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

Find the vector equation of the line joining (1, 2, 3) and (-3, 4, 3)

Question 23 - CBSE Class 12 Sample Paper for 2019 Boards - Part 2
Question 23 - CBSE Class 12 Sample Paper for 2019 Boards - Part 3

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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 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.