Exploring Parallelogram Diagonals: Vector Algebra

[lolimcool]

Homework Statement

P is the point where the diagonals of the parallelogram abcd intersect one another
let $\alpha = AB$ and $\beta = AD$ and let s and t be scalars such that $AP = sAC$ and $BP = tBD$

use vector algebra to show that
$s(\alpha + \beta) = AP = \alpha + t(\beta - \alpha)$

The Attempt at a Solution

ok so
s(AB + AD) = AP = AB + t(AD - AB)
s(AC) = AP = AB + t(BD) (s(AC) and AP have the same direction so s(AC) = AP)
AP = AB + t(BD)
AP = AB + BP
AP = AP

is that right?and could someone explain to me the principle of planar independence

 

[Mark44]

Homework Statement

P is the point where the diagonals of the parallelogram abcd intersect one another
let $\alpha = AB$ and $\beta = AD$ and let s and t be scalars such that $AP = sAC$ and $BP = tBD$

use vector algebra to show that
$s(\alpha + \beta) = AP = \alpha + t(\beta - \alpha)$

The Attempt at a Solution

ok so
s(AB + AD) = AP = AB + t(AD - AB)

The line above is essentially what you are trying to show, so you shouldn't start off by assuming that it is true.

First, you want to show that s($\alpha~+~\beta$) = AP, then show that s($\alpha~+~\beta$) = $\alpha~+t(\beta~-\alpha)$. (Or you can show that AP = $\alpha~+t(\beta~-\alpha)$.)

For the first part, you have
s($\alpha~+~\beta$) = s(AB + AD) = s(AC). From the given information, what is that last expression equal to?

s(AC) = AP = AB + t(BD) (s(AC) and AP have the same direction so s(AC) = AP)
AP = AB + t(BD)
AP = AB + BP
AP = AP

is that right?

Generally speaking, you don't want to end with a statement that is obviously true; i.e., that some quantity equals itself.

and could someone explain to me the principle of planar independence

 

[HallsofIvy]

Actually, what you have written is a valid "synthetic proof" where you start with what you want to show, then work down to an obviously true statement. The critical part is that each statement be reversible. That way a "standard proof" would start at the bottom and work upward.

That is, as "standard" proof would be:
AP= AP
AP= AB+ BP
AP = AB + t(BD)
AP = s(AC) = AB + t(BD)
AP = s(AB + AD) = AB + t(AD - AB)

By the way, you don't want to say "s(AC) and AP have the same direction so s(AC) = AP". It sounds like you are saying "if two vectors have the same direction then they are equal" which, of course, is not true. You are given that s(AC)= AP.

 

[Mark44]

Generally speaking, you don't want to end with a statement that is obviously true; i.e., that some quantity equals itself.

Actually, what you have written is a valid "synthetic proof" where you start with what you want to show, then work down to an obviously true statement. The critical part is that each statement be reversible.

And this is why I qualified what I said. I didn't want to go into details about the steps being reversible.