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
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Ex 11.3, 5 (Introduction) Find the vector and cartesian equations of the planes (a) that passes through the point (1, 0, โ€“ 2) and the normal to the plane is ๐‘–๏ทฏ + ๐‘—๏ทฏ โˆ’ ๐‘˜๏ทฏ. 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‘Ÿ๏ทฏ โˆ’(๐‘ฅ1 ๐‘–๏ทฏ + ๐‘ฆ1 ๐‘—๏ทฏ + ๐‘ง1 ๐‘˜๏ทฏ)๏ทฏ. (A ๐‘–๏ทฏ + B ๐‘—๏ทฏ + C ๐‘˜๏ทฏ) = 0 or ( ๐‘Ÿ๏ทฏ โˆ’ ๐‘Ž๏ทฏ). ๐‘›๏ทฏ = 0 Ex 11.3, 5 Find the vector and Cartesian equations of the planes (a) that passes through the point (1, 0, โ€“ 2) and the normal to the plane is ๐‘–๏ทฏ + ๐‘—๏ทฏ โˆ’ ๐‘˜๏ทฏ. Vector equation Equation of plane passing through point A whose position vector is ๐’‚๏ทฏ & perpendicular to ๐’๏ทฏ is ( ๐’“๏ทฏ โˆ’ ๐’‚๏ทฏ) . ๐’๏ทฏ = 0 Given Plane passes through (1, 0, โˆ’ 2) So, ๐’‚๏ทฏ = 1 ๐‘–๏ทฏ + 0 ๐‘—๏ทฏ โ€“ 2 ๐‘˜๏ทฏ Normal to plane = ๐‘–๏ทฏ + ๐‘—๏ทฏ โ€“ ๐‘˜๏ทฏ ๐’๏ทฏ = ๐‘–๏ทฏ + ๐‘—๏ทฏ โˆ’ ๐‘˜๏ทฏ Vector equation of plane is ( ๐’“๏ทฏ โˆ’ ๐’‚๏ทฏ) . ๐’๏ทฏ = 0 ๐‘Ÿ๏ทฏโˆ’(1 ๐‘–๏ทฏ+0 ๐‘—๏ทฏโˆ’2 ๐‘˜๏ทฏ)๏ทฏ.( ๐‘–๏ทฏ + ๐‘—๏ทฏ โˆ’ ๐‘˜๏ทฏ) = 0 ๐’“๏ทฏโˆ’( ๐’Š๏ทฏโˆ’2 ๐’Œ๏ทฏ)๏ทฏ . ( ๐’Š๏ทฏ+ ๐’‹๏ทฏ โˆ’ ๐’Œ๏ทฏ) = 0 Cartesian form (Method 1): Vector equation is ๐‘Ÿ๏ทฏโˆ’( ๐‘–๏ทฏโˆ’2 ๐‘˜๏ทฏ)๏ทฏ . ( ๐‘–๏ทฏ+ ๐‘—๏ทฏ โˆ’ ๐‘˜๏ทฏ) = 0 Put ๐’“๏ทฏ = x ๐’Š๏ทฏ + y ๐’‹๏ทฏ + z ๐’Œ๏ทฏ x ๐‘–๏ทฏ+y ๐‘—๏ทฏ+z ๐‘˜๏ทฏ๏ทฏโˆ’(1 ๐‘–๏ทฏ+0 ๐‘—๏ทฏโˆ’2 ๐‘˜๏ทฏ)๏ทฏ.(1 ๐‘–๏ทฏ + 1 ๐‘—๏ทฏ โ€“ 1 ๐‘˜๏ทฏ) = 0 xโˆ’1๏ทฏ ๐‘–๏ทฏ+ yโˆ’0๏ทฏ ๐ฝ๏ทฏ+(๐‘งโˆ’ โˆ’2๏ทฏ) ๐‘˜๏ทฏ๏ทฏ.(1 ๐‘–๏ทฏ + 1 ๐‘—๏ทฏ โ€“ 1 ๐‘˜๏ทฏ) = 0 1(x โˆ’ 1) + 1(y โˆ’ 0) โ€“ 1 (z + 2) = 0 x โˆ’ 1 + y โ€“ z โ€“ 2 = 0 x + y โ€“ z = 3 Therefore, equation of plane in Cartesian form is x + y โ€“ 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 โˆ’ x1) + B(y โˆ’ y1) + C (z โˆ’ z1) = 0 Since the plane passes through (1, 0, โ€“2) x1 = 1 , y1 = 0 , z1 = 2 And normal is ๐‘–๏ทฏ + ๐‘—๏ทฏ โ€“ ๐‘˜๏ทฏ So, Direction ratios of line perpendicular to plane = 1, 1, โ€“1 โˆด A = 1, B = 1, C = โ€“1 Therefore, equations of line in Cartesian form is 1(x โˆ’ 1) + 1(y โˆ’ 0) + 1(z โˆ’ (-2)) = 0 x + y โ€“ 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 ๐‘–๏ทฏ โ€“ 2 ๐‘—๏ทฏ + ๐‘˜๏ทฏ Vector equation Equation of plane passing through point A whose position vector is ๐’‚๏ทฏ & perpendicular to ๐’๏ทฏ is ( ๐’“๏ทฏ โˆ’ ๐’‚๏ทฏ) . ๐’๏ทฏ = 0 Given, Plane passes through (1, 4, 6) So, ๐’‚๏ทฏ = 1 ๐‘–๏ทฏ + 4 ๐‘—๏ทฏ + 6 ๐‘˜๏ทฏ Normal to plane = ๐‘–๏ทฏ โ€“ 2 ๐‘—๏ทฏ + ๐‘˜๏ทฏ ๐’๏ทฏ = ๐‘–๏ทฏ โ€“ 2 ๐‘—๏ทฏ + ๐‘˜๏ทฏ Vector equation of plane is ( ๐‘Ÿ๏ทฏ โˆ’ ๐‘Ž๏ทฏ). ๐‘›๏ทฏ = 0 ๐’“๏ทฏโˆ’( ๐’Š๏ทฏ+๐Ÿ’ ๐’‹๏ทฏ+๐Ÿ” ๐’Œ๏ทฏ)๏ทฏ . ( ๐’Š๏ทฏ โˆ’ 2 ๐’‹๏ทฏ + ๐’Œ๏ทฏ) = 0 Cartesian form (Method 1) : Vector equation is ๐‘Ÿ๏ทฏโˆ’( ๐‘–๏ทฏ+4 ๐‘—๏ทฏ+6 ๐‘˜๏ทฏ)๏ทฏ . ( ๐‘–๏ทฏ โˆ’ 2 ๐‘—๏ทฏ + ๐‘˜๏ทฏ) = 0 Put ๐’“๏ทฏ = x ๐’Š๏ทฏ + y ๐’‹๏ทฏ + z ๐’Œ๏ทฏ ๐‘ฅ ๐‘–๏ทฏ+๐‘ฆ ๐‘—๏ทฏ+๐‘ง ๐‘˜๏ทฏ๏ทฏโˆ’(1 ๐‘–๏ทฏ+4 ๐‘—๏ทฏ+6 ๐‘˜๏ทฏ)๏ทฏ. (1 ๐‘–๏ทฏ โˆ’ 2 ๐‘—๏ทฏ + 1 ๐‘˜๏ทฏ) = 0 ๐‘ฅโˆ’1๏ทฏ ๐‘–๏ทฏ + ๐‘ฆโˆ’ 4๏ทฏ ๐‘—๏ทฏ+(๐‘งโˆ’ 6) ๐‘˜๏ทฏ๏ทฏ. (1 ๐‘–๏ทฏ โˆ’ 2 ๐‘—๏ทฏ + 1 ๐‘˜๏ทฏ) = 0 1(x โˆ’ 1) + (โˆ’2)(y โˆ’ 4) +1 (z โˆ’ 6) = 0 x โˆ’ 1 โˆ’ 2(y โˆ’ 4) + z โˆ’ 6 = 0 x โˆ’ 2y + z + 1 = 0 โˆด Equation of plane in Cartesian form is x โˆ’ 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 โˆ’ x1) + B(y โˆ’ y1) + C (z โˆ’ z1) = 0 Since the plane passes through (1, 4, 6) x1 = 1 , y1 = 4 , z1 = 6 And normal is ๐‘–๏ทฏ โ€“ 2 ๐‘—๏ทฏ + ๐‘˜๏ทฏ So, Direction ratios of line perpendicular to plane = 1, โ€“2, 1 โˆด A = 1, B = โ€“2, C = 1 Therefore, equations of line in Cartesian form is 1(x โˆ’ 1) โ€“ 2 (y โˆ’ 4) + 1(x โˆ’ 6) = 0 x โˆ’ 2y + z + 1 = 0

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