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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Supplementary Exercise Q2 If 𝑎 ⃗ = 2𝑖 ̂ − 3𝑗 ̂ + 4𝑘 ̂, 𝑏 ⃗ = 𝑖 ̂ + 2𝑗 ̂ − 6𝑘 ̂ and 𝑐 ⃗ = 3𝑖 ̂ − 𝑗 ̂ + 2𝑘 ̂ , then (i) find [𝑎 ⃗ 𝑏 ⃗ 𝑐 ⃗ ] Given, 𝑎 ⃗ = 2𝑖 ̂ − 3𝑗 ̂ + 4𝑘 ̂ , 𝑏 ⃗ = 𝑖 ̂ + 2𝑗 ̂ – 𝑘 ̂ , 𝑐 ⃗ = 3𝑖 ̂ – 𝑗 ̂ + 2𝑘 ̂ [𝑎 ⃗" " 𝑏 ⃗" " 𝑐 ⃗ ] = |■8(2&−3&4@1&2&−1@3&−1&2)| = 2[(2×2)−(−1×−1) ] − (−3) [(1×2)−(3×−1) ] + 4[(1×−1)−(3×2)] = 2 [4−1]+3(2+3)+4[−1−6] = 2(3) + 3 (5) + 4(–7) = 6 + 15 – 28 = –7 Supplementary Exercise Q2 (Method 1) If 𝑎 ⃗ = 2𝑖 ̂ − 3𝑗 ̂ + 4𝑘 ̂, 𝑏 ⃗ = 𝑖 ̂ + 2𝑗 ̂ − 6𝑘 ̂ and 𝑐 ⃗ = 3𝑖 ̂ − 𝑗 ̂ + 2𝑘 ̂ , then (ii) find [𝑎 ⃗+𝑏 ⃗ 𝑏 ⃗+𝑐 ⃗ 𝑐 ⃗+𝑎 ⃗ ] Given, 𝑎 ⃗ = 2𝑖 ̂ − 3𝑗 ̂ + 4𝑘 ̂ , 𝑏 ⃗ = 𝑖 ̂ + 2𝑗 ̂ – 𝑘 ̂ , 𝑐 ⃗ = 3𝑖 ̂ – 𝑗 ̂ + 2𝑘 ̂ We need to find [■8(𝑎 ⃗" + " 𝑏 ⃗&𝑏 ⃗+𝑐 ⃗&𝑐 ⃗+𝑎 ⃗ )] 𝑎 ⃗" + " 𝑏 ⃗ = (2𝑖 ̂ − 3𝑗 ̂ + 4𝑘 ̂) + (𝑖 ̂ + 2𝑗 ̂ – 𝑘 ̂) = 3𝑖 ̂ − 𝑗 ̂ + 3𝑘 ̂ 𝑏 ⃗ + 𝑐 ⃗ = (𝑖 ̂ + 2𝑗 ̂ – 𝑘 ̂) + (3𝑖 ̂ – 𝑗 ̂ + 2𝑘 ̂) = 4𝑖 ̂ + 𝑗 ̂ + 𝑘 ̂ 𝑐 ⃗ + 𝑎 ⃗ = (3𝑖 ̂ – 𝑗 ̂ + 2𝑘 ̂) + (2𝑖 ̂ − 3𝑗 ̂ + 4𝑘 ̂) = 5𝑖 ̂ − 4𝑗 ̂ + 6𝑘 ̂ [■8(𝑎 ⃗" + " 𝑏 ⃗&𝑏 ⃗+𝑐 ⃗&𝑐 ⃗+𝑎 ⃗ )] = |■8(3&−1&3@4&1&1@5&−4&6)| = 3[(1×6)−(−4×1) ] − (−1) [(4×6)−(5×1)] + 3[(4×−4)−(5×1)] = 3 [6+4]+1(24−5)+3[−16−5] = 3(10) + 1 (19) + 3(–21) = 30 + 19 – 63 = –14 ∴ [■8(𝑎 ⃗" + " 𝑏 ⃗&𝑏 ⃗+𝑐 ⃗&𝑐 ⃗+𝑎 ⃗ )] = –14 Supplementary Exercise Q2 (Method 2) If 𝑎 ⃗ = 2𝑖 ̂ − 3𝑗 ̂ + 4𝑘 ̂, 𝑏 ⃗ = 𝑖 ̂ + 2𝑗 ̂ − 6𝑘 ̂ and 𝑐 ⃗ = 3𝑖 ̂ − 𝑗 ̂ + 2𝑘 ̂ , then (ii) find [𝑎 ⃗+𝑏 ⃗ 𝑏 ⃗+𝑐 ⃗ 𝑐 ⃗+𝑎 ⃗ ] We need to find [■8(𝑎 ⃗" + " 𝑏 ⃗&𝑏 ⃗+𝑐 ⃗&𝑐 ⃗+𝑎 ⃗ )] We know that [■8(𝒂 ⃗" + " 𝒃 ⃗&𝒃 ⃗+𝒄 ⃗&𝒄 ⃗+𝒂 ⃗ )] = 2[𝒂 ⃗" " 𝒃 ⃗" " 𝒄 ⃗ ] = 2(–7) = –14

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.