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 ๐‘Ž โƒ— = 5๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚ and ๐‘ โƒ— = ๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ 5๐‘˜ ฬ‚ , then show that the vectors ๐‘Ž โƒ— + ๐‘ โƒ— and ๐‘Ž โƒ— โˆ’ ๐‘ โƒ— are perpendicular. Two vectors ๐‘ โƒ— and ๐‘ž โƒ— are perpendicular if their scalar product is zero, i.e. ๐’‘ โƒ— . ๐’’ โƒ— = 0 Finding (๐’‚ โƒ— + ๐’ƒ โƒ—) and (๐’‚ โƒ— โˆ’ ๐’ƒ โƒ—) (๐’‚ โƒ— + ๐’ƒ โƒ—) = (5 + 1) ๐‘– ฬ‚ + (โˆ’1 + 3) ๐‘— ฬ‚ + (โˆ’3 + (โˆ’5)) ๐‘˜ ฬ‚ = 6๐’Š ฬ‚ + 2๐’‹ ฬ‚ โˆ’ 8๐’Œ ฬ‚ (๐’‚ โƒ— โˆ’ ๐’ƒ โƒ—) = (5 โˆ’ 1) ๐‘– ฬ‚ + (โˆ’1 โˆ’ 3) ๐‘— ฬ‚ + (โˆ’3 โˆ’ (โˆ’5)) ๐‘˜ ฬ‚ = 4๐’Š ฬ‚ โˆ’ 4๐’‹ ฬ‚ + 2๐’Œ ฬ‚ We have to show that (๐‘Ž โƒ— + ๐‘ โƒ—) and (๐‘Ž โƒ— โˆ’ ๐‘ โƒ—) are perpendicular to each other. So, we need to show (๐’‚ โƒ— + ๐’ƒ โƒ—) . (๐’‚ โƒ— โˆ’ ๐’ƒ โƒ—) = 0 Solving LHS (๐’‚ โƒ— + ๐’ƒ โƒ—) . (๐’‚ โƒ— โˆ’ ๐’ƒ โƒ—) = (6๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 8๐‘˜ ฬ‚) . (4๐‘– ฬ‚ โˆ’ 4๐‘— ฬ‚ + 2๐‘˜ ฬ‚) = (6 ร— 4) + (2 ร— โˆ’4) + (โˆ’8 ร— 2) = 24 โˆ’ 8 โˆ’16 = 0 Since (๐‘Ž โƒ— + ๐‘ โƒ—) . (๐‘Ž โƒ— โˆ’ ๐‘ โƒ—) = 0 Hence, (๐‘Ž โƒ— + ๐‘ โƒ—) is perpendicular to (๐‘Ž โƒ— โˆ’ ๐‘ โƒ—)

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