Prove the state

[marinaa11]

Hello, I am stuck with this task: there is a triangle given and I have to prove the state:






I really don't have any clear solution as this topic is so hard for me.. As there is a vectors' multiplication, there are formulas like and , but I don't have any numbers (values) given, maybe I should draw vectors on the coordinate plane?.. Could you explain what's the first step?
Thanks.

 

[Deleted member 4993]

Hello, I am stuck with this task: there is a triangle given and I have to prove the state:

View attachment 22701 View attachment 22702




I really don't have any clear solution as this topic is so hard for me.. As there is a vectors' multiplication, there are formulas like View attachment 22694 and View attachment 22695, but I don't have any numbers (values) given, maybe I should draw vectors on the coordinate plane?.. Could you explain what's the first step?
Thanks.

I would first note that:

CA . CB = 0

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this problem.

 

[Dr.Peterson]

Hello, I am stuck with this task: there is a triangle given and I have to prove the state:

View attachment 22701 View attachment 22702




I really don't have any clear solution as this topic is so hard for me.. As there is a vectors' multiplication, there are formulas like View attachment 22694 and View attachment 22695, but I don't have any numbers (values) given, maybe I should draw vectors on the coordinate plane?.. Could you explain what's the first step?
Thanks.

There are several ways you might start. I would label the sides lengths a, b, and c as usual, and express the three dot products in terms of a, b, c, and cosines of angle. Then this about what those cosines are equal to ...

(You don't need numbers to do algebra! In fact, letters typically make it easier.)

 

[pka]

Hello, I am stuck with this task: there is a triangle given and I have to prove the state:

View attachment 22701 View attachment 22702I really don't have any clear solution as this topic is so hard for me.. As there is a vectors' multiplication, there are formulas like View attachment 22694 and View attachment 22695, but I don't have any numbers (values) given, maybe I should draw vectors on the coordinate plane?.. Could you explain what's the first step?

Is it clear that \(\overrightarrow {CA}\cdot\overrightarrow {CB}=0~?\) If not you need to stop and go learn the basics.
\(\|\overrightarrow {CA}-\overrightarrow {CB}\|^2=\|\overrightarrow {CA}\|^2+\|\overrightarrow {CB}\|^2-2\,\overrightarrow{CA}\cdot\overrightarrow {CB}\) WHY?
Is it true that \(\overrightarrow {BA}=\overrightarrow {CA}-\overrightarrow {CB}~?\) Why and how?
Can you now finish?

 

[marinaa11]

Thanks for answers. I took a new sheet of paper and what I fully understand is why CA . CB = 0 (because vector b is perpendicular to vector a). Also, I know vectors' division rule. Now should I return to the state I have to prove as I know what CA . CB = 0 and work with Pythagoras' Theorem?...

 

[pka]

You need to use \(\|\vec{v}\|^2=\vec{v}\cdot\vec{v}\). Thus:
\( \|\vec{v}-\vec{u}\|^2=(\vec{v}-\vec{u})\cdot(\vec{v}-\vec{u})\\=\vec{v}\cdot\vec{v}-2~(\vec{u}\cdot\vec{v})+\vec{u}\cdot\vec{u}\\=\|\vec{v}\
|^2 -2~(\vec{u}\cdot\vec{v})+\|\vec{u}\|^2\)Do you follow that?

 

[marinaa11]

Thank you a lot for explanations