# Thread: Vector Triangle

1. ## Vector Triangle

As below,

I know how to find the length. but how to prove?

thanks.

2. ## Re: Vector Triangle

Originally Posted by BabyMilo
As below,
I know how to find the length. but how to prove?
This is a truly easy problem.
So you need need to show some effort on your part.
How does one find the length of a vector?

3. ## Re: Vector Triangle

$\displaystyle \sqrt 21$
$\displaystyle \sqrt 17$
$\displaystyle \sqrt 38$

4. ## Re: Vector Triangle

Originally Posted by BabyMilo
$\displaystyle \sqrt 21$
$\displaystyle \sqrt 17$
$\displaystyle \sqrt 38$
Do those three numbers form a Pythagorean triple ?
If so, then the triangle is a right triangle.

5. ## Re: Vector Triangle

Originally Posted by Plato
Do those three numbers form a Pythagorean triple ?
If so, then the triangle is a right triangle.
yes, it does.

what about the first part?
how to show they form sides of triangle?

6. ## Re: Vector Triangle

Originally Posted by BabyMilo
yes, it does.
what about the first part?
how to show they form sides of triangle?
Good grief, a triangle is a triangle is a triangle.
A Pythagorean triple gives a right triangle.

7. ## Re: Vector Triangle

Three line segments, of lengths a, b, and c, can form a triangle as long as no one length is greater than the sum of the other two (the shortest distance between two points is a straight line). But if $\displaystyle c^2= a^2+ b^2$ it is certainly true that the longest side is c and c< a+ b. A right triangle is a triangle as Plato said.

(To see that $\displaystyle c^2= a^2+ b^2$ leads to $\displaystyle c< a+ b$ [as long as neither of a or b is 0], add the positive number 2ab to both sides of $\displaystyle c^2= a^2+ b^2$. You get $\displaystyle c^2+ 2ab= a^2+ 2ab+ b^2= (a+ b)^2$. Since $\displaystyle (a+ b)^2$ is larger than $\displaystyle c^2$, a+ b is larger than c.)