Question 28 (A) - CBSE Class 12 Sample Paper for 2025 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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Question 28 (A) An ant is moving along the vector (𝑙_1 ) ⃗=ı ˆ−2ȷ ˆ+3𝑘 ˆ. Few sugar crystals are kept along the vector (𝑙_2 ) ⃗=3ı ˆ−2ȷ ˆ+𝑘 ˆ which is inclined at an angle 𝜽 with the vector (𝑙_1 ) ⃗. Then find the angle 𝜽. Also find the scalar projection of (𝑙_1 ) ⃗ on (𝑙_2 ) ⃗.To find angle, we first find dot product Now, (𝒍_𝟏 ) ⃗. (𝒍_𝟐 ) ⃗= |(𝒍_𝟏 ) ⃗ ||(𝒍_𝟐 ) ⃗ |𝒄𝒐𝒔 θ Finding dot product & magnitude separately Now, (𝒍_𝟏 ) ⃗. (𝒍_𝟐 ) ⃗ = (𝑖 ̂ – 2𝑗 ̂ + 3𝑘 ̂) . (3𝑖 ̂ − 2𝑗 ̂ + 𝑘 ̂ ) = (1 × 3) + (–2 × –2) + (3 × 1) = 3 + 4 + 3 = 10 Magnitude of (𝒍_𝟏 ) ⃗ = √(12+(−2)2+3 2) |(𝒍_𝟏 ) ⃗ | = √(1+4+9) = √𝟏𝟒 Magnitude of (𝒍_𝟐 ) ⃗ = √(32+(−2) 2+1 2) |(𝒍_𝟐 ) ⃗ | = √(9+4+1) Now, putting values in dot product (𝑙_1 ) ⃗. (𝑙_2 ) ⃗= |(𝑙_1 ) ⃗ ||(𝑙_2 ) ⃗ |𝑐𝑜𝑠 𝜃 7 = √14 × √14 × 𝑐𝑜𝑠 𝜃 Projection of 𝒂 ⃗ on 𝒃 ⃗ = 1/("|" 𝑏 ⃗"|" ) (𝑎 ⃗. 𝑏 ⃗) = 𝟕/√𝟏𝟒 So, the correct answer is (a) = √𝟏𝟒 Now, putting values in dot product (𝑙_1 ) ⃗. (𝑙_2 ) ⃗= |(𝑙_1 ) ⃗ ||(𝑙_2 ) ⃗ |𝑐𝑜𝑠 𝜃 10 = √𝟏𝟒 × √𝟏𝟒 × 𝒄𝒐𝒔 𝜽 10 = 14 𝑐𝑜𝑠 𝜃 10/14=cos⁡𝜃 5/7=cos⁡𝜃 𝒄𝒐𝒔⁡𝜽=𝟓/𝟕 ∴ 𝜃 = cos^(−𝟏)⁡〖5/7〗 So, angle between two vectors is 〖𝒄𝒐𝒔〗^(−𝟏)⁡〖𝟓/𝟕〗 Now, We need to find scalar projection of (𝑙_1 ) ⃗ on (𝑙_2 ) ⃗. Projection of 𝒂 ⃗ on 𝒃 ⃗ = 1/("|" 𝑏 ⃗"|" ) (𝑎 ⃗. 𝑏 ⃗) = 𝟕/√𝟏𝟒 So, the correct answer is (a) So, angle between two vectors is 〖𝒄𝒐𝒔〗^(−𝟏)⁡〖𝟓/𝟕〗 Now, We need to find scalar projection of (𝑙_1 ) ⃗ on (𝑙_2 ) ⃗. Now, Projection of (𝑙_1 ) ⃗ on (𝑙_2 ) ⃗ = ((𝒍_𝟏 ) ⃗ . (𝒍_𝟐 ) ⃗)/|(𝒍_𝟐 ) ⃗ | = 10/√𝟏𝟒