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

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  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 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 π‘Ž βƒ— and 𝑏 βƒ— is π‘Ÿ βƒ— = (π‘Ž ) βƒ— + πœ† (𝑏 βƒ— βˆ’ π‘Ž βƒ—) Given, Let two points be A (1, 2, 3) & B(–3, 4, 3) A (1, 2, 3) π‘Ž βƒ— = 1𝑖 Μ‚ + 2𝑗 Μ‚ + 3π‘˜ Μ‚ B (–3, 4, 3) 𝑏 βƒ— = –3𝑖 Μ‚ + 4𝑗 Μ‚ + 3π‘˜ Μ‚ So, π‘Ÿ βƒ— = (1𝑖 Μ‚ + 2𝑗 Μ‚ + 3π‘˜ Μ‚) + πœ† [("–3" 𝑖 Μ‚" + 4" 𝑗 Μ‚" + 3" π‘˜ Μ‚" " ) βˆ’ ("1" 𝑖 Μ‚" + 2" 𝑗 Μ‚" + 3" π‘˜ Μ‚)] = (1𝑖 Μ‚ + 2𝑗 Μ‚ + 3π‘˜ Μ‚) + πœ† [(βˆ’3βˆ’1) 𝑖 Μ‚+(4βˆ’2) 𝑗 Μ‚+(3βˆ’3)π‘˜ Μ‚ ] = (1𝑖 Μ‚ + 2𝑗 Μ‚ + 3π‘˜ Μ‚) + πœ† (–4𝑖 Μ‚ + 2𝑗 Μ‚) So, equation of line is π‘Ÿ βƒ— = (1𝑖 Μ‚ + 2𝑗 Μ‚ + 3π‘˜ Μ‚) + πœ† (–4𝑖 Μ‚ + 2𝑗 Μ‚) 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 π‘Ž1, b1, c1 and π‘Ž2, b2, c2 are perpendicular if π‘Ž1 π‘Ž2 + b1b2 + c1 c2 = 0 Finding direction ratios of parallel and z-axis –4π’Š Μ‚ + 2𝒋 Μ‚ Direction ratios = –4, 2, 0 ∴ π‘Ž1 = –4, b1 = 2, c1 = 0 (𝑢𝒁) βƒ— = 0π’Š Μ‚ + 0𝒋 Μ‚ + 1π’Œ Μ‚ Direction ratios = 0, 0, 1 ∴ π‘Ž2 = 0, b2 = 0, c2 = 1, Now, π‘Ž1 π‘Ž2 + b1 b2 + c1 c2 = –4 Γ— 0 + 2 Γ— 0 + 0 Γ— 1 = 0 Since π‘Ž1 π‘Ž2 + b1 b2 + c1 c2 = 0 So, parallel vector of line is perpendicular to z-axis Hence proved

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.