Chaim Ferguson

2022-10-11

Geometrical Interpretation of Vectors Addition

I don't understand what it means by $({y}_{1}-{x}_{1},{y}_{2}-{x}_{2},{y}_{3}-{x}_{3})$

This is the part of the book:

The directed line segment PQ, from the point $P=({x}_{1},{x}_{2},{x}_{3})$ to the point $Q=({y}_{l},{y}_{2},{y}_{3})$, has the same length and direction as the directed line segment from the origin $0=(0,0,0)$ to the point $({y}_{l}-{x}_{1},{y}_{2}-{x}_{2},{y}_{3}-{x}_{3})$. Furthermore, this is the only segment emanating from the origin which has the same length and direction as PQ. Thus, if one agrees to treat only vectors which emanate from the origin, there is exactly one vector associated with each given length and direction.

I don't understand what it means by $({y}_{1}-{x}_{1},{y}_{2}-{x}_{2},{y}_{3}-{x}_{3})$

This is the part of the book:

The directed line segment PQ, from the point $P=({x}_{1},{x}_{2},{x}_{3})$ to the point $Q=({y}_{l},{y}_{2},{y}_{3})$, has the same length and direction as the directed line segment from the origin $0=(0,0,0)$ to the point $({y}_{l}-{x}_{1},{y}_{2}-{x}_{2},{y}_{3}-{x}_{3})$. Furthermore, this is the only segment emanating from the origin which has the same length and direction as PQ. Thus, if one agrees to treat only vectors which emanate from the origin, there is exactly one vector associated with each given length and direction.

Phillip Fletcher

Beginner2022-10-12Added 21 answers

The point is that while the line segment $PQ$ is fixed in space, the vector (“directed line segment”) $\overrightarrow{PQ}$ is “portable” by translation in space. Think of a vector as describing a translation, rather than as being affixed in any particular position.

At the risk of stating the obvious: if

$\overrightarrow{PQ}=\left(\begin{array}{c}-1\\ 2\\ 0\end{array}\right)=\overrightarrow{OA},$

then

$\overrightarrow{PQ}=\overrightarrow{OA}.$

At the risk of stating the obvious: if

$\overrightarrow{PQ}=\left(\begin{array}{c}-1\\ 2\\ 0\end{array}\right)=\overrightarrow{OA},$

then

$\overrightarrow{PQ}=\overrightarrow{OA}.$

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