Point dividing Line Segment between Two Points in Given Ratio/Proof 1
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Theorem
Let $A$ and $B$ be points whose position vectors relative to an origin $O$ of a Euclidean space are $\mathbf a$ and $\mathbf b$.
Let $\mathbf r$ be the position vector of a point $R$ on $AB$ which divides $AB$ in the ratio $m : n$.
Then:
- $\mathbf r = \dfrac {n \mathbf a + m \mathbf b} {m + n}$
Proof
We have that:
- $\vec {A B} = \mathbf b - \mathbf a$
and so:
- $\vec {A R} = \dfrac m {m + n} \paren {\mathbf b - \mathbf a}$
Hence the position vector $\mathbf r$ of $R$ is given by:
\(\ds \mathbf r\) | \(=\) | \(\ds \vec {O R}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \vec {O A} + \vec {A R}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \mathbf a + \dfrac m {m + n} \paren {\mathbf b - \mathbf a}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {n \mathbf a + m \mathbf b} {m + n}\) |
$\blacksquare$
Sources
- 1921: C.E. Weatherburn: Elementary Vector Analysis ... (previous) ... (next): Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Centroids: $8$. Division of a line in a given ratio