Question 35 (Choice 2) - CBSE Class 12 Sample Paper for 2024 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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The given lines are non-parallel lines. Let Shortest distance = |(𝑃𝑄) ⃗ | Since (𝑃𝑄) ⃗ is shortest distance, (𝑃𝑄) ⃗ ⊥ Line 1 (𝑃𝑄) ⃗ ⊥ Line 2 Point P Since point P lies on Line 1 Position vector of P = 𝜆(𝑖 ˆ−𝑗 ˆ+𝑘 ˆ) = (𝜆) 𝑖 ̂+(−𝜆)𝑗 ̂+(𝜆)𝑘 ̂ Now, (𝑷𝑸) ⃗ = Position vector of Q − Position vector of P = [𝑖 ̂+(−1−2𝜇)𝑗 ˆ+(𝜇)𝑘 ̂ ]−[(𝜆) 𝑖 ̂+(−𝜆)𝑗 ̂+(𝜆)𝑘 ̂] = (1 − 𝜆)𝚤̂ + (− 1 − 2𝜇 + 𝜆)𝚥̂ + (𝜇 − 𝜆)𝑘 ̂ Now, (𝑷𝑸) ⃗ ⊥ Line 1 (𝑟 ⃗ = 𝜆(𝑖 ˆ−𝑗 ˆ+𝑘 ˆ) ) Thus, (𝑃𝑄) ⃗ ⊥ (𝑖 ̂−𝑗 ̂+𝑘 ̂ ) And (𝑷𝑸) ⃗ . (𝒊 ̂−𝒋 ̂+𝒌 ̂ )=𝟎 (1 − 𝜆)𝚤̂ + (− 1 − 2𝜇 + 𝜆)𝚥̂ + (𝜇 − 𝜆)𝑘 ̂. (𝒊 ̂−𝒋 ̂+𝒌 ̂ )=𝟎 (1− 𝜆)1 + (−1 −2𝜇 + 𝜆)(−1) + (𝜇 − 𝜆)1 = 0 (1− 𝜆) + (1 + 2𝜇 - 𝜆) + (𝜇 − 𝜆) = 0 (1 + 1) + (− 𝜆 − 𝜆 − 𝜆) + (2𝜇 + 𝜇) = 0 2 − 3𝜆 + 3𝜇 = 0 3𝜇 − 3𝜆 = − 2 Similarly (𝑷𝑸) ⃗ ⊥ Line 2 (𝑟 ⃗ = 𝑖 ˆ−𝑗 ˆ+𝜇(−2𝑗 ˆ+𝑘 ˆ)) Thus, (𝑃𝑄) ⃗ ⊥(−2𝑗 ˆ+𝑘 ˆ) And (𝑷𝑸) ⃗ .(−2𝑗 ˆ+𝑘 ˆ)=𝟎 (1 − 𝜆)𝚤̂ + (− 1 − 2𝜇 + 𝜆)𝚥̂ + (𝜇 − 𝜆)𝑘 ̂.(−2𝑗 ˆ+𝑘 ˆ )=0 (1 − 𝜆)(0) + (− 1 − 2𝜇 + 𝜆)(−2) + (𝜇 − 𝜆)(1) = 0 0 + (2 +4𝜇 − 2𝜆) + (𝜇 − 𝜆) = 0 (2) + (−2𝜆 − 𝜆) + (4𝜇 + 𝜇) = 0 2 − 3𝜆 + 5𝜇 = 0 5𝜇 − 3𝜆 = −2 Thus, our equations are 3𝜇 − 3𝜆 = −2 …(1) 5𝜇 − 3𝜆 = −2 …(2) Solving (1) and (2) We get 𝜇 = 0, 𝜆 = 𝟐/𝟑 Point P Position vector of P = (𝜆) 𝑖 ̂+(−𝜆)𝑗 ̂+(𝜆)𝑘 ̂ Putting 𝜆 = 𝟐/𝟑 = 𝟐/𝟑 𝒊 ̂−𝟐/𝟑 𝒋 ̂+ 𝟐/𝟑 𝒌 ̂ Point Q Position vector of Q = 𝑖 ̂+(−1−2𝜇)𝑗 ˆ+(𝜇)𝑘 ̂ Putting 𝜇 = 0 =( 𝑖) ̂+(−1 −0) 𝑗 ̂+(0)𝑘 ̂ = 𝒊 ̂−𝒋 ̂ Now, (𝑷𝑸) ⃗ = Position vector of Q − Position vector of P =[ 𝒊 ̂−(𝒋]) ̂−[ 𝟐/𝟑 𝒊 ̂−𝟐/𝟑 𝒋 ̂+ 𝟐/𝟑 𝒌 ̂] = 𝟏/𝟑 𝒊 ̂−(𝟏 )/𝟑 𝒋 ̂−𝟐/𝟑 𝒌 ̂ And, Shortest distance = |(𝑃𝑄) ⃗ | = √((1/3)^2+(−1/3)^2+(2/3)^2 ) =√( 1/9+1/9 +4/9) = √(6/9) = √(𝟐/𝟑) units