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 |(𝑨𝒙_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 |(𝑨𝒙_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 a 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 |(𝑨𝒙_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 |(𝑨𝒙_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