# Quick Easy Vectors Question

• May 11th 2009, 09:50 AM
Quick Easy Vectors Question
http://i42.tinypic.com/2hq8093.jpg

its 7a and 8a that im stuck on

thnkz
• May 11th 2009, 10:29 AM
e^(i*pi)
Quote:

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 $tan(\theta) = \frac{4}{3}$

$\theta = arctan{\frac{4}{3}}$. Since this is taken from the horizontal the vertical (bearing) will be 90 minus your answer: $90 - arctan(\frac{4}{3})$
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:

$2i+12j = k(ai + 3j)$

Equate the j component: $12 = 3k \: , \: k = 4$

Equate the i component: $2 = ka \:, \: a = \frac{2}{k}$ to get a = 0.5
• May 11th 2009, 02:48 PM
on 7a it says find the vector how come your answers a degree

and in the back of the book it says (72i + 96j)
• May 12th 2009, 11:26 AM
earboth
Quote:

http://i42.tinypic.com/2hq8093.jpg

its 7a and 8a that im stuck on

thnkz

to #7a).

The velocity vector is $\vec v = (3,4)$. The unit vector of $\vec v$ is

$\overrightarrow{v^0} = \dfrac1{|\vec v|} \cdot \vec v=\left(\dfrac35\ ,\ \dfrac45\right)$

The speed of the heli is $120\ \dfrac{km}{h}$ . Multiply the vector $\overrightarrow{v^0}$ by $120\ \dfrac{km}{h}$ to get the the velocity vector of the helicopter:

$\vec A=120\cdot \left(\dfrac35\ ,\ \dfrac45\right) = (72\ ,\ 96)$
• May 21st 2009, 07:28 AM
Quote:

Originally Posted by earboth
to #7a).

The velocity vector is $\vec v = (3,4)$. The unit vector of $\vec v$ is

$\overrightarrow{v^0} = \dfrac1{|\vec v|} \cdot \vec v=\left(\dfrac35\ ,\ \dfrac45\right)$

The speed of the heli is $120\ \dfrac{km}{h}$ . Multiply the vector $\overrightarrow{v^0}$ by $120\ \dfrac{km}{h}$ to get the the velocity vector of the helicopter:

$\vec A=120\cdot \left(\dfrac35\ ,\ \dfrac45\right) = (72\ ,\ 96)$

where did that 5 come from
• May 21st 2009, 11:35 AM
earboth
Quote:

where did that 5 come from

If you want to transform the vector $\vec v$ into a unit vector you have to multiply the vector by $\dfrac1{|\vec v|}$.

If $\vec v = (3, 4)$ then $|\vec v| = \sqrt{3^2+4^2} = 5$

The vector $\overrightarrow{v^0} = \left(\dfrac35\ ,\ \dfrac45 \right)$ 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)
• May 21st 2009, 12:08 PM
thankz

can you do question 7c and 9 for me please
• May 21st 2009, 10:19 PM
earboth
Quote:

thankz

can you do question 7c and 9 for me please

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 $\vec a = k \cdot \vec b$.

#9a)
$(t^2,\ (12-t)) = k\cdot (1,0)~\implies~ \begin{array}{rcl} t^2&=& k \\ 12-t &=&0\end{array}$

At t = 12 you'll get the position vector (144, 0) which is parallel to (1, 0)

#9b)
$(t^2,\ (12-t)) = k\cdot (1,1)~\implies~ \begin{array}{rcl} t^2&=& k \\ 12-t &=&k\end{array}$

By substitution you'll get:

$t^2 = 12-t~\implies~ t^2+t-12=0~\implies~\boxed{t=-4~\vee~t=3}$

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).