If $\theta$ is the angle between two unit vectors $\hat{a}$ and $\hat{b}$ then $|\hat{a}-\hat{b}| =$

We need to find $|\hat{a} - \hat{b}|$ where $\hat{a}$ and $\hat{b}$ are unit vectors (meaning $|\hat{a}| = 1$ and $|\hat{b}| = 1$) and $\theta$ is the angle between them.

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To find the magnitude of $(\hat{a} - \hat{b})$, we start with:

$|\hat{a} - \hat{b}|^2 = (\hat{a} - \hat{b}) \cdot (\hat{a} - \hat{b})$

Expanding the dot product:

$|\hat{a} - \hat{b}|^2 = \hat{a} \cdot \hat{a} - \hat{a} \cdot \hat{b} - \hat{b} \cdot \hat{a} + \hat{b} \cdot \hat{b}$

This is just like expanding $(a - b)^2$ in algebra, but with dot products.

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Since $\hat{a}$ and $\hat{b}$ are unit vectors:

$\hat{a} \cdot \hat{a} = |\hat{a}|^2 = 1$

$\hat{b} \cdot \hat{b} = |\hat{b}|^2 = 1$

And by the definition of dot product:

$\hat{a} \cdot \hat{b} = |\hat{a}||\hat{b}|\cos\theta = (1)(1)\cos\theta = \cos\theta$

Substituting these values:

$|\hat{a} - \hat{b}|^2 = 1 - \cos\theta - \cos\theta + 1$

$|\hat{a} - \hat{b}|^2 = 2 - 2\cos\theta$

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Using the trigonometric identity $1 - \cos\theta = 2\sin^2\left(\dfrac{\theta}{2}\right)$, we get:

$|\hat{a} - \hat{b}|^2 = 2 \times 2\sin^2\left(\dfrac{\theta}{2}\right) = 4\sin^2\left(\dfrac{\theta}{2}\right)$

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Taking the square root on both sides:

$|\hat{a} - \hat{b}| = \sqrt{4\sin^2\left(\dfrac{\theta}{2}\right)}$

Since $\theta$ is the angle between two vectors, it lies in $[0, \pi]$, which means $\dfrac{\theta}{2}$ lies in $[0, \dfrac{\pi}{2}]$, so $\sin\left(\dfrac{\theta}{2}\right) \geq 0$.

Therefore:

$|\hat{a} - \hat{b}| = 2\sin\left(\dfrac{\theta}{2}\right)$