Supplementary Exercise Q2
Last updated at April 16, 2024 by Teachoo
Supplementary examples and questions from CBSE
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Supplementary Example 5 Important
Supplementary Example 6
Supplementary Exercise Q1
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Supplementary Exercise Q3
Supplementary Exercise Q4 Important
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Supplementary Exercise Q6 Important
Supplementary Exercise Q7
Supplementary Exercise Q8
Supplementary Exercise Q9 Important
Supplementary Exercise Q10
Supplementary examples and questions from CBSE
Last updated at April 16, 2024 by Teachoo
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