Find the acute angle between the lines

(x - 4)/3 = (y + 3)/4 = (z + 1)/5 and (x - 1)/4 = (y + 1)/(-3) = (z + 10)/5

Find the acute angle between lines  (x - 4)/3 = (y + 3)/4 = (z + 1)/5

Question 25 - CBSE Class 12 Sample Paper for 2020 Boards - Part 2
Question 25 - CBSE Class 12 Sample Paper for 2020 Boards - Part 3

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Question 25 Find the acute angle between the lines (š‘„ āˆ’ 4)/3 = (š‘¦ + 3)/4 = (š‘§ + 1)/5 and (š‘„ āˆ’ 1)/4 = (š‘¦ + 1)/(āˆ’3) = (š‘§ + 10)/5 Angle between the pair of lines (š‘„ āˆ’ š‘„_1)/š‘Ž_1 = (š‘¦ āˆ’ š‘¦_1)/š‘_1 = (š‘§ āˆ’ š‘§_1)/š‘_1 and (š‘„ āˆ’ š‘„_2)/š‘Ž_2 = (š‘¦ āˆ’ š‘¦_2)/š‘_2 = (š‘§ āˆ’ š‘§_2)/š‘_2 is given by cos Īø = |(š‘Ž_1 š‘Ž_2 + š‘_1 š‘_2 + š‘_1 š‘_2)/(√(ć€–š‘Ž_1怗^2 + ć€–š‘_1怗^2 + ć€–š‘_1怗^2 ) √(ć€–š‘Ž_2怗^2 + ć€–š‘_2怗^2 + ć€–š‘_2怗^2 ))| (š’™ āˆ’ šŸ’)/šŸ‘ = (š’š + šŸ‘)/šŸ’ = (š’› + šŸ)/šŸ“ Comparing with (š‘„ āˆ’ š‘„_1)/š‘Ž_1 = (š‘¦ āˆ’ š‘¦_1)/š‘_1 = (š‘§ āˆ’ š‘§_1)/š‘_1 š‘Ž1 = 3, b1 = 4, c1 = 4 (š’™ āˆ’ šŸ)/šŸ’ = (š’š + šŸ)/(āˆ’šŸ‘) = (š’› + šŸšŸŽ)/šŸ“ Comparing with (š‘„ āˆ’ š‘„_2)/š‘Ž_2 = (š‘¦ āˆ’ š‘¦_2)/š‘_2 = (š‘§ āˆ’ š‘§_2)/š‘_2 š‘Ž2 = 4, š‘2 = –3, š‘2 = 5 Now, cos Īø = |(š‘Ž_1 š‘Ž_2 + š‘_1 š‘_2 + š‘_1 š‘_2)/(√(ć€–š‘Ž_1怗^2 + ć€–š‘_1怗^2 + ć€–š‘_1怗^2 ) √(ć€–š‘Ž_2怗^2 + ć€–š‘_2怗^2 + ć€–š‘_2怗^2 ))| = |(3 Ɨ 4 + 4 Ɨ (āˆ’3) + 5 Ɨ 5)/(√(3^2 + 4^2 + 5^2 ) √(4^2 +(āˆ’3)^2 + 5^2 ))| = |(12 āˆ’ 12 + 25)/(√(9 + 16 + 25) √(16 + 9 + 25))| = |25/(√50 √50)| = |25/50| = |1/2| = 1/2 So, cos Īø = 1/2 ∓ Īø = 60° = š…/šŸ‘ Therefore, required angle is š…/šŸ‘ Note: Please write angle in radians and not degree

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