# Find the Distance Between Two Vectors if the Lengths and the Dot Product are Given

## Problem 254

Let $\mathbf{a}$ and $\mathbf{b}$ be vectors in $\R^n$ such that their length are
$\|\mathbf{a}\|=\|\mathbf{b}\|=1$ and the inner product
$\mathbf{a}\cdot \mathbf{b}=\mathbf{a}^{\trans}\mathbf{b}=-\frac{1}{2}.$

Then determine the length $\|\mathbf{a}-\mathbf{b}\|$.
(Note that this length is the distance between $\mathbf{a}$ and $\mathbf{b}$.)

## Solution.

Recall that the length of a vector $\mathbf{x}$ is defined to be
$\|\mathbf{x}\|=\sqrt{\mathbf{x}^{\trans}\mathbf{x}},$ where $\mathbf{x}^{\trans}$ is the transpose of $\mathbf{x}$.

Also, recall that the inner product of two vectors $\mathbf{x}, \mathbf{y}$ are commutative.
Namely we have
$\mathbf{x}\cdot \mathbf{y}=\mathbf{x}^{\trans}\mathbf{y}=\mathbf{y}^{\trans}\mathbf{x}=\mathbf{y} \cdot \mathbf{x}.$

Applying the second fact with given vectors $\mathbf{a}, \mathbf{b}$, we obtain
$\mathbf{a}^{\trans}\mathbf{b}=\mathbf{b}^{\trans}\mathbf{a}= -\frac{1}{2}.$

Now we compute $\|\mathbf{a}-\mathbf{b}\|^2$ as follows.
We have
\begin{align*}
\|\mathbf{a}-\mathbf{b}\|^2&=(\mathbf{a}-\mathbf{b})^{\trans}(\mathbf{a}-\mathbf{b}) \qquad \text{ (by definition of the length)}\\
&=(\mathbf{a}^{\trans}-\mathbf{b}^{\trans})(\mathbf{a}-\mathbf{b})\\
&=\mathbf{a}^{\trans}\mathbf{a}-\mathbf{a}^{\trans}\mathbf{b}-\mathbf{b}^{\trans}\mathbf{a}+\mathbf{b}^{\trans}\mathbf{b}\\
&=\|\mathbf{a}\|^2-\mathbf{a}^{\trans}\mathbf{b}-\mathbf{b}^{\trans}\mathbf{a}+\|\mathbf{b}\|^2\\
&=1-\left(-\frac{1}{2} \right)-\left(-\frac{1}{2} \right)+1\\
&=3.
\end{align*}

Since the length is nonnegative, we take the square root of the above equality and obtain
$\|\mathbf{a}-\mathbf{b}\|=\sqrt{3}.$

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