Ex 11.2, 10 - Class 12th - 3D Geometry - Find angle between - Angle between two lines - Vector

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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise
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Ex 11.2, 10 Find the angle between the following pairs of lines: (i) 𝑟﷯ = 2 𝑖﷯− 5 𝑗﷯ + 𝑘﷯ + 𝜆(3 𝑖﷯ + 2 𝑗﷯ + 6 𝑘﷯) and 𝑟﷯ = 7 𝑖﷯ – 6 𝑘﷯ + 𝜇( 𝑖﷯ + 2 𝑗﷯ + 2 𝑘﷯) Angle between two vectors 𝑟﷯ = 𝑎1﷯ + 𝜆 𝑏1﷯ & 𝑟﷯ = 𝑎2﷯ + 𝜇 𝑏2﷯ is given by cos θ = 𝒃𝟏﷯ . 𝒃𝟐﷯﷮ 𝒃𝟏﷯﷯ 𝒃𝟐﷯﷯﷯﷯ Given, the pair of lines is Now, 𝑏1﷯. 𝑏2﷯ = (3 𝑖﷯ + 2 𝑗﷯ + 6 𝑘﷯) . (1 𝑖﷯ + 2 𝑗﷯ + 2 𝑘﷯) = (3 × 1) + (2 × 2) + (6 × 2) = 3 + 4 + 12 = 19 Magnitude of 𝑏1﷯ = ﷮32 + 22 + 62﷯ 𝑏1﷯﷯ = ﷮9 + 4 + 36﷯ = ﷮49﷯ = 7 Magnitude of 𝑏2﷯ = ﷮12+22+22﷯ 𝑏2﷯﷯ = ﷮1+4+4﷯ = ﷮9﷯ = 3 Now, cos θ = 𝑏1﷯. 𝑏2﷯﷮ 𝑏1﷯﷯ 𝑏2﷯﷯﷯﷯ cos θ = 19﷮7 × 3 ﷯﷯ cos θ = 19﷮21 ﷯ ∴ θ = cos-1 𝟏𝟗﷮𝟐𝟏 ﷯﷯ Therefore, the angle between the given vectors is cos − 1 19﷮21 ﷯﷯ Ex 11.2, 10 Find the angle between the following pairs of lines: (ii) 𝑟﷯ = (3 𝑖﷯ + 𝑗﷯ − 2 𝑘﷯) + 𝜆 ( 𝑖﷯ − 𝑗﷯ − 2 𝑘﷯) and 𝑟﷯ = (2 𝑖﷯ − 𝑗﷯ − 56 𝑘﷯) + 𝜇 (3 𝑖﷯ – 5 𝑗﷯ − 4 𝑘﷯) Angle between two vectors 𝑟﷯ = 𝑎1﷯ + 𝜆 𝑏1﷯ & 𝑟﷯ = 𝑎2﷯ + 𝜇 𝑏2﷯ is given by cos θ = 𝒃𝟏﷯ . 𝒃𝟐﷯﷮ 𝒃𝟏﷯﷯ 𝒃𝟐﷯﷯﷯﷯ Given, the pair of lines is Now, 𝑏1﷯. 𝑏2﷯ = (1 𝑖﷯ − 1 𝑗﷯ − 2 𝑘﷯).(3 𝑖﷯ − 5 𝑗﷯ − 4 𝑘﷯) = (1 × 3) + ( − 1 × − 5) + ( − 2 × –4) = 3 + 5 + 8 = 16 Magnitude of 𝑏1﷯ = ﷮ 1﷮2﷯+ −1﷯﷮2﷯+ −2﷯﷮2﷯﷯ 𝑏1﷯﷯ = ﷮1+1+4﷯ = ﷮6﷯ Magnitude of 𝑏2﷯ = ﷮ 3﷮2﷯+ −5﷯﷮2﷯+ − 4﷯2﷯ 𝑏2﷯﷯ = ﷮9+25+16﷯ = ﷮50﷯ = ﷮25×2﷯ = 5 ﷮2﷯ Now, cos θ = 𝑏1﷯. 𝑏2﷯﷮ 𝑏1﷯﷯ 𝑏2﷯﷯﷯﷯ = 16﷮ ﷮6﷯ × 5 ﷮2﷯﷯﷯ = 16﷮ ﷮3﷯ × ﷮2﷯ × 5 × ﷮2﷯﷯﷯ = 16﷮ ﷮3﷯ × 2 × 5 ﷯﷯ = 8﷮5 ﷮3﷯ ﷯ ∴ θ = cos-1 8﷮5 ﷮3﷯ ﷯﷯ Therefore, the angle between the given vectors is cos − 1 8﷮5 ﷮3﷯ ﷯﷯

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