Ex 11.2, 14 - Find shortest distance between lines	 - Ex 11.2

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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise
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Ex 11.2, 14 Find the shortest distance between the lines 𝑟﷯ = ( 𝑖﷯ + 2 𝑗﷯ + 𝑘﷯) + 𝜆 ( 𝑖﷯ − 𝑗﷯ + 𝑘﷯) and 𝑟﷯ = (2 𝑖﷯ − 𝑗﷯ − 𝑘﷯) + 𝜇 (2 𝑖﷯ + 𝑗﷯ + 2 𝑘﷯) Shortest distance between the lines with vector equations 𝑟﷯ = ğ‘Ž1ï·¯ + 𝜆 𝑏1ï·¯and 𝑟﷯ = ğ‘Ž2ï·¯ + 𝜇 𝑏2ï·¯ is 𝒃𝟏﷯ × 𝒃𝟐﷯ ï·¯. 𝒂𝟐﷯ − 𝒂𝟏﷯ ﷯﷮ 𝒃𝟏﷯ × 𝒃𝟐﷯﷯﷯﷯ Given, Now, 𝒂𝟐﷯ − 𝒂𝟏﷯ = (2 𝑖﷯ − 1 𝑗﷯ − 1 𝑘﷯) − (1 𝑖﷯ + 2 𝑗﷯ + 1 𝑘﷯) = (2 − 1) 𝑖﷯ + (−1− 2) 𝑗﷯ + (−1 − 1) 𝑘﷯ = 1 𝒊﷯ − 3 𝒋﷯ − 2 𝒌﷯ 𝒃𝟏﷯ × 𝒃𝟐﷯ = 𝑖﷯﷮ 𝑗﷯﷮ 𝑘﷯﷮1ï·® − 1ï·®1ï·®2ï·®1ï·®2﷯﷯ = 𝑖﷯ − 1× 2﷯−(1×1)ï·¯ − 𝑗﷯ 1×2﷯−(2×1)ï·¯ + 𝑘﷯ 1×1﷯−(2×−1)ï·¯ = 𝑖﷯ −2−1ï·¯ − 𝑗﷯ 2−2ï·¯ + 𝑘﷯ 1+2ï·¯ = − 3 𝒊﷯ − 0 𝒋﷯ + 3 𝒌﷯ Magnitude of ( 𝑏1ï·¯ × 𝑏2ï·¯) = ï·®( − 3)2+ 0ï·¯2+32ï·¯ 𝒃𝟏﷯ × 𝒃𝟐﷯﷯ = ï·®9+0+9ï·¯ = ï·®18ï·¯ = ï·®9 × 2ï·¯ = 3 ﷮𝟐﷯ Also, ( 𝑏1ï·¯ × 𝑏2ï·¯) . ( ğ‘Ž2ï·¯ – ğ‘Ž1ï·¯) = (− 3 𝑖﷯−0 𝑗﷯+3 𝑘﷯).(1 𝑖﷯ − 3 𝑗﷯ − 2 𝑘﷯) = (−3×1).(0×−3) + (3 × −2) = −3 − 0 − 6 = −9 So, shortest distance = 𝑏1ï·¯ × 𝑏2ï·¯ ï·¯. ğ‘Ž2ï·¯ − ğ‘Ž1ï·¯ ﷯﷮ 𝑏1ï·¯ × 𝑏2﷯﷯﷯﷯ = − 9ï·®3 ï·®2﷯﷯﷯ = 3ï·® ï·®2﷯﷯ = 3ï·® ï·®2﷯﷯ × ï·®2﷯﷮ ï·®2﷯﷯ = 𝟑 ﷮𝟐﷯﷮𝟐﷯ Therefore, shortest distance between the given two lines is 3 ï·®2﷯﷮2ï·¯.

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