Problem on vector addition
$\newcommand{\bfA}{\mathbf{A}}$ $\newcommand{\bfB}{\mathbf{B}}$ $\newcommand{\bfC}{\mathbf{C}}$ $\newcommand{\bfF}{\mathbf{F}}$ $\newcommand{\bfI}{\mathbf{I}}$ $\newcommand{\bfa}{\mathbf{a}}$ $\newcommand{\bfb}{\mathbf{b}}$ $\newcommand{\bfc}{\mathbf{c}}$ $\newcommand{\bfd}{\mathbf{d}}$ $\newcommand{\bfe}{\mathbf{e}}$ $\newcommand{\bfi}{\mathbf{i}}$ $\newcommand{\bfj}{\mathbf{j}}$ $\newcommand{\bfk}{\mathbf{k}}$ $\newcommand{\bfn}{\mathbf{n}}$ $\newcommand{\bfr}{\mathbf{r}}$ $\newcommand{\bfu}{\mathbf{u}}$ $\newcommand{\bfv}{\mathbf{v}}$ $\newcommand{\bfw}{\mathbf{w}}$ $\newcommand{\bfx}{\mathbf{x}}$ $\newcommand{\bfy}{\mathbf{y}}$ $\newcommand{\bfz}{\mathbf{z}}$
The following parallelogram has one corner at the origin. The two neighboring corners are given by vectors $\mathbf{a}$ and $\mathbf{b}$. Express the fourth corner as a vector.
Solution
Recall that
Geometric view of vector addition
The sum of two vectors is the vector obtained by lining up the tail of one vector to the head of the other.
If we move the tail of the vector $\mathbf{a}$ to start at the point $\mathbf{b}$, we end up at the fourth corner.
The vector from the origin to that fourth corner is thus $\mathbf{a} + \mathbf{b}$.
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