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Ex 11.2, 13 (Cartesian method) Find the shortest distance between the lines (𝑥 + 1)/7 = (𝑦 + 1)/( − 6) = (𝑧 + 1)/1 and (𝑥 − 3)/1 = (𝑦 − 5)/( − 2) = (𝑧 − 7)/1 Shortest distance between two lines l1: (𝑥 − 𝑥_1)/𝑎_1 = (𝑦 − 𝑦_1)/𝑏_1 = (𝑧 − 𝑧_1)/𝑐_1 l2: (𝑥 − 𝑥_2)/𝑎_2 = (𝑦 − 𝑦_2)/𝑏_2 = (𝑧 − 𝑧_2)/𝑐_2 is ||■8(𝒙_𝟐 − 𝒙_𝟏&𝒚_𝟐 − 𝒚_𝟏&𝒛_𝟐 − 𝒛_𝟏@𝒂_𝟏&𝒃_𝟏&𝒄_𝟏@𝒂_𝟐&𝒃_𝟐&𝒄_𝟐 )|/√((𝒂_𝟏 𝒃_𝟐 − 𝒂_𝟐 𝒃_𝟏 )^𝟐 + (𝒃_𝟏 𝒄_(𝟐 )− 𝒃_𝟐 𝒄_𝟏 )^𝟐 + (𝒄_𝟏 𝒂_𝟐 −〖 𝒄〗_𝟐 𝒂_𝟏 )^𝟐 )| (𝒙 + 𝟏)/𝟕 = (𝒚 + 𝟏)/( − 𝟔) = (𝒛 + 𝟏)/𝟏 (𝑥 − (−1))/7 = (𝑦 − (−1))/( −6) = (𝑧 − (−1))/1 Comparing with l1: (𝑥 − 𝑥_1)/𝑎_1 = (𝑦 − 𝑦_1)/𝑏_1 = (𝑧 − 𝑧_1)/𝑐_1 𝒙_𝟏 = –1, 𝒚_𝟏 = –1, 𝒛_𝟏 = –1, & 𝒂_𝟏 = 7, 𝒃_𝟏 = –6, 𝒄_𝟏 = 1, (𝒙 − 𝟑)/𝟏 = (𝒚 − 𝟓)/( − 𝟐) = (𝒛 − 𝟕)/𝟏 Comparing with l2: (𝑥 − 𝑥_2)/𝑎_2 = (𝑦 − 𝑦_2)/𝑏_2 = (𝑧 − 𝑧_2)/𝑐_2 𝑥_2 = 3, 𝑦_2 = 5, 𝑧_2 = 7, & 𝑎_2 = 1, 𝑏_2 = –2, 𝑐_2 = 1, d = ||■8(𝑥_2−𝑥_1&𝑦_2 − 𝑦_1&𝑧_2 − 𝑧_1@𝑎_1&𝑏_1&𝑐_1@𝑎_2&𝑏_2&𝑐_2 )|/√((𝑎_1 𝑏_2 − 𝑎_2 𝑏_1 )^2 + (𝑏_1 𝑐_(2 )− 𝑏_2 𝑐_1 )^2 + (𝑐_1 𝑎_2 −〖 𝑐〗_2 𝑎_1 )^2 )| d = ||■8(3−(−1)&5−(−1)&7−(−1)@7&−6&1@1&−2&1)|/√((7(−2) −1(−6))^2 + (−6(1)−(−2)1)^2 + (1(1) −1(7))^2 )| d = ||■8(𝟒&𝟔&𝟖@𝟕&−𝟔&𝟏@𝟏&−𝟐&𝟏)|/√((−𝟏𝟒 + 𝟔)^𝟐 + (−𝟔 + 𝟐)^𝟐 + (𝟏 − 𝟕)^𝟐 )| d = ||■8(4&6&8@7&−6&1@1&−2&1)|/√((8)^2 + (−4)^2 + (−6)^2 )| d = ||■8(4&6&8@7&−6&1@1&−2&1)|/√116| d = |(4(−6(1) − (−2)1) − 6(7(1) − 1(1)) + 8(7(−2) − 1(−6)))/√116| d = |(4(−6 + 2)−6(7 − 1)+8(−14 + 6))/√116| d = |(−16 − 36 − 64)/√116| d = |(−𝟏𝟏𝟔)/√𝟏𝟏𝟔| d = |−√116| d = √116 d = √(4 × 29) d = 𝟐√𝟐𝟗 Ex 11.2, 13 (Vector method) Find the shortest distance between the lines (𝑥 + 1)/7 = (𝑦 + 1)/( − 6) = (𝑧 + 1)/1 and (𝑥 − 3)/1 = (𝑦 − 5)/( − 2) = (𝑧 − 7)/1 Shortest distance between two lines 𝑟 ⃗ = (𝑎"1" ) ⃗ + λ(𝑏"1" ) ⃗ and 𝑟 ⃗ = (𝑎"2" ) ⃗ + μ(𝑏"2" ) ⃗ is |(((𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ ).((𝒂𝟐) ⃗ × (𝒂𝟏) ⃗ ))/|(𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ | | (𝒙 + 𝟏 )/𝟕 = (𝒚 + 𝟏 )/(−𝟔) = (𝒛 + 𝟏 )/𝟏 (𝑥 − (−1) )/7 = (𝑦 − (−1) )/(−6) = (𝑧 − (−1) )/1 Comparing with (𝑥 − 𝑥1 )/𝑎1 = (𝑦 − 𝑦1 )/𝑏1 = (𝑧 − 𝑧1 )/𝑐1, 𝑥1 = −1, y1 = −1, 𝑧1= −1 𝑎1 = 7, b1 = − 6, 𝑐1= 1 ∴ (𝒂"1" ) ⃗ = 𝑥1𝑖 ̂ + 𝑦1𝑗 ̂ + 𝑧1𝑘 ̂ = −𝟏𝒊 ̂ − 𝟏𝒋 ̂ − 𝟏𝒌 ̂ ("b1" ) ⃗ = 𝑎1𝑖 ̂ + 𝑏1𝑗 ̂ + 𝑐1𝑘 ̂ = 𝟕𝒊 ̂ − 𝟔𝒋 ̂ +𝟏𝒌 ̂ (𝒙 − 𝟑 )/𝟏 = (𝒚 − 𝟓 )/( − 𝟐) = (𝒛 − 𝟕)/𝟏 Comparing with (𝑥 − 𝑥2 )/𝑎2 = (𝑦 − 𝑦2 )/𝑏2 = (𝑧 − 𝑧2 )/𝑐2, 𝑥2 = 3, y2 = 5, 𝑧2= 7 𝑎2 = 1, b2 = − 2, 𝑐2 = 1 ∴ (𝒂"2" ) ⃗ = 𝑥2𝑖 ̂ + 𝑦2𝑗 ̂ + 𝑧2𝑘 ̂ = 𝟑𝒊 ̂ + 𝟓𝒋 ̂ + 𝟕𝒌 ̂ ("b2" ) ⃗ = 𝑎2𝑖 ̂ + 𝑏2𝑗 ̂ + 𝑐2𝑘 ̂ = 𝟏𝒊 ̂ − 𝟐𝒋 ̂ + 𝟏𝒌 ̂ Now, ((𝒂"2" ) ⃗ − (𝒂"1" ) ⃗) = (3𝑖 ̂ + 5𝑗 + 7𝑘 ̂) − (−1𝑖 ̂ − 1𝑗 ̂ − 1𝑘 ̂) = 3𝑖 ̂ + 5𝑗 ̂ + 7𝑘 ̂ + 1𝑖 ̂ + 1𝑗 ̂ + 1𝑘 ̂ = (3 + 1) 𝑖 ̂ + (5 + 1)𝑗 ̂ + (7 + 1)𝑘 ̂ = 4𝒊 ̂ + 6𝒋 ̂ + 8𝒌 ̂ (𝒃"1" ) ⃗ × (𝒃"2" ) ⃗ = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@7& −6&1@1& −2&1)| = 𝑖 ̂[(−6×1)−(−2×1)] − 𝑗 ̂[(−7×1)−(1×1)] + k[(7×−2)−(1×−6)] = 𝑖 ̂[−6+2] − 𝑗 ̂ [(7−1)] + 𝑘 ̂ [−14+6] = −4𝒊 ̂ − 6𝒋 ̂ − 8𝒌 ̂ Magnitude of ((𝑏"1" ) ⃗×(𝑏"2" ) ⃗ ) = √((−4)2 + (−6)2 + (−8)2) |(𝒃"1" ) ⃗" " ×" " (𝒃"2" ) ⃗ | = √116 = √(4 × 29) = 2√𝟐𝟗 Also, ((𝒃"1" ) ⃗ ×" " (𝒃"2" ) ⃗).((𝒂"1" ) ⃗" "−" " (𝒂"2" ) ⃗) = (−4𝑖 ̂ − 6𝑗 ̂ − 8𝑘 ̂).(4𝑖 ̂ + 6𝑗 ̂ + 8𝑘 ̂) = (−4 × 4) + (−6 × 6) + (−8 + 8) = −16 + (−36) + (−64) = −116 ∴ Shortest distance = |(((𝑏"1" ) ⃗ × (𝑏"2" ) ⃗ ).((𝑎"2" ) ⃗ − (𝑎"1" ) ⃗) )/|(𝑏"1" ) ⃗ × (𝑏"2" ) ⃗ | | = |(−116 )/(2√29)| = |(−58 )/√29| = |(−2 × 29 )/√29| = 𝟐√𝟐𝟗 Therefore, the shortest distance between the two given lines is 2√29.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo