Plane

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

### Transcript

Question 5 Find the vector and cartesian equations of the planes (b) That passes through the point (1, 4, 6) and the normal vector to the plane is ๐ ฬ โ 2๐ ฬ + ๐ ฬ Vector equation 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 (1, 4, 6) So, ๐ โ = 1๐ ฬ + 4๐ ฬ + 6๐ ฬ Normal to plane = ๐ ฬ โ 2๐ ฬ + ๐ ฬ ๐ โ = ๐ ฬ โ 2๐ ฬ + ๐ ฬ Vector equation of plane is (๐ โ โ ๐ โ).๐ โ = 0 [๐ โโ(๐ ฬ+๐๐ ฬ+๐๐ ฬ)] . (๐ ฬ โ 2๐ ฬ + ๐ ฬ) = 0 Now, Finding Cartesian form using two methods Cartesian form (Method 1) : Vector equation is [๐ โโ(๐ ฬ+4๐ ฬ+6๐ ฬ)] . (๐ ฬ โ 2๐ ฬ + ๐ ฬ) = 0 Put ๐ โ = x๐ ฬ + y๐ ฬ + z๐ ฬ [(๐ฅ๐ ฬ+๐ฆ๐ ฬ+๐ง๐ ฬ )โ(1๐ ฬ+4๐ ฬ+6๐ ฬ)]. (1๐ ฬ โ 2๐ ฬ + 1๐ ฬ) = 0 [(๐ฅโ1) ๐ ฬ +(๐ฆโ4) ๐ ฬ+(๐งโ6)๐ ฬ ]. (1๐ ฬ โ 2๐ ฬ + 1๐ ฬ) = 0 1(x โ 1) + (โ2)(y โ 4) + 1 (z โ 6) = 0 x โ 1 โ 2(y โ 4) + z โ 6 = 0 x โ 2y + z + 1 = 0 โด Equation of plane in Cartesian form is x โ 2y + z + 1 = 0. 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 (1, 4, 6) x1 = 1, y1 = 4, z1 = 6 And normal is ๐ ฬ โ 2๐ ฬ + ๐ ฬ So, Direction ratios of line perpendicular to plane = 1, โ2, 1 โด A = 1, B = โ2, C = 1 Therefore, equation of line in Cartesian form is 1(x โ 1) โ 2 (y โ 4) + 1(x โ 6) = 0 x โ 2y + z + 1 = 0