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Ex 10.3, 11 - Chapter 10 Class 12 Vector Algebra
Last updated at April 16, 2024 by Teachoo
Chapter 10 Class 12 Vector Algebra
Concept wise
- Scalar or vector
- Graphical displacement
- Type of vector
- Multiplication of a vector by a scalar
- Equal vectors
- Unit vector
- Direction cosines and ratios
- Addition(resultant) of vectors
- Joining two points
- Section formula
- Right Angled triangle
- Collinearity Of two vectors
- Collinearity of 3 points or 3 position vectors
- Scalar product - Defination
- Scalar product - Projection
-
Scalar product - Solving
- Vector product - Defination
- Vector product - Area
- Vector product - Solving
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.
