Example 15 - Show vectors a + b and a - b are perpendicular - Scalar product - Solving

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  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise
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Example 15 If ๐‘Ž๏ทฏ = 5 ๐‘–๏ทฏ โˆ’ ๐‘—๏ทฏ โˆ’ 3 ๐‘˜๏ทฏ and ๐‘๏ทฏ = ๐‘–๏ทฏ + 3 ๐‘—๏ทฏ โˆ’ 5 ๐‘˜๏ทฏ , then show that the vectors ๐‘Ž๏ทฏ + ๐‘๏ทฏ and ๐‘Ž๏ทฏ โˆ’ ๐‘๏ทฏ are perpendicular. Two vectors ๐‘๏ทฏ and ๐‘ž๏ทฏ are perpendicular if their scalar product is zero, i.e. ๐’‘๏ทฏ . ๐’’๏ทฏ = 0 ๐‘Ž๏ทฏ = 5 ๐‘–๏ทฏ โˆ’ ๐‘—๏ทฏ โˆ’ 3 ๐‘˜๏ทฏ ๐‘๏ทฏ = ๐‘–๏ทฏ + 3 ๐‘—๏ทฏ โˆ’ 5 ๐‘˜๏ทฏ ( ๐‘Ž๏ทฏ + ๐‘๏ทฏ) = (5 + 1) ๐‘–๏ทฏ + (โˆ’1 + 3) ๐‘—๏ทฏ + (โˆ’3 + (โˆ’5)) ๐‘˜๏ทฏ = 6 ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ + 8 ๐‘˜๏ทฏ ( ๐‘Ž๏ทฏ โˆ’ ๐‘๏ทฏ) = (5 โˆ’ 1) ๐‘–๏ทฏ + (โˆ’1 โˆ’ 3) ๐‘—๏ทฏ + (โˆ’3 โˆ’ (โˆ’5)) ๐‘˜๏ทฏ = 4 ๐‘–๏ทฏ โˆ’ 4 ๐‘—๏ทฏ + 2 ๐‘˜๏ทฏ Now, we have to show that ( ๐‘Ž๏ทฏ + ๐‘๏ทฏ) and ( ๐‘Ž๏ทฏ โˆ’ ๐‘๏ทฏ) are perpendicular to each other. So, we need to show ( ๐‘Ž๏ทฏ + ๐‘๏ทฏ) . ( ๐‘Ž๏ทฏ โˆ’ ๐‘๏ทฏ) = 0 Taking LHS ( ๐’‚๏ทฏ + ๐’ƒ๏ทฏ) . ( ๐’‚๏ทฏ โˆ’ ๐’ƒ๏ทฏ) = (6 ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ โˆ’ 8 ๐‘˜๏ทฏ) . (4 ๐‘–๏ทฏ โˆ’ 4 ๐‘—๏ทฏ + 2 ๐‘˜๏ทฏ) = (6 ร— 4) + (2 ร— โˆ’4) + (โˆ’8 ร— 2) = 24 + (โˆ’8) + (-16) = 24 + (โˆ’24) = 0 Since ( ๐‘Ž๏ทฏ + ๐‘๏ทฏ) . ( ๐‘Ž๏ทฏ โˆ’ ๐‘๏ทฏ) = 0 Hence, ( ๐‘Ž๏ทฏ + ๐‘๏ทฏ) is perpendicular to ( ๐‘Ž๏ทฏ โˆ’ ๐‘๏ทฏ)

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