if a(cap) and b(cap) are unit vectors at an angle theta show that ∣ a ( cap ) − b ( cap ) ∣ = 2 |a(\text{cap}) - b(\text{cap})| = 2 ∣ a ( cap ) − b ( cap ) ∣ = 2 (sin theta/2)
a ⃗ \vec{a} a and b ⃗ \vec{b} b - are unit vectors
θ \theta θ - the angle between them
Law of cosines:
∣ a ⃗ − b ⃗ ∣ 2 = ∣ a ⃗ ∣ 2 + ∣ b ⃗ ∣ 2 − 2 ∣ a ⃗ ∣ ∣ b ⃗ ∣ cos θ , \left|\vec{a} -\vec{b}\right|^2 = |\vec{a} |^2 +\left|\vec{b}\right|^2 -2|\vec{a} |\left|\vec{b}\right|\cos \theta , ∣ ∣ a − b ∣ ∣ 2 = ∣ a ∣ 2 + ∣ ∣ b ∣ ∣ 2 − 2∣ a ∣ ∣ ∣ b ∣ ∣ cos θ ,
a ⃗ , b ⃗ \vec{a}, \vec{b} a , b - some vectors, θ \theta θ - the angle between them
a ⃗ \vec{a} a and b ⃗ \vec{b} b - are unit vectors, therefore ∣ a ⃗ ∣ = ∣ b ⃗ ∣ = 1 |\vec{a}| = |\vec{b}| = 1 ∣ a ∣ = ∣ b ∣ = 1
Law of cosines:
∣ a ⃗ − b ⃗ ∣ 2 = 1 + 1 − 2 cos θ = 2 ( 1 − cos θ ) = ∣ 1 − cos θ 2 = sin θ 2 ∣ = 4 sin 2 θ 2 \left|\vec{a} -\vec{b}\right|^2 = 1 + 1 - 2\cos \theta = 2(1 - \cos \theta) = \left|\sqrt{\frac{1 - \cos\theta}{2}} = \sin \frac{\theta}{2}\right| = 4\sin^2\frac{\theta}{2} ∣ ∣ a − b ∣ ∣ 2 = 1 + 1 − 2 cos θ = 2 ( 1 − cos θ ) = ∣ ∣ 2 1 − c o s θ = sin 2 θ ∣ ∣ = 4 sin 2 2 θ
Or:
∣ a ⃗ − b ⃗ ∣ = 2 sin θ 2 \left|\vec{a} -\vec{b}\right| = 2\sin \frac{\theta}{2} ∣ ∣ a − b ∣ ∣ = 2 sin 2 θ