# Resultant vector problem

$\newcommand{\v}[1]{\vec #1}\newcommand{\i}{\hat i}\newcommand{\j}{\hat j}$ Problem statement (1,2)

A shopper at the supermarket follows the path indicated by vectors $\v A, \v B, \v C, \v D$ in the figure. Given that the vectors have magnitudes $A=51\:\mathrm{ft}, B=45\:\mathrm{ft}, C=35\:\mathrm{ft}, D=13\:\mathrm{ft}$, find the total displacement of the shopper using (a) the graphical method and (b) the component method of vector addition. Give the direction of displacement relative to $\v A$

\begin{align}A &=&0&\i + &51&\j \\B &=&45&\i + &51&\j\\C &= &10&\i + &-35 &\j\\ D &= &10&\i + &-13 &\j \\\text{Resultant} &=&54&\i+&65&\j\end{align} Essentially I am adding all the components to get the resultant vector but it is not leading me to the right answer. What am I doing wrong here?

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Your mistake is in a misunderstanding of what the graphical representation of the vectors means. To get the components of a vector that is drawn on a diagram, imagine moving the tail of the vector to the origin while keeping the orientation of the vector fixed, and then its components are the projections onto the $x$ and $y$ axes. In particular, when a horizontal vector is drawn, this means that its vertical component is zero, and when a vertical vector is drawn, this means that its horizontal component is zero.
So for example, you write that vector $\mathbf B$ has a $y$-component of $51\,\mathrm{ft}$, but that's because you're taking into account the position of its "tail," which is not what the diagram means. Instead, imagine moving $\mathbf B$ to the origin without changing its orientation, then it will point along the positive $x$-axis, but it will have no $y$-component, and you should get $$\mathbf B = (45\,\mathrm{ft})\,\mathbf i.$$ I'll let you do the rest. I hope that helps!
Your resultant vector is correct. It's just the wording of the problem that's a bit confusing. It says "relative to the direction of vector $\mathbf A$," so they just wanted to know the net "vertical" distance traveled since $\mathbf A$ points up. Nice work though! – joshphysics Feb 11 '13 at 4:43