# IB Vectors

• Feb 9th 2008, 11:55 AM
overduex
IB Vectors
Find the position vectors that join the origin to the points with coordinates A(2, -1) and B(-3, 2). Express your answers as column vectors. Hence find line AB.

All help is welcome :)
• Feb 9th 2008, 12:13 PM
earboth
Quote:

Originally Posted by overduex
Find the position vectors that join the origin to the points with coordinates A(2, -1) and B(-3, 2). Express your answers as column vectors. Hence find line AB.

Vector $\displaystyle \vec a$ points at the point A and $\displaystyle \vec b$ points at the point B:

$\displaystyle \vec a=\left(\begin{array}{c}2\\-1\end{array}\right)$

$\displaystyle \vec b=\left(\begin{array}{c}-3\\2\end{array}\right)$

The direction of the line AB is determined by $\displaystyle \vec a - \vec b = \left(\begin{array}{c}5\\-3\end{array}\right)$

The line AB passes either through A or B. Therefore the equation of the line is:

$\displaystyle \vec x = \left(\begin{array}{c}x\\y\end{array}\right) = \left(\begin{array}{c}2\\-1\end{array}\right) + r \cdot \left(\begin{array}{c}5\\-3\end{array}\right)~,~r \in \mathbb{R}$
• Feb 9th 2008, 12:23 PM
mr fantastic
Quote:

Originally Posted by overduex
Find the position vectors that join the origin to the points with coordinates A(2, -1) and B(-3, 2). Express your answers as column vectors. Hence find line AB.

All help is welcome :)
$\displaystyle \vec{OA} = \left( \begin{array}{c} 2 \\ -1 \end{array} \right)$ and $\displaystyle \vec{OB} = \left( \begin{array}{c} -3 \\ 2 \end{array} \right)$.
$\displaystyle \vec{AB} = \vec{AO} + \vec{OB} = -\vec{OA} + \vec{OB} = - \left( \begin{array}{c} 2 \\ -1 \end{array} \right) + \left( \begin{array}{c} -3 \\ 2 \end{array} \right) = \left( \begin{array}{c} -2 \\ 1 \end{array} \right) + \left( \begin{array}{c} -3 \\ 2 \end{array} \right)$
$\displaystyle = \left( \begin{array}{c} -5 \\ 3 \end{array} \right)$.