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Example 11 - Chapter 11 Class 12 - Find shortest distance - Shortest distance between two skew lines

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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise
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Example 11 Find the shortest distance between the lines l1 and l2 whose vector equations are 𝑟﷯ = 𝑖﷯ + 𝑗﷯ + 𝜆(2 𝑖﷯ − 𝑗﷯ + 𝑘﷯ ) and 𝑟﷯ = 2 𝑖﷯ + 𝑗﷯ – 𝑘﷯ + 𝜇 (3 𝑖﷯ – 5 𝑗﷯ + 2 𝑘﷯ ) Shortest distance between lines with vector equations 𝑟﷯ = 𝑎1﷯ + 𝜆 𝑏1﷯ and 𝑟﷯ = 𝑎2﷯ + 𝜇 𝑏2﷯ is 𝑏1﷯ × 𝑏2﷯ ﷯. 𝑎2﷯ − 𝑎1﷯ ﷯﷮ 𝑏1﷯ × 𝑏2﷯﷯﷯﷯ Given, Now 𝒂𝟐﷯ − 𝒂𝟏﷯ = (2 𝑖﷯ + 1 𝑗﷯ − 1 𝑘﷯) − (1 𝑖﷯ + 1 𝑗﷯ + 0 𝑘﷯) = (2 − 1) 𝑖﷯ + (1 − 1) 𝑗﷯ + ( − 1 − 0) 𝑘﷯ = 1 𝒊﷯ + 0 𝒋﷯ − 1 𝒌﷯ 𝒃𝟏﷯ × 𝒃𝟐﷯ = 𝑖﷯﷮ 𝑗﷯﷮ 𝑘﷯﷮2﷮ − 1﷮1﷮3﷮ − 5﷮2﷯﷯ = 𝑖﷯ − 1×2﷯−(− 5×1)﷯ − 𝑗﷯ 2×2﷯−(− 3×1)﷯ + 𝑘﷯ 2×−5﷯−(3×− 1)﷯ = 𝑖﷯ − 2+5﷯ − 𝑗﷯ 4−3﷯ + 𝑘﷯ − 10+3﷯ = 𝑖﷯ (3) − 𝑗﷯ (1) + 𝑘﷯(− 7) = 3 𝒊﷯ − 𝒋﷯ − 7 𝒌﷯ Magnitude of ( 𝑏1﷯ × 𝑏2﷯) = ﷮32+ − 1﷯2+ − 7﷯﷮2﷯﷯ 𝒃𝟏﷯× 𝒃𝟐﷯﷯ = ﷮9+1+49﷯ = ﷮𝟓𝟗﷯ Also, ( 𝒃𝟏﷯ × 𝒃𝟐﷯) .( 𝒂𝟐﷯ − 𝒂𝟏﷯) = (3 𝑖﷯ − 𝑗﷯ − 7 𝑘﷯) . (1 𝑖﷯ + 0 𝑗﷯ − 1 𝑘﷯) = (3 × 1) + (− 1 × 0) + (−7 × −1) = 3 + 0 + 7 = 10 ∴ Shortest distance = 𝑏1﷯ × 𝑏2﷯ ﷯. 𝑎2﷯ − 𝑎1﷯ ﷯﷮ 𝑏1﷯ × 𝑏2﷯﷯﷯﷯ = 10﷮ ﷮59﷯﷯﷯ = 𝟏𝟎﷮ ﷮𝟓𝟗﷯﷯ Therefore the shortest distance between the given lines is 10﷮ ﷮59﷯﷯.

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