Check sibling questions

Ex 10.3, 11 - Show |a|b + |b|a is perpendicular to |a|b - |b|a

Ex 10.3, 11 - Chapter 10 Class 12 Vector Algebra - Part 2

This video is only available for Teachoo black users

Solve all your doubts with Teachoo Black (new monthly pack available now!)


Transcript

Ex 10.3, 11 Show that |𝑎 ⃗ | 𝑏 ⃗+ |𝑏 ⃗ | 𝑎 ⃗ is perpendicular to |𝑎 ⃗ | 𝑏 ⃗− |𝑏 ⃗ | 𝑎 ⃗, for any two nonzero vectors 𝑎 ⃗ and 𝑏 ⃗ If two vectors 𝑝 ⃗ and 𝑞 ⃗ are perpendicular to each other , then their scalar (dot) product is zero, i.e. 𝒑 ⃗ . 𝒒 ⃗ = 0 Hence, to show (|𝑎 ⃗ | 𝑏 ⃗ + |𝑏 ⃗ | 𝑎 ⃗) is perpendicular to (|𝑎 ⃗ | 𝑏 ⃗ − |𝑏 ⃗ | 𝑎 ⃗), We need to prove (|𝑎 ⃗ | 𝑏 ⃗ + |𝑏 ⃗ | 𝑎 ⃗) . (|𝑎 ⃗ | 𝑏 ⃗ − |𝑏 ⃗ | 𝑎 ⃗) = 0 Solving L.HS. (|𝑎 ⃗ | 𝑏 ⃗ + |𝑏 ⃗ | 𝑎 ⃗) . (|𝑎 ⃗ | 𝑏 ⃗ − |𝑏 ⃗ | 𝑎 ⃗) = (|𝑎 ⃗ | 𝑏 ⃗ ) . (|𝑎 ⃗ |" " 𝑏 ⃗ ) − (|𝑎 ⃗ | 𝑏 ⃗ ) . (|𝑏 ⃗ | 𝑎 ⃗ ) + (|𝑏 ⃗ | 𝑎 ⃗ ). (|𝑎 ⃗ | 𝑏 ⃗ ) – (|𝑏 ⃗ | 𝑎 ⃗ ) . (|𝑏 ⃗ | 𝑎 ⃗ ) = |𝑎 ⃗ |2 𝑏 ⃗ . 𝑏 ⃗ − |𝑎 ⃗ ||𝑏 ⃗ | 𝒃 ⃗ . 𝒂 ⃗ + |𝑏 ⃗ | |𝑎 ⃗ | 𝑎 ⃗. 𝑏 ⃗ − |𝑏 ⃗ |2𝑎 ⃗ . 𝑎 ⃗ = |𝑎 ⃗ |2 𝑏 ⃗ . 𝑏 ⃗ − |𝑎 ⃗ ||𝑏 ⃗ | 𝒂 ⃗ . 𝒃 ⃗ + |𝑎 ⃗ ||𝑏 ⃗ | 𝑎 ⃗. 𝑏 ⃗ − |𝑏 ⃗ |2𝑎 ⃗ . 𝑎 ⃗ = |𝑎 ⃗ |2 𝒃 ⃗ . 𝒃 ⃗ − |𝑏 ⃗ |2𝒂 ⃗ . 𝒂 ⃗ = |𝑎 ⃗ |2 |𝒃 ⃗ |𝟐 −|𝑏 ⃗ |2 |𝒂 ⃗ |𝟐 = 0 = RHS Hence proved.

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.