This question is similar to Chapter 11 Class 12 Three Dimensional Geometry - Miscellaneous

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https://www.teachoo.com/3553/755/Misc-20---Find-vector-equation-of-line-perpendicular-to-two/category/Miscellaneous/

 

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Question 28 (B) Find the vector and the cartesian equation of the line that passes through (−1, 2, 7) and is perpendicular to the lines 𝑟 ⃗=2ı ˆ+ȷ ˆ−3𝑘 ˆ+𝜆(ı ˆ+2ȷ ˆ+5𝑘 ˆ) and 𝑟 ⃗=3ı ˆ+3ȷ ˆ−7𝑘 ˆ+𝜇(3ı ˆ−2ȷ ˆ+5𝑘 ˆ).The vector equation of a line passing through a point with position vector 𝑎 ⃗ and parallel to a vector 𝑏 ⃗ is 𝒓 ⃗ = 𝒂 ⃗ + 𝜆𝒃 ⃗ The line passes through (–1, 2, 7) So, 𝒂 ⃗ = –1𝒊 ̂ + 2𝒋 ̂ + 7𝒌 ̂ Now, finding 𝑏 ⃗ Given, line 𝑏 ⃗ is perpendicular to 𝑟 ⃗=2𝚤 ˆ+𝚥 ˆ−3𝑘 ˆ+𝜆(𝚤 ˆ+2𝚥 ˆ+5𝑘 ˆ) and 𝑟 ⃗=3𝚤 ˆ+3𝚥 ˆ−7𝑘 ˆ+𝜇(3𝚤 ˆ−2𝚥 ˆ+5𝑘 ˆ) So, 𝒃 ⃗ is the cross product of these two parallel vectors ∴ 𝑏 ⃗ = (𝚤 ˆ+2𝚥 ˆ+5𝑘 ˆ) × (3𝚤 ˆ−2𝚥 ˆ+5𝑘 ˆ) = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@1&2&5@3&−2&5)| = 𝑖 ̂ (2 × 5 – (-2) × 5) − 𝑗 ̂ (1 × 5 − 3 × 5) + 𝑘 ̂ (1 × (-2) − 3 × 2) = 𝑖 ̂ (10 + 10) − 𝑗 ̂ (5 – 15) + 𝑘 ̂ (–2 – 6) = 20𝒊 ̂ + 10𝒋 ̂ – 8𝒌 ̂ = 2(10𝑖 ̂ + 5𝑗 ̂ – 4𝑘 ̂) Putting value of 𝑎 ⃗ & 𝑏 ⃗ in formula 𝑟 ⃗ = 𝑎 ⃗ + 𝜆𝑏 ⃗ ∴ 𝒓 ⃗ = (–𝑖 ̂ + 2𝑗 ̂ + 7𝑘 ̂) + 𝜆 × 2(10𝑖 ̂ + 5𝑗 ̂ – 4𝑘 ̂) = (–𝑖 ̂ + 2𝑗 ̂ + 7𝑘 ̂) + 2𝜆 (10𝑖 ̂ + 5𝑗 ̂ – 4𝑘 ̂) Putting 𝜇 = 2𝜆 = (–𝒊 ̂ + 2𝒋 ̂ + 7𝒌 ̂) + 𝝁 (10𝒊 ̂ + 5𝒋 ̂ – 4𝒌 ̂) Therefore, the equation of line is (–𝒊 ̂ + 2𝒋 ̂ + 7𝒌 ̂) + 𝝁 (10𝒊 ̂ + 5𝒋 ̂ – 4𝒌 ̂) Converting to cartesian form (𝑥 −(−1))/10=(𝑦 − 2)/5=(𝑧 − 7)/(−4) (𝒙 + 𝟏)/𝟏𝟎=(𝒚 − 𝟐)/𝟓=(𝒛 − 𝟕)/(−𝟒)

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.