Question 23

Find the vector equation of the line joining (1, 2, 3) and (–3, 4, 3) and show that it is perpendicular to the z-axis   1. Class 12
2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards
3. CBSE Class 12 Sample Paper for 2019 Boards

Transcript

Question 23 Find the vector equation of the line joining (1, 2, 3) and (–3, 4, 3) and show that it is perpendicular to the z-axis Vector equation of a line passing though two points with position vectors 𝑎 ⃗ and 𝑏 ⃗ is 𝑟 ⃗ = (𝑎 ) ⃗ + 𝜆 (𝑏 ⃗ − 𝑎 ⃗) Given, Let two points be A (1, 2, 3) & B(–3, 4, 3) A (1, 2, 3) 𝑎 ⃗ = 1𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂ B (–3, 4, 3) 𝑏 ⃗ = –3𝑖 ̂ + 4𝑗 ̂ + 3𝑘 ̂ So, 𝑟 ⃗ = (1𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂) + 𝜆 [("–3" 𝑖 ̂" + 4" 𝑗 ̂" + 3" 𝑘 ̂" " ) − ("1" 𝑖 ̂" + 2" 𝑗 ̂" + 3" 𝑘 ̂)] = (1𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂) + 𝜆 [(−3−1) 𝑖 ̂+(4−2) 𝑗 ̂+(3−3)𝑘 ̂ ] = (1𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂) + 𝜆 (–4𝑖 ̂ + 2𝑗 ̂) So, equation of line is 𝑟 ⃗ = (1𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂) + 𝜆 (–4𝑖 ̂ + 2𝑗 ̂) Also, we have to prove that line is perpendicular to z-axis So, parallel vector of line will be perpendicular to z-axis Theory : Two lines with direction ratios 𝑎1, b1, c1 and 𝑎2, b2, c2 are perpendicular if 𝑎1 𝑎2 + b1b2 + c1 c2 = 0 Finding direction ratios of parallel and z-axis –4𝒊 ̂ + 2𝒋 ̂ Direction ratios = –4, 2, 0 ∴ 𝑎1 = –4, b1 = 2, c1 = 0 (𝑶𝒁) ⃗ = 0𝒊 ̂ + 0𝒋 ̂ + 1𝒌 ̂ Direction ratios = 0, 0, 1 ∴ 𝑎2 = 0, b2 = 0, c2 = 1, Now, 𝑎1 𝑎2 + b1 b2 + c1 c2 = –4 × 0 + 2 × 0 + 0 × 1 = 0 Since 𝑎1 𝑎2 + b1 b2 + c1 c2 = 0 So, parallel vector of line is perpendicular to z-axis Hence proved

CBSE Class 12 Sample Paper for 2019 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards 