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Question 23
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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 Dec. 12, 2018 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

CBSE Class 12 Sample Paper for 2019 Boards

Paper Summary

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Class 12

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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.