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Here is a quick guide to understanding vectors, to help you do your physics homework or assignment:
In this section we will have a look at vectors. Vectors only impart magnitude and direction. Quantities which have such properties are forces, velocities, distance, and other such things. For instance, forces have both a magnitude (the amount of force applied) and a direction (the direction the force is acting in).
Vectors are represented graphically as a line segment, with the length of the line being directly related to the magnitude, and its orientation being the direction. Consider the following sketch:
Each of the vectors in the sketch have the same magnitude (length) and direction (orientation). They are, therefore, equivalent vectors. A vectors positioning on the graph has no relevance to its magnitude and direction, but it can be useful to place them starting at the origin when problem solving.
A vector  expressed as a quantity has the form:
where
 and 
and 'a' and 'b' are constants
Finding a vector given two points
To find the vector from point a = (x,y) to point b = (u,z)
Which is a vector starting at 'a' and pointing towards and ending at 'b'
Example
Find the vector pointing towards (3,5) and starting at (1,2)
Solution:
Magnitude of a vector
given a vector
The magnitude of a vector is defined to be
Example
Find the magnitude of the vector we found in the first example
^2+(3)^2} =\sqrt{4+9}=\sqrt{13} "||\vec{v}||=\sqrt{(2)^2+(3)^2} =\sqrt{4+9}=\sqrt{13}")
This means that the magnitude of this vector is 
Unit vector
The unit vector is a vector with magnitude equal to 1, to find the unit vector of a given vector use:
This will change any vector to a unit vector  of length 1
Example
Change the vector from the previous example to a unit vector.
To do this, we simply have to divide the the vector by it's magnitude.
for the vector
its unit vector is
just to check and see if this vector truly has a magnitude of 1, lets find the magnitude
^2+(\frac{3}{\sqrt{13}})^2} =\sqrt{\frac{4}{13}+\frac{9}{13}}=\sqrt{1}=1 "||\vec{u}||=\sqrt{(\frac{2}{\sqrt{13}})^2+(\frac{3}{\sqrt{13}})^2} =\sqrt{\frac{4}{13}+\frac{9}{13}}=\sqrt{1}=1")
Scalar Multiplication
Given a vector multiplied by a constant
all components are multiplied by the constant
Example
,3(3)>=<6,9> "3<2,3> = <3(2),3(3)>=<6,9>")
or
=6\vec{i}+9\vec{j} "3(2\vec{i}+3\vec{j})=6\vec{i}+9\vec{j}")
Vector Addition and Subtraction
Given two vectors
then
which has both of the following geometrical interpretations:
and
also,
which has the following geometrical interpretation:
Example
Find the sum and difference of the given vectors
Solution:
+(-7)> =<4,-9> "\vec{a}+\vec{b}=<3+1,(-2)+(-7)> =<4,-9>")
and
-(-7)> =<2,5> "\vec{a}-\vec{b}=<3-1,(-2)-(-7)> =<2,5>")
Dot Product
Given vectors
their dot product is defined as
which means that the dot product of two vectors produces a scalar.
Also, the following relationship is true for two vectors and the angle  between them:
Examples
1.
Find the dot product of the given vectors
Solution:
their product is
 (4)+ (5)(6)=-8+30=22 "\vec{a} \, \cdot \, \vec{b} =(-2) (4)+ (5)(6)=-8+30=22")
2. Find the angle between the vectors in the previous example
we know that
we also know from the previous example that
now,
^2+(5)^2}=\sqrt{29} "||\vec{a}|| = \sqrt{(-2)^2+(5)^2}=\sqrt{29}")
^2+(6)^2}=\sqrt{52} "||\vec{b}|| = \sqrt{(4)^2+(6)^2}=\sqrt{52}")
solving the dot product formula for we get
(\sqrt{52})}=59^o "\theta = \arccos{\frac{\vec{a} \, \cdot \, \vec{b}}{||\vec{a}||\,||\vec{b}||} = \arccos{\frac{22}{(\sqrt{29})(\sqrt{52})}=59^o")
hence the angle between the vectors is 59 degrees.
Finally, notice that if the angle between the vectors is 90 degrees, the value of the dot product will be zero since cos(90) = 0
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