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, – 2) and the normal to the plane is 𝑖﷯ + 𝑗﷯ − 𝑘﷯. Vector equation of a plane passing through a point (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is 𝑟﷯ −(𝑥1 𝑖﷯ + 𝑦1 𝑗﷯ + 𝑧1 𝑘﷯)﷯. (A 𝑖﷯ + B 𝑗﷯ + C 𝑘﷯) = 0 or ( 𝑟﷯ − 𝑎﷯). 𝑛﷯ = 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 𝑖﷯ + 𝑗﷯ − 𝑘﷯. Vector equation Equation of plane passing through point A whose position vector is 𝒂﷯ & perpendicular to 𝒏﷯ is ( 𝒓﷯ − 𝒂﷯) . 𝒏﷯ = 0 Given Plane passes through (1, 0, − 2) So, 𝒂﷯ = 1 𝑖﷯ + 0 𝑗﷯ – 2 𝑘﷯ Normal to plane = 𝑖﷯ + 𝑗﷯ – 𝑘﷯ 𝒏﷯ = 𝑖﷯ + 𝑗﷯ − 𝑘﷯ Vector equation of plane is ( 𝒓﷯ − 𝒂﷯) . 𝒏﷯ = 0 𝑟﷯−(1 𝑖﷯+0 𝑗﷯−2 𝑘﷯)﷯.( 𝑖﷯ + 𝑗﷯ − 𝑘﷯) = 0 𝒓﷯−( 𝒊﷯−2 𝒌﷯)﷯ . ( 𝒊﷯+ 𝒋﷯ − 𝒌﷯) = 0 Cartesian form (Method 1): Vector equation is 𝑟﷯−( 𝑖﷯−2 𝑘﷯)﷯ . ( 𝑖﷯+ 𝑗﷯ − 𝑘﷯) = 0 Put 𝒓﷯ = x 𝒊﷯ + y 𝒋﷯ + z 𝒌﷯ x 𝑖﷯+y 𝑗﷯+z 𝑘﷯﷯−(1 𝑖﷯+0 𝑗﷯−2 𝑘﷯)﷯.(1 𝑖﷯ + 1 𝑗﷯ – 1 𝑘﷯) = 0 x−1﷯ 𝑖﷯+ y−0﷯ 𝐽﷯+(𝑧− −2﷯) 𝑘﷯﷯.(1 𝑖﷯ + 1 𝑗﷯ – 1 𝑘﷯) = 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 𝑖﷯ + 𝑗﷯ – 𝑘﷯ 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 𝑖﷯ – 2 𝑗﷯ + 𝑘﷯ Vector equation Equation of plane passing through point A whose position vector is 𝒂﷯ & perpendicular to 𝒏﷯ is ( 𝒓﷯ − 𝒂﷯) . 𝒏﷯ = 0 Given, Plane passes through (1, 4, 6) So, 𝒂﷯ = 1 𝑖﷯ + 4 𝑗﷯ + 6 𝑘﷯ Normal to plane = 𝑖﷯ – 2 𝑗﷯ + 𝑘﷯ 𝒏﷯ = 𝑖﷯ – 2 𝑗﷯ + 𝑘﷯ Vector equation of plane is ( 𝑟﷯ − 𝑎﷯). 𝑛﷯ = 0 𝒓﷯−( 𝒊﷯+𝟒 𝒋﷯+𝟔 𝒌﷯)﷯ . ( 𝒊﷯ − 2 𝒋﷯ + 𝒌﷯) = 0 Cartesian form (Method 1) : Vector equation is 𝑟﷯−( 𝑖﷯+4 𝑗﷯+6 𝑘﷯)﷯ . ( 𝑖﷯ − 2 𝑗﷯ + 𝑘﷯) = 0 Put 𝒓﷯ = x 𝒊﷯ + y 𝒋﷯ + z 𝒌﷯ 𝑥 𝑖﷯+𝑦 𝑗﷯+𝑧 𝑘﷯﷯−(1 𝑖﷯+4 𝑗﷯+6 𝑘﷯)﷯. (1 𝑖﷯ − 2 𝑗﷯ + 1 𝑘﷯) = 0 𝑥−1﷯ 𝑖﷯ + 𝑦− 4﷯ 𝑗﷯+(𝑧− 6) 𝑘﷯﷯. (1 𝑖﷯ − 2 𝑗﷯ + 1 𝑘﷯) = 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 𝑖﷯ – 2 𝑗﷯ + 𝑘﷯ 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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