) (x โ (โ4)) = 0 2 (x โ 5) + 3(y โ 2) โ 1 (z + 4) = 0 2x โ 10 + 3y โ 6 โ z โ 4 = 0 2x + 3y โ z โ 20 = 0 2x + 3y โ z = 20 Therefore equation of plane in Cartesian form is 2x + 3y โ z = 20