Assertion (A): The acute angle between the line

r = i ̂+j ̂+2k ̂+λ(i ̂-j ̂ ) and the x-axis is π/4

Reason(R): The acute angle šœƒ between the lines

Ā r = x 1 i Ģ‚+y 1 j Ģ‚+z 1 k Ģ‚+Ī»(a 1 i Ģ‚+b 1 j Ģ‚+c 1 k Ģ‚ )Ā  and

Ā rĀ  = x 1 i Ģ‚+y 1 j Ģ‚+z 1 k Ģ‚+μ(a 1 i Ģ‚+b 1 j Ģ‚+c 1 k Ģ‚ )is given by

š‘š‘œš‘ šœƒ = |a 1 a 2 + b 1 b 2 + c 1 c 2 |/(√a 1 2 + b 1 2 + c 1 2 √a 2 2 + b2 2 . + c 2 2

Ā 

Assertion (A): The acute angle between the line r = i + j + 2k + Ī»(i-j - CBSE Class 12 Sample Paper for 2023 Boards

part 2 - Question 20 [Assertion Reasoning] - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 3 - Question 20 [Assertion Reasoning] - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 4 - Question 20 [Assertion Reasoning] - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 5 - Question 20 [Assertion Reasoning] - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12 part 6 - Question 20 [Assertion Reasoning] - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12 part 7 - Question 20 [Assertion Reasoning] - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

Ā 

Ā 

Remove Ads Share on WhatsApp

Transcript

Question 20 Assertion (A): The acute angle between the line š‘Ÿ āƒ— = š‘– Ģ‚+š‘— Ģ‚+2š‘˜ Ģ‚+šœ†(š‘– Ģ‚āˆ’š‘— Ģ‚ ) and the x-axis is šœ‹/4 Reason(R): The acute angle šœƒ between the lines š‘Ÿ āƒ— = š‘„_1 š‘– Ģ‚+š‘¦_1 š‘— Ģ‚+š‘§_1 š‘˜ Ģ‚+šœ†(š‘Ž_1 š‘– Ģ‚+š‘_1 š‘— Ģ‚+š‘_1 š‘˜ Ģ‚ ) and š‘Ÿ āƒ— = š‘„_2 š‘– Ģ‚+š‘¦_2 š‘— Ģ‚+š‘§_2 š‘˜ Ģ‚+šœ‡(š‘Ž_2 š‘– Ģ‚+š‘_2 š‘— Ģ‚+š‘_2 š‘˜ Ģ‚ )is given by š‘š‘œš‘ šœƒ = |š‘Ž_1 š‘Ž_2 + š‘_1 š‘_2 + 怖 š‘ć€—_1 š‘_2 |/(√(ć€–š‘Ž1怗^2 + ć€–š‘1怗^2 + ć€–š‘1怗^2 ) √(ć€–š‘Ž2怗^2 + ć€–š‘2怗^2. + ć€–š‘2怗^2 )) Checking Assertion Assertion (A): The acute angle between the line š‘Ÿ āƒ— = š‘– Ģ‚+š‘— Ģ‚+2š‘˜ Ģ‚+šœ†(š‘– Ģ‚āˆ’š‘— Ģ‚ ) and the x-axis is šœ‹/4 Equation of x-axis Let’s consider two points on x-axis – (a, 0, 0), and (0, 0, 0) Vector equation of a line passing though two points with position vectors š‘Ž āƒ— and š‘ āƒ— is š’“ āƒ— = (š’‚ ) āƒ— + šœ† (š’ƒ āƒ— āˆ’ š’‚ āƒ—) Here, (a, 0, 0) š’‚ āƒ— = aš‘– Ģ‚ + 0š‘— Ģ‚ + 0š‘˜ Ģ‚ (0, 0, 0) š’ƒ āƒ— = 0š‘– Ģ‚ + 0š‘— Ģ‚ + 0š‘˜ Ģ‚ Thus, equation of line is š’“ āƒ— = (aš‘– Ģ‚ + 0š‘— Ģ‚ + 0š‘˜ Ģ‚) + šœ† [(0š‘– Ģ‚+0š‘—+0š‘˜ Ģ‚ ) āˆ’ (š‘Žš‘– Ģ‚ +0š‘— Ģ‚ + 20)] = aš‘– Ģ‚ + šœ† [āˆ’š‘Žš‘– Ģ‚ ] = (a āˆ’ šœ†a)š’Š Ģ‚ Since (a āˆ’ šœ†a) is a constant, let (a āˆ’ šœ†a) = k = kš’Š Ģ‚ Now, we need to find angle between the line š‘Ÿ āƒ— = š‘– Ģ‚+š‘— Ģ‚+2š‘˜ Ģ‚+šœ†(š‘– Ģ‚āˆ’š‘— Ģ‚ ) and the x-axis i.e. Angle between š’“ āƒ— = š’Š Ģ‚+š’‹ Ģ‚+šŸš’Œ Ģ‚+š€(š’Š Ģ‚āˆ’š’‹ Ģ‚ ) and š’“ āƒ— = kš’Š Ģ‚ Using formula from Reasoning š‘š‘œš‘  šœƒ = |š’‚_šŸ š’‚_šŸ + š’ƒ_šŸ š’ƒ_šŸ + 怖 š’„ć€—_šŸ š’„_šŸ |/(√(ć€–š’‚šŸć€—^šŸ + ć€–š’ƒšŸć€—^šŸ + ć€–š’„šŸć€—^šŸ ) √(ć€–š’‚šŸć€—^šŸ + ć€–š’ƒšŸć€—^šŸ. + ć€–š’„šŸć€—^šŸ )) Thus, equation of line is š’“ āƒ— = (aš‘– Ģ‚ + 0š‘— Ģ‚ + 0š‘˜ Ģ‚) + šœ† [(0š‘– Ģ‚+0š‘—+0š‘˜ Ģ‚ ) āˆ’ (š‘Žš‘– Ģ‚ +0š‘— Ģ‚ + 20)] = aš‘– Ģ‚ + šœ† [āˆ’š‘Žš‘– Ģ‚ ] = (a āˆ’ šœ†a)š’Š Ģ‚ Since (a āˆ’ šœ†a) is a constant, let (a āˆ’ šœ†a) = k = kš’Š Ģ‚ Now, we need to find angle between the line š‘Ÿ āƒ— = š‘– Ģ‚+š‘— Ģ‚+2š‘˜ Ģ‚+šœ†(š‘– Ģ‚āˆ’š‘— Ģ‚ ) and the x-axis i.e. Angle between š’“ āƒ— = š’Š Ģ‚+š’‹ Ģ‚+šŸš’Œ Ģ‚+š€(š’Š Ģ‚āˆ’š’‹ Ģ‚ ) and š’“ āƒ— = kš’Š Ģ‚ Using formula from Reasoning š‘š‘œš‘  šœƒ = |š’‚_šŸ š’‚_šŸ + š’ƒ_šŸ š’ƒ_šŸ + 怖 š’„ć€—_šŸ š’„_šŸ |/(√(ć€–š’‚šŸć€—^šŸ + ć€–š’ƒšŸć€—^šŸ + ć€–š’„šŸć€—^šŸ ) √(ć€–š’‚šŸć€—^šŸ + ć€–š’ƒšŸć€—^šŸ. + ć€–š’„šŸć€—^šŸ )) š’“ āƒ— = š’Š Ģ‚+š’‹ Ģ‚+šŸš’Œ Ģ‚+š€(š’Š Ģ‚āˆ’š’‹ Ģ‚ ) Comparing with š‘Ÿ āƒ— = š‘„_1 š‘– Ģ‚+š‘¦_1 š‘— Ģ‚+š‘§_1 š‘˜ Ģ‚+šœ†(š‘Ž_1 š‘– Ģ‚+š‘_1 š‘— Ģ‚+š‘_1 š‘˜ Ģ‚ ) š’‚1 = 1, b1 = āˆ’1, c1 = 0 š’“ āƒ—" = k" š’Š Ģ‚ Comparing with š‘Ÿ āƒ— = š‘„_2 š‘– Ģ‚+š‘¦_2 š‘— Ģ‚+š‘§_2 š‘˜ Ģ‚+šœ†(š‘Ž_2 š‘– Ģ‚+š‘_2 š‘— Ģ‚+š‘_2 š‘˜ Ģ‚ ) š’‚2 = 1, š’ƒ2 = 0, š’„2 = 0 Now, cos Īø = |(š‘Ž_1 š‘Ž_2 + š‘_1 š‘_2 +怖 š‘ć€—_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 +怖怖 š‘ć€—_2怗^2+ ć€–š‘_2怗^2 ))| = |((1 Ɨ 1) + (āˆ’1 Ɨ 0) + (0 Ɨ 0))/(√(1^2 +(āˆ’1)^2 + 0^2 ) Ɨ √(1^2 + 0^2 + 0^2 ))| = |1/(√(1 + 1) Ɨ √1)| = |1/√2| = šŸ/āˆššŸ So, cos Īø = 1/√2 ∓ Īø = šœ‹/šŸ’ Therefore, the angle between the given pair of line is šœ‹/šŸ’ So, Assertion is true Checking Reason Reason(R): The acute angle šœƒ between the lines š‘Ÿ āƒ— = š‘„_1 š‘– Ģ‚+š‘¦_1 š‘— Ģ‚+š‘§_1 š‘˜ Ģ‚+šœ†(š‘Ž_1 š‘– Ģ‚+š‘_1 š‘— Ģ‚+š‘_1 š‘˜ Ģ‚ ) and š‘Ÿ āƒ— = š‘„_2 š‘– Ģ‚+š‘¦_2 š‘— Ģ‚+š‘§_2 š‘˜ Ģ‚+šœ‡(š‘Ž_2 š‘– Ģ‚+š‘_2 š‘— Ģ‚+š‘_2 š‘˜ Ģ‚ )is given by š‘š‘œš‘ šœƒ = |š‘Ž_1 š‘Ž_2 + š‘_1 š‘_2 + 怖 š‘ć€—_1 š‘_2 |/(√(ć€–š‘Ž1怗^2 + ć€–š‘1怗^2 + ć€–š‘1怗^2 ) √(ć€–š‘Ž2怗^2 + ć€–š‘2怗^2. + ć€–š‘2怗^2 )) This is a formula and it is a correct formula Thus, Reasoning is true Is Reason a Correct explanation for Assertion? Sine we used formula mentioned in Reasoning to find Assertion Therefore, Reasoning is a correct explanation for Assertion So, Assertion is true Reasoning is true And, Reasoning is a correct explanation for Assertion So, the correct answer is (a)

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo