# Thread: Vector Geometry -> Prove Right Triangle Efficiently

1. ## Vector Geometry -> Prove Right Triangle Efficiently

Ok,

You have points A(-1,3,2) B(-1,5,2) C(1,5,-2) .. Prove that triangle ABC is a right triangle.

We know dot product and cross product and all the minor stuff that you need to do those, but cross product is the most advanced stuff we know. Mind you, I don't think we need cross product for this (in my opinion anyways). The way I can think of doing it is proving that AB(dot)AC or CA(dot)CB or BA(dot)BC equals zero... If it is zero, those vectors are 90 degrees, and ABC is a right triangle. But this is a lot of work! I don't mind doing it, but I have a feeling there is an easier way.

Anyone have any ideas?

2. Are any pair of the vectors, $\displaystyle \overrightarrow {AB} ,~\overrightarrow {AC} ,~\text{or}~\overrightarrow {BC}$ perpendicular?
(Dot product equal zero.)

3. Yea... That's what I mean... That's how I would've figured the question out... my issue with that is that it means testing out a lot of diffent vectors with the dot product in order to find the answer. I mean, sure it works, but I was wondering if there is a more efficient way that I am just not thinking of.

Thanks

4. Originally Posted by mike_302
Yea... That's what I mean... That's how I would've figured the question out... my issue with that is that it means testing out a lot of diffent vectors with the dot product in order to find the answer. I mean, sure it works, but I was wondering if there is a more efficient way that I am just not thinking of.
Do you really call finding the dot product of three different pairs of vectors a lot?

5. lol, well.. relative to what I had a feeling could be done... I guess thats the way then... I just felt there was a little too much guessing in the whole thing that could lead to too many failed attempts... but I guess not.

thanks!

6. Originally Posted by mike_302
lol, well.. relative to what I had a feeling could be done... I guess thats the way then... I just felt there was a little too much guessing in the whole thing that could lead to too many failed attempts
Well you could see if $\displaystyle \left\| {\overrightarrow {AB} } \right\|,~\left\| {\overrightarrow {AC} } \right\|,~\left\| {\overrightarrow {BC} } \right\|$ form a Pythagorean triple.