Heretofore, you have been dealing with scalar quantities. These are things like numbers, mass, length, that just have a size (magnitude). We will now formally introduce the vector. A vector is a mathematical object that has both a magnitude and a direction. Vectors are used to describe things like position, forces, weight, and hence play a big part in physics. In order to describe the direction of a vector, we need a frame of reference. In the physical world, this will quite often be horizontal and vertical axes (wrt the center of the earth). In math, we will be using the Cartesian axes, either the number line (x), x & y, or x, y, & z.
Vectors in R1
You have already been working with vectors, even though you may not have realized it. When you talk about a point on a number line, you are actually talking about a vector. If you have a number A, then it has a magnitude, which is |A|, and a direction, which is it's sign.
Vectors in R2
A vector in R2 can expressed as a magnitude and an angle from one of the axes, for example, 5N at 30 degrees up from horizontal. In R2, a vector can also be expressed in the same form as a line segment. For example, if you have two points P(1,1) and Q(2,2) you can use them to describe a vector $\vec{PQ}$. Up until now, you have been used to using an equation to describe a line in R2: Ax + By + C = 0 A vector in R2 can be also used to describe a position on a line, and so you can write an equation of a line using an initial point and a direction vector. $$\vec{v} = (x_0,y_0) + t(B,-A)$$ You should notice that the direction vector uses the same variables as the equation of the line. This is no co-incidence. Let us re-arrange the equation of the line y = (-A/B)x - C/B To go from an initial point on the line, say the y-intercept (x=0), to x=1, we move up -A and right B (remember that in this form of the equation, the number in front of the x is the slope, or rise over run). And this gives us our direction vector. Incidentally, you can also get your initial point from this equation as (0, -C/B). You could just as easily use the x-intercept, in which case it would be (-C/A, 0). This brings us to another way of describing a line in R2, using a free variable.
Vectors in R3
Vectors in R3, look and work very much like vectors in R2, just with one extra variable. If you are expressing it as an angle and magnitude, you need 3 angles to giive the direction, one from the x-axis, one from the y-axis and one from the z-axis. Theses are usually called $\alpha, \beta, and \gamma$ Lets start with equations in R3 Ax + By + Cz +D = 0 This equation describes a plane in R3. There is no way to describe a line using this form of expression.
Useful Formula
Projection of $\vec{u}$ on $\vec{v}$ is $\frac{(\vec{u}\cdotp\vec{v})\vec{v}}{|\vec{v}|^2}$ It's magnitude is $\frac{|\vec{u}\cdotp\vec{v}|}{|\vec{v}|}$ The distance from a point $P(x_1,y_1)$ to a line l: $Ax + By + C = 0$ is $d(P,l) = \frac{|Ax_1 + By_1 +C|}{\sqrt{A^2 + B^2}}$ extending to n dimensions
Worksheets
Net Force Word Problems Vector Addition and Subtraction problems Moving About Worksheets