Ex 11.3, 5 - Class 12 3D Geomerty - Find vector and cartesian - Equation of plane - Prependicular to Vector & Passing Through Point

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

Transcript

Ex 11.3, 5 (Introduction) Find the vector and cartesian equations of the planes (a) that passes through the point (1, 0, 2) and the normal to the plane is + . Vector equation of a plane passing through a point (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is ( 1 + 1 + 1 ) . (A + B + C ) = 0 or ( ). = 0 Ex 11.3, 5 Find the vector and Cartesian equations of the planes (a) that passes through the point (1, 0, 2) and the normal to the plane is + . Vector equation Equation of plane passing through point A whose position vector is & perpendicular to is ( ) . = 0 Given Plane passes through (1, 0, 2) So, = 1 + 0 2 Normal to plane = + = + Vector equation of plane is ( ) . = 0 (1 +0 2 ) .( + ) = 0 ( 2 ) . ( + ) = 0 Cartesian form (Method 1): Vector equation is ( 2 ) . ( + ) = 0 Put = x + y + z x +y +z (1 +0 2 ) .(1 + 1 1 ) = 0 x 1 + y 0 +( 2 ) .(1 + 1 1 ) = 0 1(x 1) + 1(y 0) 1 (z + 2) = 0 x 1 + y z 2 = 0 x + y z = 3 Therefore, equation of plane in Cartesian form is x + y z = 3 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 x1) + B(y y1) + C (z z1) = 0 Since the plane passes through (1, 0, 2) x1 = 1 , y1 = 0 , z1 = 2 And normal is + So, Direction ratios of line perpendicular to plane = 1, 1, 1 A = 1, B = 1, C = 1 Therefore, equations of line in Cartesian form is 1(x 1) + 1(y 0) + 1(z (-2)) = 0 x + y z = 3 Ex 11.3, 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 2 + Vector equation Equation of plane passing through point A whose position vector is & perpendicular to is ( ) . = 0 Given, Plane passes through (1, 4, 6) So, = 1 + 4 + 6 Normal to plane = 2 + = 2 + Vector equation of plane is ( ). = 0 ( + + ) . ( 2 + ) = 0 Cartesian form (Method 1) : Vector equation is ( +4 +6 ) . ( 2 + ) = 0 Put = x + y + z + + (1 +4 +6 ) . (1 2 + 1 ) = 0 1 + 4 +( 6) . (1 2 + 1 ) = 0 1(x 1) + ( 2)(y 4) +1 (z 6) = 0 x 1 2(y 4) + z 6 = 0 x 2y + z + 1 = 0 Equation of plane in Cartesian form is x 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 x1) + B(y y1) + C (z z1) = 0 Since the plane passes through (1, 4, 6) x1 = 1 , y1 = 4 , z1 = 6 And normal is 2 + So, Direction ratios of line perpendicular to plane = 1, 2, 1 A = 1, B = 2, C = 1 Therefore, equations of line in Cartesian form is 1(x 1) 2 (y 4) + 1(x 6) = 0 x 2y + z + 1 = 0

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