1. Chapter 11 Class 12 Three Dimensional Geometry
2. Serial order wise

Transcript

Misc 9 Find the shortest distance between lines 𝑟﷯ = 6 𝑖﷯ + 2 𝑗﷯ + 2 𝑘﷯ + 𝜆 ( 𝑖﷯ – 2 𝑗﷯ + 2 𝑘﷯) and 𝑟﷯ = –4 𝑖﷯ – 𝑘﷯ + 𝜇 (3 𝑖﷯ – 2 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﷯﷯﷯﷯ Now, ( 𝒂𝟐﷯ − 𝒂𝟏﷯) = (−4 𝑖﷯ + 0 𝑗﷯ − 1 𝑘﷯) − (6 𝑖﷯ + 2 𝑗﷯ + 2 𝑘﷯) = (−4 − 6) 𝑖﷯ + (0 − 2) 𝑗﷯ + (−1 − 2) 𝑘﷯ = − 10 𝒊﷯ − 2 𝒋﷯ − 3 𝒌﷯ ( 𝒃𝟏﷯ × 𝒃𝟐﷯) = 𝑖﷯﷮ 𝑗﷯﷮ 𝑘﷯﷮1﷮ −2﷮2﷮3﷮−2﷮−2﷯﷯ = 𝑖﷯ −2×−2﷯−(−2×2)﷯ − 𝑗﷯ 1×−2﷯−(3×2)﷯ + 𝑘﷯ 1×−2﷯−(3×−2)﷯ = 𝑖﷯ 4+4﷯ − 𝑗﷯ −2−6﷯ + 𝑘﷯ −2+6﷯ = 𝑖﷯ (8) − 𝑗﷯ (−8) + 𝑘﷯(4) = 8 𝒊﷯ + 8 𝒋﷯ + 4 𝒌﷯ Magnitude of 𝑏1﷯ × 𝑏2﷯ = ﷮ 8﷮2﷯+ 8﷮2﷯+ 4﷮2﷯﷯ 𝒃𝟏﷯ × 𝒃𝟐﷯﷯ = ﷮64+64+16﷯ = ﷮144﷯ = 𝟏𝟐 Also, 𝒃𝟏﷯× 𝒃𝟐﷯﷯ . 𝒂𝟐﷯ − 𝒂𝟏﷯﷯ = (8 𝑖﷯ + 8 𝑗﷯ + 4 𝑘﷯).(− 10 𝑖﷯ − 2 𝑗﷯ − 3 𝑘﷯) = (8 × − 10) + (8 × − 2) + (4 × − 3) = − 80 + (−16) + (-12) = − 108 Shortest distance = 𝑏1﷯ × 𝑏2﷯﷯ . 𝑎2﷯ − 𝑎1﷯﷯﷮ 𝑏1﷯ × 𝑏2﷯﷯﷯﷯ = −108﷮12﷯﷯ = −9﷯ = 9 Therefore, the shortest distance between the given two lines is 9.

Serial order wise