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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 March 30, 2023 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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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.