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For vectors $\mathbf{R}, \mathbf{r_1}$ and $\mathbf{r_2}$, if $\mathbf{R}$ is much longer than both $\mathbf{r_1}$ and $\mathbf{r_2}$ we have \begin{equation} \Vert \mathbf{R} - \mathbf{r_1} \Vert - \Vert \mathbf{R} - \mathbf{r_2} \Vert \approx \frac{\mathbf{R}}{\Vert \mathbf{R} \Vert} \cdot (\mathbf{r_2} -\mathbf{r_1}), \end{equation} where '$\cdot$' stands for the inner product.

This relation can be seen geometrically by assuming the angle of incidence from $\mathbf{R}$ to both $\mathbf{r_1}$ and $\mathbf{r_2}$ to be the same. Can we show this result in an algebraic manner by some approximation methods?

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Algebraically,

$$\begin{align} \lVert \mathbf{R} - \mathbf{r}_1\rVert - \lVert \mathbf{R}-\mathbf{r}_2\rVert &= \frac{\lVert \mathbf{R} - \mathbf{r}_1\rVert^2 - \lVert \mathbf{R}-\mathbf{r}_2\rVert^2}{\lVert \mathbf{R} - \mathbf{r}_1\rVert + \lVert \mathbf{R}-\mathbf{r}_2\rVert}\\ &= \frac{2\mathbf{R}\cdot(\mathbf{r}_2-\mathbf{r}_1) + \lVert\mathbf{r}_1\rVert^2 - \lVert\mathbf{r}_2\rVert^2}{\lVert \mathbf{R} - \mathbf{r}_1\rVert + \lVert \mathbf{R}-\mathbf{r}_2\rVert}. \end{align}$$

You can then unscrupulously say that $\lVert\mathbf{r}_i\rVert$ is so much smaller than $\lVert\mathbf{R}\rVert$ that the last two terms in the numerator don't matter, and the denominator can be replaced by $2\lVert\mathbf{R}\rVert$ to obtain the approximation. You can also be a bit more scrupulous and obtain the next order terms in the approximation, if you use the Taylor approximation for $\lVert \mathbf{R}-\mathbf{r}_i\rVert$.

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