1. Chapter 10 Class 12 Vector Algebra
2. Serial order wise

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Supplementary Exercise Q2 If 𝑎﷯ = 2 𝑖﷯ − 3 𝑗﷯ + 4 𝑘﷯, 𝑏﷯ = 𝑖﷯ + 2 𝑗﷯ − 6 𝑘﷯ and 𝑐﷯ = 3 𝑖﷯ − 𝑗﷯ + 2 𝑘﷯ , then (i) find 𝑎﷯ 𝑏﷯ 𝑐﷯﷯ Given, 𝑎﷯ = 2 𝑖﷯ − 3 𝑗﷯ + 4 𝑘﷯ , 𝑏﷯ = 𝑖﷯ + 2 𝑗﷯ – 𝑘﷯ , 𝑐﷯ = 3 𝑖﷯ – 𝑗﷯ + 2 𝑘﷯ 𝑎﷯ 𝑏﷯ 𝑐﷯﷯ = 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﷯ = 2 𝑖﷯ − 3 𝑗﷯ + 4 𝑘﷯, 𝑏﷯ = 𝑖﷯ + 2 𝑗﷯ − 6 𝑘﷯ and 𝑐﷯ = 3 𝑖﷯ − 𝑗﷯ + 2 𝑘﷯ , then (ii) find 𝑎﷯+ 𝑏﷯ 𝑏﷯+ 𝑐﷯ 𝑐﷯+ 𝑎﷯﷯ Given, 𝑎﷯ = 2 𝑖﷯ − 3 𝑗﷯ + 4 𝑘﷯ , 𝑏﷯ = 𝑖﷯ + 2 𝑗﷯ – 𝑘﷯ , 𝑐﷯ = 3 𝑖﷯ – 𝑗﷯ + 2 𝑘﷯ We need to find 𝑎﷯ + 𝑏﷯﷮ 𝑏﷯+ 𝑐﷯﷮ 𝑐﷯+ 𝑎﷯﷯﷯ 𝑎﷯ + 𝑏﷯ = (2 𝑖﷯ − 3 𝑗﷯ + 4 𝑘﷯) + ( 𝑖﷯ + 2 𝑗﷯ – 𝑘﷯) = 3 𝑖﷯ − 𝑗﷯ + 3 𝑘﷯ 𝑏﷯ + 𝑐﷯ = ( 𝑖﷯ + 2 𝑗﷯ – 𝑘﷯) + (3 𝑖﷯ – 𝑗﷯ + 2 𝑘﷯) = 4 𝑖﷯ + 𝑗﷯ + 𝑘﷯ 𝑐﷯ + 𝑎﷯ = (3 𝑖﷯ – 𝑗﷯ + 2 𝑘﷯) + (2 𝑖﷯ − 3 𝑗﷯ + 4 𝑘﷯) = 5 𝑖﷯ − 4 𝑗﷯ + 6 𝑘﷯ 𝑎﷯ + 𝑏﷯﷮ 𝑏﷯+ 𝑐﷯﷮ 𝑐﷯+ 𝑎﷯﷯﷯ = 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 ∴ 𝑎﷯ + 𝑏﷯﷮ 𝑏﷯+ 𝑐﷯﷮ 𝑐﷯+ 𝑎﷯﷯﷯ = –14 Supplementary Exercise Q2 (Method 2) If 𝑎﷯ = 2 𝑖﷯ − 3 𝑗﷯ + 4 𝑘﷯, 𝑏﷯ = 𝑖﷯ + 2 𝑗﷯ − 6 𝑘﷯ and 𝑐﷯ = 3 𝑖﷯ − 𝑗﷯ + 2 𝑘﷯ , then (ii) find 𝑎﷯+ 𝑏﷯ 𝑏﷯+ 𝑐﷯ 𝑐﷯+ 𝑎﷯﷯ We need to find 𝑎﷯ + 𝑏﷯﷮ 𝑏﷯+ 𝑐﷯﷮ 𝑐﷯+ 𝑎﷯﷯﷯ We know that 𝒂﷯ + 𝒃﷯﷮ 𝒃﷯+ 𝒄﷯﷮ 𝒄﷯+ 𝒂﷯﷯﷯ = 2 𝒂﷯ 𝒃﷯ 𝒄﷯﷯ = 2(–7) = –14