# Given U is a vector with an initial point of (6,-4) and a terminal point of (-2, 8). How do you write u as a linear combination of the standard unit vector i and j?

Feb 4, 2017

$\vec{U} = - 8 \vec{i} + 12 \vec{j}$

#### Explanation:

$\text{label "(6,-4)" point"A; "and "(-2,8)" point} B$

so the position vectors for each point is as follows ( writing them as column vectors)

$\vec{O A} = \left(\begin{matrix}6 \\ - 4\end{matrix}\right)$

$\vec{O B} = \left(\begin{matrix}- 2 \\ 8\end{matrix}\right)$

$\vec{U} \text{ has initial point "A " and terminal point } B$

$\text{in vector notation } \vec{U} = \vec{A B}$

so:

vecU=vec(AB)=vec(AO)+vec(OB

vecU=vec(AB)=-vec(OA)+vec(OB

$\vec{U} = \vec{A B} = - \left(\begin{matrix}6 \\ - 4\end{matrix}\right) + \left(\begin{matrix}- 2 \\ 8\end{matrix}\right)$

$\vec{U} = \vec{A B} = \left(\begin{matrix}- 8 \\ 12\end{matrix}\right)$

in $\vec{i}$$\text{& "vecj "terms}$

$\vec{U} = - 8 \vec{i} + 12 \vec{j}$

Feb 4, 2017

$- 8 \vec{i} + 12 \vec{j}$

#### Explanation:

If $\vec{u} = x \vec{i} + y \vec{j}$, the conventional notation is

$\vec{u} = < x , y >$, in Cartesian form, and

$= r < \cos \theta , \sin \theta >$, in polar form.

Here, $\vec{u} = \vec{A B}$, where the position vectors

$\vec{O A} = < 6 , - 4 > \mathmr{and} \vec{O B} = < - 2 , 8 >$, giving

$\vec{u} = \vec{A B} = \vec{O B} - \vec{O A}$

=<-2,8> - <6, -4> = <((-2-6), (8-(-4))> = <-8, 12>

$= - 8 \vec{i} + 12 \vec{j}$