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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Example 10 Find the distance between the lines 𝑙_1 and 𝑙_2 given by 𝑟 ⃗ = 𝑖 ̂ + 2𝑗 ̂ – 4𝑘 ̂ + 𝜆 (2𝒊 ̂ + 3𝒋 ̂ + 6𝒌 ̂ ) and 𝑟 ⃗ = 3𝑖 ̂ + 3𝑗 ̂ − 5𝑘 ̂ + μ (2𝒊 ̂ + 3𝒋 ̂ + 6𝒌 ̂)Distance between two parallel lines with vector equations 𝑟 ⃗ = (𝑎_1 ) ⃗ + 𝜆𝒃 ⃗ and 𝑟 ⃗ = (𝑎_2 ) ⃗ + 𝜇𝒃 ⃗ is |(𝒃 ⃗ × ((𝒂_𝟐 ) ⃗ − (𝒂_𝟏 ) ⃗))/|𝒃 ⃗ | | 𝑟 ⃗ = (𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂) + 𝜆 (2𝒊 ̂ + 3𝒋 ̂ + 6𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎1) ⃗ + 𝜆 𝑏 ⃗, (𝑎1) ⃗ = 1𝑖 ̂ + 2𝑗 ̂ – 4𝑘 ̂ & 𝑏 ⃗ = 2𝑖 ̂ + 3𝑗 ̂ + 6𝑘 ̂ 𝑟 ⃗ = (3𝑖 ̂ + 3𝑗 ̂ − 5𝑘 ̂) + 𝜇 (2𝒊 ̂ + 3𝒋 ̂ + 6𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎2) ⃗ + 𝜇𝑏 ⃗, (𝑎2) ⃗ = 3𝑖 ̂ + 3𝑗 ̂ − 5𝑘 ̂ & 𝑏 ⃗ = 2𝑖 ̂ + 3𝑗 ̂ + 6𝑘 ̂ Now, ((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗) = (3𝑖 ̂ + 3𝑗 ̂ − 5𝑘 ̂) − (1𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂) = (3 − 1) 𝑖 ̂ + (3 − 2)𝑗 ̂ + ( − 5 + 4)𝑘 ̂ = 2𝒊 ̂ + 1𝒋 ̂ − 1𝒌 ̂ Magnitude of 𝑏 ⃗ = √(22 + 32 + 62) |𝒃 ⃗ | = √(4+9+36) = √49 = 7 Also, 𝒃 ⃗ × ((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗) = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@2&3&6@2&1&−1)| = 𝑖 ̂ [(3×−1)−(1×6)] − 𝑗 ̂ [(2×−1)−(2×6)] + 𝑘 ̂ [(2×1)−(2×3)] = 𝑖 ̂ [−3−6] − 𝑗 ̂ [−2−12] + 𝑘 ̂ [2−6] = 𝑖 ̂ (–9) − 𝑗 ̂ (–14) + 𝑘 ̂(−4) = −𝟗𝒊 ̂ + 14𝒋 ̂ − 4𝒌 ̂ Now, |𝒃 ⃗" × (" (𝒂𝟐) ⃗" − " (𝒂𝟏) ⃗")" | = √((−9)^2+(14)^2+(−4)^2 ) = √(81+196+16) = √𝟐𝟗𝟑 So, Distance = |(𝑏 ⃗ × ((𝑎_2 ) ⃗ − (𝑎_1 ) ⃗))/|𝑏 ⃗ | | = |√293/7| = √𝟐𝟗𝟑/𝟕 Therefore, the distance between the given two parallel lines is √293/7.

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.