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CBSE Class 12 Sample Paper for 2020 Boards

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Question 24 (OR 2nd Question) - CBSE Class 12 Sample Paper for 2020 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Feb. 21, 2020 by

If a + b + c = 0 and |a| = 3, |b| = 5, |c| = 7, then find the value of a.bΒ  + b.cΒ  + 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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Transcript

Question 24 (OR 2nd Question) if π‘Ž βƒ— + 𝑏 βƒ— + 𝑐 βƒ— = 0 and |π‘Ž βƒ— | = 3, |𝑏 βƒ— | = 5, |𝑐 βƒ— | = 7, then find the value of π‘Ž βƒ—. 𝑏 βƒ— + 𝑏 βƒ—. 𝑐 βƒ— + 𝑐 βƒ—. π‘Ž βƒ— . Given, π‘Ž βƒ— + 𝑏 βƒ— + 𝑐 βƒ— = 0 βƒ— So, |𝒂 βƒ—" + " 𝒃 βƒ—" + " 𝒄 βƒ— | = |𝟎 βƒ— | = 0 Now, |𝒂 βƒ—+𝒃 βƒ—+𝒄 βƒ— |2 = (𝒂 βƒ— + 𝒃 βƒ— + 𝒄 βƒ—) . (𝒂 βƒ— + 𝒃 βƒ— + 𝒄 βƒ—) = π‘Ž βƒ—. π‘Ž βƒ— + π‘Ž βƒ— . 𝑏 βƒ— + 𝒂 βƒ— . 𝒄 βƒ— + 𝒃 βƒ— . 𝒂 βƒ— + 𝑏 βƒ— . 𝑏 βƒ— + 𝑏 βƒ— . 𝑐 βƒ— + 𝑐 βƒ— . π‘Ž βƒ— + 𝒄 βƒ— . 𝒃 βƒ— + 𝑐 βƒ— . 𝑐 βƒ— = π‘Ž βƒ—. π‘Ž βƒ— + π‘Ž βƒ— . 𝑏 βƒ— + 𝒄 βƒ— . 𝒂 βƒ— + 𝒂 βƒ— . 𝒃 βƒ— + 𝑏 βƒ— . 𝑏 βƒ— + 𝑏 βƒ— . 𝑐 βƒ— + 𝑐 βƒ— . π‘Ž βƒ— + 𝒃 βƒ— . 𝒄 βƒ— + 𝑐 βƒ— . 𝑐 βƒ— = π‘Ž βƒ— . π‘Ž βƒ— + 𝑏 βƒ— . 𝑏 βƒ— + 𝑐 βƒ— . 𝑐 βƒ— + 2π‘Ž βƒ—. 𝑏 βƒ— + 2𝑏 βƒ—. 𝑐 βƒ— + 2𝑐 βƒ—. π‘Ž βƒ— = 𝒂 βƒ— . 𝒂 βƒ— + 𝒃 βƒ— . 𝒃 βƒ— + 𝒄 βƒ— . 𝒄 βƒ— + 2(π‘Ž βƒ—. 𝑏 βƒ— + 𝑏 βƒ—. 𝑐 βƒ— + 𝑐 βƒ—. π‘Ž βƒ—) Prop : π‘Ž βƒ— . π‘Ž βƒ— = |π‘Ž βƒ— |2 = |𝒂 βƒ— |𝟐 + |𝒃 βƒ— |𝟐 + |𝒄 βƒ— |𝟐 + 2 (π‘Ž βƒ—. 𝑏 βƒ— + 𝑏 βƒ—. 𝑐 βƒ— + 𝑐 βƒ— . π‘Ž βƒ—) = 32 + 52 + 72 + 2(π‘Ž βƒ—. 𝑏 βƒ— + 𝑏 βƒ—. 𝑐 βƒ— + 𝑐 βƒ—. π‘Ž βƒ—) = 9 + 25 + 49 + 2(π‘Ž βƒ—. 𝑏 βƒ— + 𝑏 βƒ—. 𝑐 βƒ— + 𝑐 βƒ—. π‘Ž βƒ—) = 83 + 2 (π‘Ž βƒ—. 𝑏 βƒ— + 𝑏 βƒ—. 𝑐 βƒ— + 𝑐 βƒ—. π‘Ž βƒ—) ∴ |π‘Ž βƒ—+𝑏 βƒ—+𝑐 βƒ—|2 = 83 + 2 (π‘Ž βƒ—. 𝑏 βƒ— + 𝑏 βƒ—. 𝑐 βƒ— + 𝑐 βƒ—. π‘Ž βƒ—) Now, |π‘Ž βƒ—" + " 𝑏 βƒ—" + " 𝑐 βƒ— | = 0 |π‘Ž βƒ—" + " 𝑏 βƒ—" + " 𝑐 βƒ— |^2 = 0 83 + 2 (π‘Ž βƒ—. 𝑏 βƒ— + 𝑏 βƒ—. 𝑐 βƒ— + 𝑐 βƒ—. π‘Ž βƒ—) = 0 2(π‘Ž βƒ—. 𝑏 βƒ— + 𝑏 βƒ—. 𝑐 βƒ— + 𝑐 βƒ—. π‘Ž βƒ—) = βˆ’83 (𝒂 βƒ—. 𝒃 βƒ— + 𝒃 βƒ—. 𝒄 βƒ— + 𝒄 βƒ—. 𝒂 βƒ—) = (βˆ’πŸ–πŸ‘)/𝟐

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.