Vectors

• Jun 5th 2010, 06:21 AM
Punch
Vectors
http://i952.photobucket.com/albums/a...g/P6050015.jpg

sry i but i really have no idea how to start the question... i have problems with both parts of the question as they are similar...

i am not asking u to spoonfeed me, perhaps u can solve the first part and let me solve the second part as they are similar ^^

• Jun 5th 2010, 07:10 AM
sa-ri-ga-ma
Shift y from the present position to (2,1) and (4,3).

Complete the parallelogram. The resultant of x and y will be (1,4) and (4,3)

Parallel to this resultant from P is (0,5) and (3,4)
• Jun 6th 2010, 06:00 AM
Punch
thanks! mind if i ask u for part ii how do i handle the fraction?
• Jun 6th 2010, 06:34 AM
sa-ri-ga-ma
Quote:

Originally Posted by Punch
thanks! mind if i ask u for part ii how do i handle the fraction?

3y/2 will be (4, 1) and (7, 4)

Shift x to the right so that x touches y.
• Jun 6th 2010, 08:49 AM
Movement (displacement) vectors
Hello Punch

From the diagram, we can see that $\textbf{x}$ represents a movement which can be described as:
Move 1 unit to the right and 3 units down.
We can write this as
$\textbf{x}=\binom{1}{-3}$
Similarly $\text{y}$ represents the movement:
Move 2 units to the right and 2 units up
or:
$\textbf{y}=\binom{2}{2}$
We add two movement (or displacement) vectors together using an ordinary $+$ sign, by thinking of $+$ as representing the phrase 'and then'. So $\textbf{x} + \textbf{y}$ means
Move 1 unit to the right and 3 units down
and then
Move 2 units to the right and 2 units up
which is obviously equivalent to
Move 3 units to the right and 1 unit down
We sometimes write all of this as:
$\textbf{x} + \textbf{y}=\binom{1}{-3} + \binom{2}{2}$
$= \binom{3}{-1}$
Do you see how it works?

Now $\vec{PR}$ stands for:
Move from $P$ to $R$
So if $\vec{PR}=\textbf{x} + \textbf{y}$ and we must start at the point $P\; (0,5)$, and carry out the movement $\binom{3}{-1}$. When we do this, we get to the point $(3,4)$. So that's where $R$ is: $(3,4)$.

Now $\frac32\textbf{y}$ is the same movement as $\textbf y$, but multiplied by $\frac32$. In other words:
$\frac32\textbf{y}=\frac32\times\binom22$
$=\binom33$
So:
$\frac32\textbf{y}-\textbf x=\binom33 - \binom{1}{-3}$
$=\binom{2}{6}$ (Check this carefully!)
and this is the movement $\vec{AB}$. So, to find the position of the point $B$, start at $A$ and carry out the movement $\binom26$. Can you see that $B$ is at $(6,6)$?