Linear Algebra

Angle Between Vectors

cos(θ) = u·v / (|u||v|)

  • High school
  • about 4 min
  • 4 of 5 in the family
Not started
53.1°uv

u = (3, 1)

v = (1, 3)

u⃗⋅v⃗=3⋅1+1⋅3=6\vec{u} \cdot \vec{v} = 3 \cdot 1 + 1 \cdot 3 = 6
cos⁡θ=u⃗⋅v⃗∣u⃗∣∣v⃗∣=63.16⋅3.16=0.6000\cos\theta = \frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|} = \frac{6}{3.16 \cdot 3.16} = 0.6000
θ=53.13°=0.9273 rad\theta = 53.13° = 0.9273\text{ rad}

Angle between vectors is found via dot product: θ = arccos(u·v / |u||v|). Perpendicular vectors have θ = 90°, parallel θ = 0°.

Where it is taught