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Question

Mathematics Question on Vector Algebra

If a⃗,b⃗,c⃗\vec{a} , \vec{b} ,\vec{c} are unit vectors and θ\theta is the angle between them, then ∣a⃗−b⃗∣=| \vec{a} - \vec{b}| =

A

2cos⁡θ22 \cos \frac{\theta}{2}

B

2sin⁡θ22 \sin\frac{\theta}{2}

C

2cos⁡ θ2 \cos \, \theta

D

2sin⁡ θ2 \sin\, \theta

Answer

2sin⁡θ22 \sin\frac{\theta}{2}

Explanation

Solution

a⃗,b⃗,c⃗\vec{a} ,\vec{b} ,\vec{c} are unit vectors
Now, ∣a⃗−b⃗∣2=∣a⃗∣2+∣b⃗∣2−2a⃗.b⃗\left|\vec{a} - \vec{b}\right|^{2} = \left|\vec{a}\right|^{2} +\left|\vec{b}\right|^{2} - 2\vec{a} .\vec{b}
??????????=1+1−2∣a⃗∣∣b⃗∣cos⁡θ= 1 + 1 - 2\left|\vec{a}\right|\left|\vec{b}\right| \cos \theta
??????????=2−2cos⁡θ=2(1−cos⁡θ)= 2-2 \cos \theta =2\left(1 -\cos \theta\right)
??????????=4sin⁡2θ2 = 4 \sin^{2} \frac{\theta}{2}
Hence, ∣a⃗−b⃗∣=2sin⁡θ2 \left|\vec{a} - \vec{b}\right| = 2\sin \frac{\theta}{2}