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:
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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 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.
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!
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