Whether you realize it or not, you actually use vectors very frequently in common daily activities. Vectors are quantities that differ from the quantities we have mostly used in mathematics so far called scalars.
Quantities that require only magnitude are called scalar quantities, or more simple just scalars.
Quantities that require both magnitude and direction are called vector quantities, or more simply just vectors.
Daily examples of vectors are things like walking to a certain place you have the pace that you are walking and the direction you are going. Breathing, your muscles exert a force that has both size and direction.
A vector is a directed line segment.
Let's look at two distinct points on the plane, $A$A and $B$B.
Then we draw in a directed line segment (a line segment with an arrow at one end), from $A$A to $B$B.
$A$A (first point) is called the initial point
$B$B (second point) is called the terminal point.
Sometimes the arrow appears at the end of the line segment like in the image above, or in the line segment like this
This is called a vector. We notate the vector using a number of different notations.
This notation indicates the initial point and the terminal point by the direction of the arrow and the order the points are written. This notation is also called vector displacement notation as it demonstrates the displacement of a point B, from A.
a The vector can be written as vector a (where the $a$a is printed in bold text) or if you were writing by hand you would put a squiggle underneath like this,
If the vector has a fixed point, like the origin, $O$O, then it is called a position vector. Position vectors can also be represented using matrix notation. For example,
can be represented
and
can be represented
Note that the first element in the column matrix indicates the $x$x movement (right is positive and left is negative) and the second element indicates the change in $y$y, (up positive and down negative).
This applet will give you some practice at making vectors, (uses column vector notation)
As with most things we have learned about in mathematics, a negative sing indicates direction. A negative vector relates to a vector that has the same magnitude (size), but goes in the opposite direction.
This applet will give you some practice at making negative vectors.
A zero vector has a zero magnitude and any direction, effectively it is a point on the plane.
Which of the following are vector quantities?
a force of 8 N acting horizontally right
a displacement of 4 m along the line joining A and B
a mass of 1 kg
a time of 1 seconds
Remember that a vector quantity has to have both magnitude (size) AND direction.
All these options have magnitude.
Only $(A)$(A) and $(B)$(B) have direction.
$(A)$(A) is in the direction given by "horizontally right", and $(B)$(B) is in the direction "along the line joining A and B".
Write the vector represented on the plane as a column vector.
$\editable{}$ | ||||
$\editable{}$ |
The first component in the column vector is the $x$x-component of the vector. This vector travels $4$4 units in the positive $x$x-direction.
The second component in the column vector is the $y$y-component. This vector travels $4$4 units in the positive $y$y-direction.
So the column vector required is
Write the vector represented on the plane as a column vector.
$\editable{}$ | ||||
$\editable{}$ |
We use the word magnitude to mean the size, or length, of the vector. The magnitude of a vector is indicated using the absolute value notation. So magnitude of of vector is indicated by .
How can we work out the length of a vector? Well we use the Pythagorean Theorem.
To find the length of this vector
We first need the horizontal and vertical components
Using the Pythagorean Theorem
$\text{length}^2$length2 | $=$= | $4^2+3^2$42+32 |
$\text{length}^2$length2 | $=$= | $16+9$16+9 |
$\text{length}$length | $=$= | $\sqrt{25}$√25 |
$\text{length}$length | $=$= | $5$5 |
If we are just given the coordinates of the initial and terminal points we would work out the magnitude like this.
Find the magnitude of the vector with initial point
$$ and terminal point $$
The horizontal component is the distance between the x-coordinates. Which is $$
The vertical component is the distance between the y-coordinates. Which is $$
So $=\sqrt{3^2+9^2}=\sqrt{90}=3\sqrt{10}$=√32+92=√90=3√10
Vectors are only equal if they have the same magnitude and direction. They need not have the same initial and terminal points, just the same size and direction.
So these vectors are all equal
And these are not
A unit vector is a vector of of magnitude (length) 1. There is specific notation we can use for a unit vector that uses i and j's and we will discuss this in later entries. For now we just need to know that if the length is 1, it is a unit vector in that particular direction.
Consider the vector $\left(2\sqrt{3},2\right)$(2√3,2).
Find the magnitude of the vector.
Find the direction of the vector.
Give your answer in degrees.
Let $G$G and $H$H be the points $G$G$\left(11,3\right)$(11,3) and $H$H$\left(12,-2\right)$(12,−2).
Find the vector $\vec{HG}$→HG in component form:
$\vec{HG}$→HG$=$=$\left(\editable{},\editable{}\right)$(,)
What is the exact length of the vector $\vec{HG}$→HG?
Find the magnitude of the vector |
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What do we know about a vector so far?
Each vector has
Knowing some of the above conditions will allow us to calculate the others, and because a vector also has links to trigonometry we can also use the trigonometric ratios to help us with the calculations.
Given that the vector a, projects from initial point $\left(1,3\right)$(1,3), at an angle of $45^\circ$45° find the terminal point if the magnitude is $3.5$3.5 units.
The information given to us here, results in the following right triangle image.
The terminal point will have the coordinates $$
Using trigonometry we can see $\cos45^\circ=\frac{x}{3.5}$cos45°=x3.5, so $x=3.5\times\cos45^\circ$x=3.5×cos45° and similarly $\sin45^\circ=\frac{y}{3.5}$sin45°=y3.5, so $y=3.5\times\sin45^\circ$y=3.5×sin45°. Evaluating these to $2$2 decimal places we get $x=2.47$x=2.47 and $y=2.47$y=2.47. The fact that both x and y are equal make sense because an angle of $45^\circ$45° creates an isosceles triangle.
Now we can work out the terminal point, $$
This applet will help you to visualize the $x$x component and $y$y component. Remember that it uses the principles of right-angled trigonometry.
Consider the vector $v$v plotted on the graph.
What is the horizontal component, $a$a, of vector $v$v?
What is the vertical component, $b$b, of vector $v$v?
Consider the vector $v$v with magnitude $161$161 and direction angle $156.5^\circ$156.5°.
Find the horizontal component of $v$v.
Give your answer to the nearest tenth.
Find the vertical component of $v$v.
Give your answer to the nearest tenth.
A vector with magnitude $6$6 is shown in the diagram.
Write the vector in the form $\left(a,b\right)$(a,b), leaving $a$a and $b$b as exact values.