Q1. Find a·b for a = 2i + 3j + k and b = i − j + 2k.
a·b = (2)(1) + (3)(−1) + (1)(2)
= 2 − 3 + 2
= 1
Vector Algebra — Class 12
Vectors carry both size and direction. Compute the dot product, cross product magnitude and the angle between any two 3D vectors instantly.
Magnitude of a vectorFor a = a₁i + a₂j + a₃k, |a| = √(a₁² + a₂² + a₃²).
Dot (scalar) producta·b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b|cosθ. Zero when the vectors are perpendicular.
Cross (vector) producta×b is a vector perpendicular to both a and b, with magnitude |a||b|sinθ. Its magnitude equals the area of the parallelogram formed by a and b.
Angle between vectorscosθ = (a·b)/(|a||b|).
Vector calculator Interactive
Enter two 3D vectors a = (a₁,a₂,a₃) and b = (b₁,b₂,b₃) to get their dot product, magnitudes and the angle between them.
Dot product a·b—
|a|—
|b|—
Angle θ (degrees)—
If the dot product is 0, the vectors are perpendicular (θ = 90°) — a quick way to check.
Solved examples
Formula sheet
| Quantity | Formula |
|---|---|
| Magnitude | |a| = √(a₁²+a₂²+a₃²) |
| Dot product | a·b = a₁b₁+a₂b₂+a₃b₃ |
| Angle between vectors | cosθ = (a·b)/(|a||b|) |
| Cross-product magnitude | |a×b| = |a||b|sinθ |
Common mistakes & exam wins
- Dot product gives a SCALAR (a number); cross product gives a VECTOR.
- a·b = 0 means the vectors are perpendicular.
- |a×b| equals the area of the parallelogram with sides a and b — half of it is the area of the triangle.
Frequently asked questions
What is the dot product of two vectors?
a·b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b|cosθ, a scalar that is zero when the vectors are perpendicular.
What is the cross product of two vectors?
A vector a×b perpendicular to both, with magnitude |a||b|sinθ, equal to the area of the parallelogram they span.
How do you find the angle between two vectors?
cosθ = (a·b)/(|a||b|), then take the inverse cosine.