Find the direction ratio and direction cosines of a line parallel to the line whose equations are 6š‘„ − 12 = 3š‘¦ + 9 = 2š‘§ − 2

[Sample paper] Find the direction ratio and direction cosines of a - CBSE Class 12 Sample Paper for 2023 Boards
part 2 - Question 23 (Choice 2) - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

 

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Question 23 (Choice 2) Find the direction ratio and direction cosines of a line parallel to the line whose equations are 6š‘„ āˆ’ 12 = 3š‘¦ + 9 = 2š‘§ āˆ’ 2 Given equation of line 6š‘„ āˆ’ 12 = 3š‘¦ + 9 = 2š‘§ āˆ’ 2 6(š‘„ āˆ’ 2) = 3(š‘¦ + 3) = 2(š‘§ āˆ’ 1) Dividing both sides by 6 (6(š‘„ āˆ’ 2))/6=(3(š‘¦ + 3))/6=(2(š‘§ āˆ’ 1))/6 ((š’™ āˆ’ šŸ))/šŸ=((š’š + šŸ‘))/šŸ=((š’› āˆ’ šŸ))/šŸ‘ Thus, Direction ratios of the line parallel to the line = 1, 2, 3 ∓ š‘Ž = 1, b = 2, c = 3 Also, √(š’‚^šŸ + š’ƒ^šŸ + š’„^šŸ ) = √(12 +22 +32) = √(1 +4 +9) = āˆššŸšŸ’ Direction cosines = š‘Ž/√(š‘Ž^2 + š‘^2 + š‘^2 ) , š‘/√(š‘Ž^2 + š‘^2 + š‘^2 ) , š‘/√(š‘Ž^2 + š‘^2 + š‘^2 ) = šŸ/āˆššŸšŸ’ , šŸ/āˆššŸšŸ’ , šŸ‘/āˆššŸšŸ’

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