Example 20 - Show |a + b| < |a| + |b| (triangle inequality) - Examples

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  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise
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Example 20 For any two vectors and , we always have | + | | | + | | (triangle inequality). To Prove: | + | | | + | | We first prove trivially, Therefore, the inequality | + | | | + | | is satisfied trivially. Now, let us assume 0 & 0 | + |2 = ( + ) . ( + ) = . + . + . + . = . + . + . + . = . + 2 . + . = | |2 + 2 . + | |2 = | |2 + 2| || | cos + | |2 | + |2 = | |2 + 2| || | cos + | |2 We know that cos 1 Multiplying 2| || | on both sides 2| || | cos 2| || | Adding | |2 + | |2 on both sides, | |2 + | |2 + 2| || | cos | |2 + | |2 + 2 | | | | | + |2 (| | + | |) 2 Taking square root both sides | + | (| | + | |) Hence proved.

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