CBSE Class 12 Sample Paper for 2025 Boards
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CBSE Class 12 Sample Paper for 2025 Boards
Last updated at August 14, 2026 by Teachoo
This question is similar to Chapter 11 Class 12 Three Dimensional Geometry - Miscellaneous
Please check the question here
Transcript
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)
