http://i42.tinypic.com/2hq8093.jpg
its 7a and 8a that im stuck on
thnkz
http://i42.tinypic.com/2hq8093.jpg
its 7a and 8a that im stuck on
thnkz
7a) If you draw a diagram with i on the x axis and j on the y axis then you can use the tan to find the argument.
In your case it is $\displaystyle tan(\theta) = \frac{4}{3}$
$\displaystyle \theta = arctan{\frac{4}{3}}$. Since this is taken from the horizontal the vertical (bearing) will be 90 minus your answer: $\displaystyle 90 - arctan(\frac{4}{3})$
The speed component is given already
8a) If two vectors are parallel then b = ka where b and a are vectors and k is a constant:
$\displaystyle 2i+12j = k(ai + 3j)$
Equate the j component: $\displaystyle 12 = 3k \: , \: k = 4$
Equate the i component: $\displaystyle 2 = ka \:, \: a = \frac{2}{k}$ to get a = 0.5
to #7a).
The velocity vector is $\displaystyle \vec v = (3,4)$. The unit vector of $\displaystyle \vec v$ is
$\displaystyle \overrightarrow{v^0} = \dfrac1{|\vec v|} \cdot \vec v=\left(\dfrac35\ ,\ \dfrac45\right)$
The speed of the heli is $\displaystyle 120\ \dfrac{km}{h}$ . Multiply the vector $\displaystyle \overrightarrow{v^0}$ by $\displaystyle 120\ \dfrac{km}{h}$ to get the the velocity vector of the helicopter:
$\displaystyle \vec A=120\cdot \left(\dfrac35\ ,\ \dfrac45\right) = (72\ ,\ 96)$
If you want to transform the vector $\displaystyle \vec v$ into a unit vector you have to multiply the vector by $\displaystyle \dfrac1{|\vec v|}$.
If $\displaystyle \vec v = (3, 4)$ then $\displaystyle |\vec v| = \sqrt{3^2+4^2} = 5$
The vector $\displaystyle \overrightarrow{v^0} = \left(\dfrac35\ ,\ \dfrac45 \right)$ has the same direction as (3, 4) but a length of 1.
If you multiply this vector by 120 km/h you'll get a vector with a length of 120 and the direction of (3, 4)
I'm going to do #9.
(To help you more efficiently you should show some work of your own so we can see where you need some additional assistance)
Two vectors are parallel if $\displaystyle \vec a = k \cdot \vec b$.
#9a)
$\displaystyle (t^2,\ (12-t)) = k\cdot (1,0)~\implies~ \begin{array}{rcl} t^2&=& k \\ 12-t &=&0\end{array}$
At t = 12 you'll get the position vector (144, 0) which is parallel to (1, 0)
#9b)
$\displaystyle (t^2,\ (12-t)) = k\cdot (1,1)~\implies~ \begin{array}{rcl} t^2&=& k \\ 12-t &=&k\end{array}$
By substitution you'll get:
$\displaystyle t^2 = 12-t~\implies~ t^2+t-12=0~\implies~\boxed{t=-4~\vee~t=3}$
t = -4 will yield the vector (16, -16) which is parallel to the vector (1, 1) but pointing into the opposite direction.
t = 3 will yield the vector (9, 9) which is parallel to (1, 1).