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
Last updated at Oct. 1, 2019 by Teachoo
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
Transcript
Question 23 Find the vector equation of the line joining (1, 2, 3) and (β3, 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 β and π β is π β = (π ) β + π (π β β π β) Given, Let two points be A (1, 2, 3) & B(β3, 4, 3) A (1, 2, 3) π β = 1π Μ + 2π Μ + 3π Μ B (β3, 4, 3) π β = β3π Μ + 4π Μ + 3π Μ So, π β = (1π Μ + 2π Μ + 3π Μ) + π [("β3" π Μ" + 4" π Μ" + 3" π Μ" " ) β ("1" π Μ" + 2" π Μ" + 3" π Μ)] = (1π Μ + 2π Μ + 3π Μ) + π [(β3β1) π Μ+(4β2) π Μ+(3β3)π Μ ] = (1π Μ + 2π Μ + 3π Μ) + π (β4π Μ + 2π Μ) So, equation of line is π β = (1π Μ + 2π Μ + 3π Μ) + π (β4π Μ + 2π 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 π1, b1, c1 and π2, b2, c2 are perpendicular if π1 π2 + b1b2 + c1 c2 = 0 Finding direction ratios of parallel and z-axis β4π Μ + 2π Μ Direction ratios = β4, 2, 0 β΄ π1 = β4, b1 = 2, c1 = 0 (πΆπ) β = 0π Μ + 0π Μ + 1π Μ Direction ratios = 0, 0, 1 β΄ π2 = 0, b2 = 0, c2 = 1, Now, π1 π2 + b1 b2 + c1 c2 = β4 Γ 0 + 2 Γ 0 + 0 Γ 1 = 0 Since π1 π2 + b1 b2 + c1 c2 = 0 So, parallel vector of line is perpendicular to z-axis Hence proved
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CBSE Class 12 Sample Paper for 2019 Boards
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