Last updated at May 29, 2018 by Teachoo

Transcript

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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Example 20 You are here

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Example 26 (Supplementary NCERT)

Example 27 (Supplementary NCERT)

Example 28 (Supplementary NCERT)

Example 29 (Supplementary NCERT)

Example 30 (Supplementary NCERT)

Example 31 (Supplementary NCERT)

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.