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:

## Saturday, November 6, 2010

### 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!

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 + ~~bd~~ – ~~bd~~ – ~~ac~~ – 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 vectorsThe 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 P

_{1}: y = x^{2}- 3x + 1, but the teacher asks you to find graphically the roots of P_{2}: y = x^{2}- 4x + 3. What a bother – now you need to plot another parabola. Or do you?Johnny can rearrange P

_{2}to get an equation of the form: sloped line = parabola already plotted. 0 = x ^{2 }- 4x + 3 | Given P _{2} |

+x =( x ^{2} - 4x + 3) +x | Add 'x' to both sides |

x = x ^{2} - 3x + 3 | Simplify by collecting ‘x’ terms on the right |

x - 2 = (x ^{2} - 4x + 3) - 2 | Add '-2' to both sides, i.e., subtract |

x - 2 = x ^{2} - 3x + 1 | Simplify by collecting constant terms on the right |

line = given P _{1} | |

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.

As a check, Mary factorises y= x

^{2}- 4x + 3y = (x – 3) (x – 1)

When y = 0,

x – 3 = 0, so x = 3

or x – 1 = 0 so x = 1

Done!

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