CBSE Class 12 Sample Paper for 2020 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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

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Question 25 Find the acute angle between the lines (π₯ β 4)/3 = (π¦ + 3)/4 = (π§ + 1)/5 and (π₯ β 1)/4 = (π¦ + 1)/(β3) = (π§ + 10)/5 Angle between the pair of lines (π₯ β π₯_1)/π_1 = (π¦ β π¦_1)/π_1 = (π§ β π§_1)/π_1 and (π₯ β π₯_2)/π_2 = (π¦ β π¦_2)/π_2 = (π§ β π§_2)/π_2 is given by cos ΞΈ = |(π_1 π_2 + π_1 π_2 + π_1 π_2)/(β(γπ_1γ^2 + γπ_1γ^2 + γπ_1γ^2 ) β(γπ_2γ^2 + γπ_2γ^2 + γπ_2γ^2 ))| (π β π)/π = (π + π)/π = (π + π)/π Comparing with (π₯ β π₯_1)/π_1 = (π¦ β π¦_1)/π_1 = (π§ β π§_1)/π_1 π1 = 3, b1 = 4, c1 = 4 (π β π)/π = (π + π)/(βπ) = (π + ππ)/π Comparing with (π₯ β π₯_2)/π_2 = (π¦ β π¦_2)/π_2 = (π§ β π§_2)/π_2 π2 = 4, π2 = β3, π2 = 5 Now, cos ΞΈ = |(π_1 π_2 + π_1 π_2 + π_1 π_2)/(β(γπ_1γ^2 + γπ_1γ^2 + γπ_1γ^2 ) β(γπ_2γ^2 + γπ_2γ^2 + γπ_2γ^2 ))| = |(3 Γ 4 + 4 Γ (β3) + 5 Γ 5)/(β(3^2 + 4^2 + 5^2 ) β(4^2 +(β3)^2 + 5^2 ))| = |(12 β 12 + 25)/(β(9 + 16 + 25) β(16 + 9 + 25))| = |25/(β50 β50)| = |25/50| = |1/2| = 1/2 So, cos ΞΈ = 1/2 β΄ ΞΈ = 60Β° = π/π Therefore, required angle is π/π Note: Please write angle in radians and not degree