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Example 15 - Show vectors a + b and a - b are perpendicular

Example 15 - Chapter 10 Class 12 Vector Algebra - Part 2

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Example 15 If π‘Ž βƒ— = 5𝑖 Μ‚ βˆ’ 𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚ and 𝑏 βƒ— = 𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚ , then show that the vectors π‘Ž βƒ— + 𝑏 βƒ— and π‘Ž βƒ— βˆ’ 𝑏 βƒ— are perpendicular. Two vectors 𝑝 βƒ— and π‘ž βƒ— are perpendicular if their scalar product is zero, i.e. 𝒑 βƒ— . 𝒒 βƒ— = 0 Finding (𝒂 βƒ— + 𝒃 βƒ—) and (𝒂 βƒ— βˆ’ 𝒃 βƒ—) (𝒂 βƒ— + 𝒃 βƒ—) = (5 + 1) 𝑖 Μ‚ + (βˆ’1 + 3) 𝑗 Μ‚ + (βˆ’3 + (βˆ’5)) π‘˜ Μ‚ = 6π’Š Μ‚ + 2𝒋 Μ‚ βˆ’ 8π’Œ Μ‚ (𝒂 βƒ— βˆ’ 𝒃 βƒ—) = (5 βˆ’ 1) 𝑖 Μ‚ + (βˆ’1 βˆ’ 3) 𝑗 Μ‚ + (βˆ’3 βˆ’ (βˆ’5)) π‘˜ Μ‚ = 4π’Š Μ‚ βˆ’ 4𝒋 Μ‚ + 2π’Œ Μ‚ We have to show that (π‘Ž βƒ— + 𝑏 βƒ—) and (π‘Ž βƒ— βˆ’ 𝑏 βƒ—) are perpendicular to each other. So, we need to show (𝒂 βƒ— + 𝒃 βƒ—) . (𝒂 βƒ— βˆ’ 𝒃 βƒ—) = 0 Solving LHS (𝒂 βƒ— + 𝒃 βƒ—) . (𝒂 βƒ— βˆ’ 𝒃 βƒ—) = (6𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 8π‘˜ Μ‚) . (4𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ + 2π‘˜ Μ‚) = (6 Γ— 4) + (2 Γ— βˆ’4) + (βˆ’8 Γ— 2) = 24 βˆ’ 8 βˆ’16 = 0 Since (π‘Ž βƒ— + 𝑏 βƒ—) . (π‘Ž βƒ— βˆ’ 𝑏 βƒ—) = 0 Hence, (π‘Ž βƒ— + 𝑏 βƒ—) is perpendicular to (π‘Ž βƒ— βˆ’ 𝑏 βƒ—)

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