## Saturday, November 6, 2010

### Column Vectors

What is a vector? By definition it is something that has a magnitude (length) and a direction (or bearing) but no fixed position. Hard to imagine? Then try the applet below:

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)

### Area of a Parallelogram – The Hard Way

Nixon as a young boy, long before he became President, was a Quaker and evidently his parents told him to never take the easy way of doing any task. How well he applied that advice is best left for historians. This is a maths investigation.

The easy way to find the area of a parallelogram is simply multiply the base times the height, just as you would do for a rectangle. A justification - though perhaps not a 'proof'- is offered pictorially below. Please note that by height, I most definitely do not mean the slant height, but the height that is perpendicular to the base.

In a slightly more difficult method we make use of the substitution:
height = slant height × sin Θ.

The overall formula for the area of the partallelogram is now the still very reasonable:
Area = base × (slant height × sin theta). Again the justification – for those who know a little trigonometry- is a picture.

And then, there is the following complication!

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)

The parallelogeram is formed by the vector addition u + v. The coordinates of u are (a, c) and the coordinates of v are (b, d). The entire shape is a rectangle. To find the area of the parallelogram we will cut away two triangles and two trapeziums. Note that the diagonal of the parallelogram separates the diagram into two congruent halves.

parallelogram = rectangle – 2 congruent triangles – 2 congruent trapeziums
= (a + b) × (c + d) – 2 × ½ (bd) – 2 × ½ [(a + b) + b ] c
(ac + ad + bc + bd) – bd – (a + 2b)c
= ac + ad + bc + bdbdac – 2bc
= ad – bc (Q.E.D. Is sometimes put here. Originally Q.E.D. was short for quod erat demonstrandum, which means "That which was to be demonstrated," although I prefer the somewhat cheeky, “Quite Easily Done.”)

The matrix M is made from u and v written as column vectors
The determinant of the matrix is, det M = product of main diagonal – product of off main
det M = ad – bc, which is exactly the area of the parallelogram, thus justifying one description of the determinant of a matrix as being or representing an area scaling factor.

### Parabolas and Intersecting Lines

Suppose you have already plotted P1: y = x2 - 3x + 1, but the teacher asks you to find graphically the roots of P2: y = x2 - 4x + 3. What a bother – now you need to plot another parabola. Or do you?
Johnny can rearrange P2 to get an equation of the form: sloped line = parabola already plotted.

 0 = x2 - 4x + 3 Given P2 +x =( x2 - 4x + 3) +x Add 'x' to both sides x = x2 - 3x + 3 Simplify by collecting ‘x’ terms on the right x - 2 = (x2 - 4x + 3) - 2 Add '-2' to both sides, i.e., subtract x - 2 = x2 - 3x + 1 Simplify by collecting constant terms on the right line = given P1

Now Johnny has a sloped line = parabola that he has already graphed
The sloped line has m = 1 and c = -2, so it is easy to draw.
Reading from the graph, the solution to the teacher’s question is x = 1 or x = 3.

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)

As a check, Mary factorises y= x2 - 4x + 3
y = (x – 3) (x – 1)
When y = 0,
x – 3 = 0, so x = 3
or x – 1 = 0 so x = 1
Done!