# Vectors Given that vec(OP)=((-3),(-2)), vec(OQ)=((5),(-2)) and T is a point on the line vec(PQ) such that 5vec(PT)=3vec(TO) express as column vectors? (I) vec(PQ) (II) vec(TQ)

Jul 4, 2018

$\left(I\right) : \vec{P Q} = \left(\begin{matrix}8 \\ 0\end{matrix}\right) .$

$\left(I I\right) : \vec{T Q} = \left(\begin{matrix}\frac{55}{8} \\ - \frac{3}{4}\end{matrix}\right) .$

#### Explanation:

For ease of writing, let us denote the position vectors of $P \mathmr{and} Q$

by $\left(- 3 , - 2\right) \mathmr{and} \left(5 , - 2\right)$, resp.

Let, $T = T \left(x , y\right) \text{ be on line "PQ," such that, } 5 \vec{P T} = 3 \vec{T O}$.

Now, $5 \vec{P T} = 5 \left(\left(x , y\right) - \left(- 3 , - 2\right)\right) = 5 \left(x + 3 , y + 2\right) ,$

$\text{&, } 3 \vec{T O} = 3 \left(- x , - y\right)$.

$\therefore 5 \vec{P T} = 3 \vec{T O} \Rightarrow 5 \left(x + 3 , y + 2\right) = 3 \left(- x , - y\right)$.

$\Rightarrow 5 x + 15 = - 3 x , \mathmr{and} , 5 y + 10 = - 3 y$.

$\Rightarrow x = - \frac{15}{8} , \mathmr{and} , y = - \frac{10}{8}$.

Hence, $T = T \left(- \frac{15}{8} , - \frac{10}{8}\right)$.

So, $\left(I\right) : \vec{P Q} = \left(5 , - 2\right) - \left(- 3 , - 2\right) = \left(8 , 0\right) .$

$\left(I I\right) : \vec{T Q} = \left(5 , - 2\right) - \left(- \frac{15}{8} , - \frac{10}{8}\right) = \left(\frac{55}{8} , - \frac{3}{4}\right) .$

Jul 4, 2018

The vectors are $\vec{P Q} = \left(\begin{matrix}8 \\ 0\end{matrix}\right)$ and $\vec{T Q} = \left(\begin{matrix}\frac{31}{8} \\ - \frac{3}{4}\end{matrix}\right)$

#### Explanation:

Apply Chasles' relation

$\vec{O P} = \left(\begin{matrix}- 3 \\ - 2\end{matrix}\right)$

$\vec{O Q} = \left(\begin{matrix}5 \\ - 2\end{matrix}\right)$

$\vec{P O} = - \vec{O P}$

$\vec{P Q} = \vec{P O} + \vec{O Q}$

$= \vec{O Q} - \vec{O P}$

$= \left(\begin{matrix}5 \\ - 2\end{matrix}\right) - \left(\begin{matrix}- 3 \\ - 2\end{matrix}\right)$

$= \left(\begin{matrix}5 + 3 \\ - 2 + 2\end{matrix}\right)$

$= \left(\begin{matrix}8 \\ 0\end{matrix}\right)$

$5 \vec{P T} = 3 \vec{T O}$

$5 \left(\vec{P Q} + \vec{Q T}\right) = 3 \left(\vec{T Q} + \vec{Q O}\right)$

$5 \vec{P Q} - 5 \vec{T Q} = 3 \vec{T Q} - 3 \vec{O Q}$

$8 \vec{T Q} = 5 \vec{P Q} + 3 \vec{O Q}$

$\vec{T Q} = \frac{1}{8} \left(5 \vec{P Q} + 3 \vec{O Q}\right)$

$= \frac{1}{8} \left(5 \cdot \left(\begin{matrix}8 \\ 0\end{matrix}\right) + 3 \cdot \left(\begin{matrix}5 \\ - 2\end{matrix}\right)\right)$

$= \frac{1}{8} \left(\begin{matrix}55 \\ - 6\end{matrix}\right)$

$= \left(\begin{matrix}\frac{55}{8} \\ - \frac{3}{4}\end{matrix}\right)$