Ex 11.3, 14 - Chapter 11 Class 12 Three Dimensional Geometry

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Ex 11.3, 14 In the following cases, find the distance of each of the given points from the corresponding given plane. The distance of the point (x1, y1, z1) from the plane Ax + By + Cz = D is 𝑨 𝒙﷮𝟏﷯ + 𝑩𝒚﷮𝟏﷯ + 𝑪𝒛﷮𝟏﷯ − 𝑫﷮ ﷮ 𝑨﷮𝟐﷯ + 𝑩﷮𝟐﷯ + 𝑪﷮𝟐﷯﷯﷯﷯ Given, the point is (0, 0, 0) So, 𝑥﷮1﷯ = 0, 𝑦﷮1﷯ = 0, 𝑧﷮1﷯ = 0 and the equation of plane is 3x − 4y + 12z = 3 Comparing with Ax + By + Cz = D, A = 3, B = −4, C = 12, D = 3 Now, Distance of point from the plane is = 3 × 0﷯ + −4 × 0﷯ + 12 × 0﷯ − 3﷮ ﷮ 3﷮2﷯ + −4﷯﷮2﷯ + 12﷮2﷯﷯﷯﷯ = 0 + 0 + 0 − 3﷮ ﷮9 + 16 + 144﷯﷯﷯ = 3﷮ ﷮169﷯﷯﷯ = 𝟑﷮𝟏𝟑﷯ Ex 11.3, 14 In the following cases, find the distance of each of the given points from the corresponding given plane. The distance of the point (x1, y1, z1) from the plane Ax + By + Cz = D is 𝑨 𝒙﷮𝟏﷯ + 𝑩𝒚﷮𝟏﷯ + 𝑪𝒛﷮𝟏﷯ − 𝑫﷮ ﷮ 𝑨﷮𝟐﷯ + 𝑩﷮𝟐﷯ + 𝑪﷮𝟐﷯﷯﷯﷯ Given, the point is (3, −2, 1) So, 𝑥﷮1﷯ = 3, 𝑦﷮1﷯ = −2, 𝑧﷮1﷯ = 1 And the equation of the plane is 2x − y + 2z + 3 = 0 2x − y + 2z = −3 −(2x − y + 2z) = 3 −2x + y − 2z = 3 Comparing with Ax + By + Cz = D, A = −2, B = 1, C = −2, D = 3 Now, Distance of the point form the plane = −2 × 3﷯ + 1 × −2﷯ + −2 × 1﷯ − 3﷮ ﷮ −2﷯﷮2﷯ + 1﷮2﷯ + 2﷯﷮2﷯﷯﷯﷯ = −6﷯ + −2﷯ + −2﷯ − 3﷮ ﷮4 + 1 + 4﷯﷯﷯ = −13﷮ ﷮9﷯﷯﷯ = −13﷮3﷯﷯ = 𝟏𝟑﷮𝟑﷯ Ex 11.3, 14 In the following cases, find the distance of each of the given points from the corresponding given plane. The distance of the point (x1, y1, z1) from the plane Ax + By + Cz = D is 𝑨 𝒙﷮𝟏﷯ + 𝑩𝒚﷮𝟏﷯ + 𝑪𝒛﷮𝟏﷯ − 𝑫﷮ ﷮ 𝑨﷮𝟐﷯ + 𝑩﷮𝟐﷯ + 𝑪﷮𝟐﷯﷯﷯﷯ Given, the point is (2, 3, −5) So, 𝑥﷮1﷯= 2, 𝑦﷮1﷯= 3, 𝑧﷮1﷯= −5 and the equation of the plane is 1x + 2y − 2z = 9 Comparing with Ax + By + Cz = D, A = 1, B = 2, C = −2, D = 9 Now, Distance of the point from the plane = 1 × 2﷯ + 2 × 3﷯ + −2 × −5﷯ − 9﷮ ﷮ 1﷮2﷯ + 2﷮2﷯ + (−2)﷮2﷯﷯﷯﷯ = 2 + 6 + 10 − 9﷮ ﷮1 + 4 + 4﷯﷯﷯ = 18 − 9﷮ ﷮9﷯﷯﷯ = 9﷮3﷯ = 3 Ex 11.3, 14 In the following cases, find the distance of each of the given points from the corresponding given plane. The distance of the point (x1, y1, z1) from the plane Ax + By + Cz = D is 𝑨 𝒙﷮𝟏﷯ + 𝑩𝒚﷮𝟏﷯ + 𝑪𝒛﷮𝟏﷯ − 𝑫﷮ ﷮ 𝑨﷮𝟐﷯ + 𝑩﷮𝟐﷯ + 𝑪﷮𝟐﷯﷯﷯﷯ Given, the point is (−6, 0, 0) So, 𝑥﷮1﷯ = −6, 𝑦﷮1﷯ = 0, 𝑧﷮1﷯ = 0 and the equation of plane is 2x − 3y + 6z − 2 = 0 2x − 3y + 6z = 2 Comparing with Ax + By + Cz = D, A = 2, B = −3, C = 6 D = 3 Now, Distance of the point from the plane = 2 × −6﷯ + −3 × 0﷯ + 6 × 0﷯− 2 ﷮ ﷮ 2﷮2﷯+ −3﷯﷮2﷯+ 6﷮2﷯﷯﷯﷯ = −12 + 0 + 0 − 2﷮ ﷮4 + 9 + 36﷯﷯﷯ = −14﷮ ﷮49﷯﷯﷯ = −14﷮7﷯﷯ = −2﷯ = 2