If a + b + c = 0 and |a| = 3, |b| = 5, |c| = 7, then find the value of a.bĀ  + b.cĀ  + c.a

If a + b + c = 0 and |a| = 3, |b| = 5, |c| = 7, then find value of

Question 24 (OR 2nd Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 2
Question 24 (OR 2nd Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 3

Note : - This is similar to Example 29 of NCERT – Chapter 10 Class 12

Check the answer here https://www.teachoo.com/3461/748/Example-29---Let-a---b---c--0--find-a.b---b.c---c.a-if--a---1/category/Examples/

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Question 24 (OR 2nd Question) if š‘Ž āƒ— + š‘ āƒ— + š‘ āƒ— = 0 and |š‘Ž āƒ— | = 3, |š‘ āƒ— | = 5, |š‘ āƒ— | = 7, then find the value of š‘Ž āƒ—. š‘ āƒ— + š‘ āƒ—. š‘ āƒ— + š‘ āƒ—. š‘Ž āƒ— . Given, š‘Ž āƒ— + š‘ āƒ— + š‘ āƒ— = 0 āƒ— So, |š’‚ āƒ—" + " š’ƒ āƒ—" + " š’„ āƒ— | = |šŸŽ āƒ— | = 0 Now, |š’‚ āƒ—+š’ƒ āƒ—+š’„ āƒ— |2 = (š’‚ āƒ— + š’ƒ āƒ— + š’„ āƒ—) . (š’‚ āƒ— + š’ƒ āƒ— + š’„ āƒ—) = š‘Ž āƒ—. š‘Ž āƒ— + š‘Ž āƒ— . š‘ āƒ— + š’‚ āƒ— . š’„ āƒ— + š’ƒ āƒ— . š’‚ āƒ— + š‘ āƒ— . š‘ āƒ— + š‘ āƒ— . š‘ āƒ— + š‘ āƒ— . š‘Ž āƒ— + š’„ āƒ— . š’ƒ āƒ— + š‘ āƒ— . š‘ āƒ— = š‘Ž āƒ—. š‘Ž āƒ— + š‘Ž āƒ— . š‘ āƒ— + š’„ āƒ— . š’‚ āƒ— + š’‚ āƒ— . š’ƒ āƒ— + š‘ āƒ— . š‘ āƒ— + š‘ āƒ— . š‘ āƒ— + š‘ āƒ— . š‘Ž āƒ— + š’ƒ āƒ— . š’„ āƒ— + š‘ āƒ— . š‘ āƒ— = š‘Ž āƒ— . š‘Ž āƒ— + š‘ āƒ— . š‘ āƒ— + š‘ āƒ— . š‘ āƒ— + 2š‘Ž āƒ—. š‘ āƒ— + 2š‘ āƒ—. š‘ āƒ— + 2š‘ āƒ—. š‘Ž āƒ— = š’‚ āƒ— . š’‚ āƒ— + š’ƒ āƒ— . š’ƒ āƒ— + š’„ āƒ— . š’„ āƒ— + 2(š‘Ž āƒ—. š‘ āƒ— + š‘ āƒ—. š‘ āƒ— + š‘ āƒ—. š‘Ž āƒ—) Prop : š‘Ž āƒ— . š‘Ž āƒ— = |š‘Ž āƒ— |2 = |š’‚ āƒ— |šŸ + |š’ƒ āƒ— |šŸ + |š’„ āƒ— |šŸ + 2 (š‘Ž āƒ—. š‘ āƒ— + š‘ āƒ—. š‘ āƒ— + š‘ āƒ— . š‘Ž āƒ—) = 32 + 52 + 72 + 2(š‘Ž āƒ—. š‘ āƒ— + š‘ āƒ—. š‘ āƒ— + š‘ āƒ—. š‘Ž āƒ—) = 9 + 25 + 49 + 2(š‘Ž āƒ—. š‘ āƒ— + š‘ āƒ—. š‘ āƒ— + š‘ āƒ—. š‘Ž āƒ—) = 83 + 2 (š‘Ž āƒ—. š‘ āƒ— + š‘ āƒ—. š‘ āƒ— + š‘ āƒ—. š‘Ž āƒ—) ∓ |š‘Ž āƒ—+š‘ āƒ—+š‘ āƒ—|2 = 83 + 2 (š‘Ž āƒ—. š‘ āƒ— + š‘ āƒ—. š‘ āƒ— + š‘ āƒ—. š‘Ž āƒ—) Now, |š‘Ž āƒ—" + " š‘ āƒ—" + " š‘ āƒ— | = 0 |š‘Ž āƒ—" + " š‘ āƒ—" + " š‘ āƒ— |^2 = 0 83 + 2 (š‘Ž āƒ—. š‘ āƒ— + š‘ āƒ—. š‘ āƒ— + š‘ āƒ—. š‘Ž āƒ—) = 0 2(š‘Ž āƒ—. š‘ āƒ— + š‘ āƒ—. š‘ āƒ— + š‘ āƒ—. š‘Ž āƒ—) = āˆ’83 (š’‚ āƒ—. š’ƒ āƒ— + š’ƒ āƒ—. š’„ āƒ— + š’„ āƒ—. š’‚ āƒ—) = (āˆ’šŸ–šŸ‘)/šŸ

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