Vector Algebra
Board: CBSE | Class: Class 12
Comprehensive study notes for Vector Algebra by Ajay Yadav.
Key Concepts
Vectors and Scalars
Vector: quantity with magnitude and direction (displacement, velocity). Scalar: quantity with only magnitude (mass, time). Position vector of P(x,y,z): r = xi + yj + zk.
Types of Vectors
Zero vector: 0. Unit vector: รข = a/|a|.Collinear:parallelorantiparallel.Equal:samemagnitudeanddirection.
AdditionandScalarMultiplication
Triangle/parallelogramlaw.Properties:commutative,associative.Scalarmultiplication:ka.
DotProduct
a·b=|a||b|cosθ = a1b1+a2b2+a₃b₃. Properties: a·b = b·a. If a⊥b then a·b=0. a·a = |a|².
Cross Product
a×b = |a||b|sinθ n̂. Direction: right-hand rule. a×b = -b×a. If a∣b then a×b=0. Magnitude = area of parallelogram.
Scalar Triple Product
[a b c] = a·(b×c). Volume of parallelepiped formed by a,b,c. Coplanar if [a b c]=0.
Important Formulas
| Magnitude | |a| = √a1²+a2²+a₃² |
| Dot product | a·b = |a||b|cosθ = a1b1+a2b2+a₃b₃ |
| Cross product | a×b = |a||b|sinθ n̂ |
| Scalar triple | [a b c] = a·(b×c) |
Solved Examples
Example 1: If a=i+2j+3k, find |a|.
Solution: |a| = √1+4+9 = √14.
Example 2: Find dot product of a=2i+3j+k and b=i-2j+4k.
Solution: a·b = 2(1)+3(-2)+1(4) = 2-6+4 = 0. So a⊥b.
Example 3: Find cross product of a=i+j+k and b=2i+3j+4k.
Solution: a×b = i(4-3)-j(4-2)+k(3-2) = i – 2j + k = (1,-2,1).
Practice Questions
- Find unit vector along a=3i-4j+5k.
- If a=2i+3j-k, b=i+j+2k, find a·b and angle between them.
- Find a×b for a=i+2j, b=2j+3k.
- Find area of parallelogram with sides a=i+j, b=j+k.
- Check if vectors a,b,c are coplanar: a=i+j+k, b=i+2j+3k, c=2i+3j+4k.
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Video Lessons
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