
Product of Vectors: Product of vectors or multiplication of vectors can be performed by the following two ways:
- Scalar Product or Dot Product
- Vector Product or Cross
Product
Scalar Product or Dot Product

Scalar Product in Component Form

Angle between two vectors

Properties of scalar product

Perpendicular vectors

Parallel Vectors

Mutually Perpendicular Unit Vectors

Scalar Product in Terms of Unit Vectors
![Scalar Product in Terms of Unit Vectors Let (" a " ) ⃗ = a1(" i " ) ⃗ + a2(" j " ) ⃗ and (" b " ) ⃗ = b1(" i " ) ⃗ + b2(" j " ) ⃗ be two vectors, then (" a " ) ⃗.(" b " ) ⃗ = (a1(" i " ) ⃗ + a2(" j " ) ⃗).( b1(" i " ) ⃗ + b2(" j " ) ⃗) = a1b1 (" i " ) ⃗.(" i " ) ⃗ + a1b2 (" i " ) ⃗.(" j " ) ⃗ + a2b1 (" j " ) ⃗.(" i " ) ⃗ + a2b2 (" j " ) ⃗.(" j " ) ⃗ = a1b1 + 0 + 0 + a2b2 [∵(" i " ) ⃗.(" j " ) ⃗ = (" j " ) ⃗.(" i " ) ⃗ = 0] = a1b1 + a2b2](https://1.bp.blogspot.com/-cqjBzwBu0LI/X8kOi4C5U8I/AAAAAAAAGfY/mAizvEpe9Boc5Xeqjv5K4EVsLxVTjn2mQCLcBGAsYHQ/s16000/Scalar%2Bproduct%2Bin%2Bterms%2Bof%2Bunit%2Bvectors.png)
Worked Out Examples:








![Example 9: Example 9: In the given figure, ABC is a triangle in which ∠ABC = 120° and AB = AC. Show that AB = "1" /√("3" ) AC. Solution: We have, ("AB" ) ⃗ + ("BC" ) ⃗ = ("AC" ) ⃗ [By triangle law] or, (("AB" ) ⃗ + ("BC" ) ⃗)2 = (("AC" ) ⃗)2 [Squaring] or, (("AB" ) ⃗)2 + 2("AB" ) ⃗ . ("BC" ) ⃗ + (("BC" ) ⃗)2 = (("AC" ) ⃗)2 or, AB2 + 2|("AB" ) ⃗|.|("BC" ) ⃗| cos60° + BC2 = AC2 [█("∵ Angle made by " ("AB" ) ⃗" and " ("BC" ) ⃗" is 180° – 120° " @"= 60° and AB = BC" @)] or, AB2 + 2AB.AB × "1" /"2" + AB2 = AC2 or, 2AB2 + 2 AB2 × "1" /"2" = AC2 or, 3AB2 = AC2 or, AB2 = "1" /"3" AC2 or, AB = "1" /√("3" ) AC Proved.](https://1.bp.blogspot.com/-2gPFu7v3zc8/X8kRt42cjHI/AAAAAAAAGgo/uzAv94j4IiUsxF9mdsws891bucjmyl6kwCLcBGAsYHQ/s16000/example%2B9.png)
![Example 10: Example 10: If ABC be a right angled triangle with ∠BAC = 90°, show by vector method that AB2 + AC2 = BC2. Solution: By triangle law of vector addition, ("BA" ) ⃗ + ("AC" ) ⃗ = ("BC" ) ⃗ i.e. (("BA" ) ⃗ + ("AC" ) ⃗)2 = (("BC" ) ⃗)2 [Squaring] or, (("BA" ) ⃗)2 + 2 ("BA" ) ⃗ . ("AC" ) ⃗ + (("AC" ) ⃗)2 = (("BC" ) ⃗)2 [Since ("BA" ) ⃗ ⊥ ("AC" ) ⃗, ("BA" ) ⃗ . ("AC" ) ⃗ = 0] or, BA2 + 0 + AC2 = BC2 [∵ BA2 = AB2] or, AB2 + AC2 = BC2 Proved.](https://1.bp.blogspot.com/-uNkiY9kzZHo/X8kSAuVpfcI/AAAAAAAAGgw/JT152o5-Fxg6SSWnW9yXbApPQrZ8IEcFACLcBGAsYHQ/s16000/example%2B10.png)
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