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 š‘Ž āƒ— = 5š‘– Ģ‚ āˆ’ š‘— Ģ‚ āˆ’ 3š‘˜ Ģ‚ and š‘ āƒ— = š‘– Ģ‚ + 3š‘— Ģ‚ āˆ’ 5š‘˜ Ģ‚ , then show that the vectors š‘Ž āƒ— + š‘ āƒ— and š‘Ž āƒ— āˆ’ š‘ āƒ— are perpendicular. Two vectors š‘ āƒ— and š‘ž āƒ— are perpendicular if their scalar product is zero, i.e. š’‘ āƒ— . š’’ āƒ— = 0 Finding (š’‚ āƒ— + š’ƒ āƒ—) and (š’‚ āƒ— āˆ’ š’ƒ āƒ—) (š’‚ āƒ— + š’ƒ āƒ—) = (5 + 1) š‘– Ģ‚ + (āˆ’1 + 3) š‘— Ģ‚ + (āˆ’3 + (āˆ’5)) š‘˜ Ģ‚ = 6š’Š Ģ‚ + 2š’‹ Ģ‚ āˆ’ 8š’Œ Ģ‚ (š’‚ āƒ— āˆ’ š’ƒ āƒ—) = (5 āˆ’ 1) š‘– Ģ‚ + (āˆ’1 āˆ’ 3) š‘— Ģ‚ + (āˆ’3 āˆ’ (āˆ’5)) š‘˜ Ģ‚ = 4š’Š Ģ‚ āˆ’ 4š’‹ Ģ‚ + 2š’Œ Ģ‚ We have to show that (š‘Ž āƒ— + š‘ āƒ—) and (š‘Ž āƒ— āˆ’ š‘ āƒ—) are perpendicular to each other. So, we need to show (š’‚ āƒ— + š’ƒ āƒ—) . (š’‚ āƒ— āˆ’ š’ƒ āƒ—) = 0 Solving LHS (š’‚ āƒ— + š’ƒ āƒ—) . (š’‚ āƒ— āˆ’ š’ƒ āƒ—) = (6š‘– Ģ‚ + 2š‘— Ģ‚ āˆ’ 8š‘˜ Ģ‚) . (4š‘– Ģ‚ āˆ’ 4š‘— Ģ‚ + 2š‘˜ Ģ‚) = (6 Ɨ 4) + (2 Ɨ āˆ’4) + (āˆ’8 Ɨ 2) = 24 āˆ’ 8 āˆ’16 = 0 Since (š‘Ž āƒ— + š‘ āƒ—) . (š‘Ž āƒ— āˆ’ š‘ āƒ—) = 0 Hence, (š‘Ž āƒ— + š‘ āƒ—) is perpendicular to (š‘Ž āƒ— āˆ’ š‘ āƒ—)

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