CBSE Class 12 Sample Paper for 2024 Boards

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

## An aeroplane is flying along the line r =λ( i  ˆ - j  ˆ + k  ˆ ); where ' λ ' is a scalar and another aeroplane is flying along the line r =i ˆ-j ˆ+μ(-2j ˆ+ k  ˆ ); where ' μ ' is a scalar. At what points on the lines should they reach, so that the distance between them is the shortest? Find the shortest possible distance between them.

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ο»ΏThe given lines are non-parallel lines. Let Shortest distance = |(ππ) β | Since (ππ) β is shortest distance, (ππ) β β₯ Line 1 (ππ) β β₯ Line 2 Point P Since point P lies on Line 1 Position vector of P = π(π Λβπ Λ+π Λ) = (π) π Μ+(βπ)π Μ+(π)π Μ Now, (π·πΈ) β = Position vector of Q β Position vector of P = [π Μ+(β1β2π)π Λ+(π)π Μ ]β[(π) π Μ+(βπ)π Μ+(π)π Μ] = ο»Ώ(1 β π)π€Μ + (β 1 β 2π + π)π₯Μ + (π β π)π Μ Now, (π·πΈ) β β₯ Line 1 (π β = π(π Λβπ Λ+π Λ) ) Thus, (ππ) β β₯ (π Μβπ Μ+π Μ ) And (π·πΈ) β . (π Μβπ Μ+π Μ )=π ο»Ώ(1 β π)π€Μ + (β 1 β 2π + π)π₯Μ + (π β π)π Μ. (π Μβπ Μ+π Μ )=π (1β π)1 + (β1 β2π + π)(β1) + (π β π)1 = 0 (1β π) + (1 + 2π - π) + (π β π) = 0 (1 + 1) + (β π β π β π) + (2π + π) = 0 2 β 3π + 3π = 0 3π β 3π = β 2 Similarly (π·πΈ) β β₯ Line 2 (π β = π Λβπ Λ+π(β2π Λ+π Λ)) Thus, (ππ) β β₯(β2π Λ+π Λ) And (π·πΈ) β .(β2π Λ+π Λ)=π (1 β π)π€Μ + (β 1 β 2π + π)π₯Μ + (π β π)π Μ.(β2π Λ+π Λ )=0 (1 β π)(0) + (β 1 β 2π + π)(β2) + (π β π)(1) = 0 0 + (2 +4π β 2π) + (π β π) = 0 (2) + (β2π β π) + (4π + π) = 0 2 β 3π + 5π = 0 5π β 3π = β2 Thus, our equations are 3π β 3π = β2 β¦(1) 5π β 3π = β2 β¦(2) Solving (1) and (2) We get π = 0, π = π/π Point P Position vector of P = (π) π Μ+(βπ)π Μ+(π)π Μ Putting π = π/π = π/π π Μβπ/π π Μ+ π/π π Μ Point Q Position vector of Q = π Μ+(β1β2π)π Λ+(π)π Μ Putting π = 0 =( π) Μ+(β1 β0) π Μ+(0)π Μ = π Μβπ Μ Now, (π·πΈ) β = Position vector of Q β Position vector of P =[ π Μβ(π]) Μβ[ π/π π Μβπ/π π Μ+ π/π π Μ] = π/π π Μβ(π )/π π Μβπ/π π Μ And, Shortest distance = |(ππ) β | = β((1/3)^2+(β1/3)^2+(2/3)^2 ) =β( 1/9+1/9 +4/9) = β(6/9) = β(π/π) units