Vectors

Parallelogram law of vector addition

For 5 and 3 at 60 degrees, R = 7, at 21.79° from A.

Vectors

Drag the glowing tips, or type values. Angles are measured from the x axis, and θ is the angle from A to B. Pick a result card to see it built in the picture.

Angle between60°cos θ = A·B / AB

Unit vectors1iB = 0.5i + 0.87j

Vector A of size 5 and vector B of size 3 with 60 degrees between them. The resultant is 7, 21.79 degrees from A. The dot product is 7.5 and the cross product has size 12.99.

Show the formulas

Resultant: R = √(A² + B² + 2AB cos θ) = √(5² + 3² + 2 × 5 × 3 × cos 60°) = 7
Direction: tan β = B sin θ / (A + B cos θ) = 3 × sin 60° / (5 + 3 × cos 60°), so β = 21.79° from A
Difference: |A - B| = √(A² + B² - 2AB cos θ) = √(5² + 3² - 2 × 5 × 3 × cos 60°) = 4.36
Dot product: A·B = AB cos θ = 5 × 3 × cos 60° = 7.5
Cross product: |A×B| = AB sin θ = 5 × 3 × sin 60° = 12.99
Components: A·B = AxBx + AyBy + AzBz = 5 × 1.5 + 0 × 2.6 + 0 × 0 = 7.5
Determinant: A×B = (AyBz - AzBy) i + (AzBx - AxBz) j + (AxBy - AyBx) k = 12.99k
Unit vector: Â = A / |A| = 1i

Working

Step by step

Given A = 5, B = 3 and the angle between them θ = 60°.
R² = A² + B² + 2AB cos θ = 5² + 3² + 2 × 5 × 3 × cos 60° = 49
R = √49 = 7
tan β = B sin θ / (A + B cos θ) = 3 × sin 60° / (5 + 3 × cos 60°) = 0.4
β = 21.79° from A, towards B.

What the angle does

R for 5 and 3 at other angles

0° 80° from A
30° 7.7411.17° from A
45° 7.4316.59° from A
60° 721.79° from A
90° 5.8330.96° from A
120° 4.3636.59° from A
150° 2.8331.98° from A
180° 20° from A

Derivation

Why it works

Draw A and B from the same point and complete the parallelogram with the lines parallel to each.
The diagonal that starts at the common point is R. The other diagonal is A - B.
In the right triangle made by dropping a perpendicular on the line of A, the base is A + B cos θ and the height is B sin θ.
So R² = (A + B cos θ)² + (B sin θ)² = A² + B² + 2AB cos θ.
The direction follows from the same triangle: tan β = B sin θ / (A + B cos θ).

Questions