# Two boats defined by vectors? (See picture)

## May 11, 2017

$\textcolor{b l u e}{A \left(3 , - 4\right) , B \left(4 , 3\right)}$.
we get for A we have we get $\sqrt{{3}^{2} + {2}^{2}} = \sqrt{13}$
similarly for B we get $\sqrt{13}$ but with different direction .

#### Explanation:

Nice question ,
Given stuff

A :{$x \left(t\right) = 3 - t$ , $y \left(t\right) = 2 t - 4$}
B :{$x \left(t\right) = 4 - 3 t$ , $y \left(t\right) = 3 - 2 t$}

$1.$ The initial position of the boats can be find out by simply substituting the values of $t = 0$ which yield there initial position
for $\textcolor{b l u e}{A \left(3 , - 4\right) , B \left(4 , 3\right)}$.
$2.$ To find the velocity we know that the velocity is the first derivative of the position vector ,
for A,
$x \left(t\right) = 3 - t$
$x ' \left(t\right) = 3$

$y \left(t\right) = 2 t - 4$
$y ' \left(t\right) = 2$

For B similarly we get
$x ' \left(t\right) = - 3$
$y ' \left(t\right) = - 2$

for the resultant we know that the formula for resultant vector is
$\vec{\text{resultant}} = \sqrt{{x}^{2} + {y}^{2} - 2 x y \cos \theta}$

we get for A we have we get $\sqrt{{3}^{2} + {2}^{2}} = \sqrt{13}$
similarly for B we get $\sqrt{13}$ but with different direction .

$3.$ for finding out the angle between them we need to make its trajectory a vector , we know its initial coordinates, now we find its coordinates after 1 second then we use it as a vector ,

position vector after 1 sec of A boat and B boat are color(red)((2,-2), color(red)((1,1) respectively ,so we make there vector which is $\left(- \hat{i} + 2 \hat{j}\right)$ and $\left(- 3 \hat{i} - 2 \hat{j}\right)$ after that we have formula
$\cos \theta = \frac{\hat{p} \cdot \hat{q}}{| p | \cdot | q |}$

to get the angle between the vectors we get $\theta = {\cos}^{- 1} \left(- \frac{1}{\sqrt{65}}\right)$

$4.$