I attached my proof below and I am curious if there is any truth to it.my solution
Asked
Active
Viewed 48 times
-2
-
4Please don't use pictures for text or formulas. See here, why. Use MathJax. Here is a tutorial. – Dietrich Burde May 02 '23 at 15:08
-
I didn't check your solution's validity, but it looks overly complicated. Hint: all you need is $\vec a \cdot \vec a = |\vec a|^2$. – Deepak May 02 '23 at 15:12
-
I know the correct solution at this point, I am asking as it was an exam question, and I am curious if that answer has any validity. – vitaliymsh May 02 '23 at 15:16
-
Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community May 02 '23 at 15:17
-
1What does "$u$ is orthogonal" mean? – kimchi lover May 02 '23 at 15:18
-
Does this answer your question? Linear Algebra: proof – Another User May 02 '23 at 15:28
-
Thank you, this answers my question! – vitaliymsh May 02 '23 at 15:32
1 Answers
1
We have $\|u-v\|=\|u+v\|$ iff $$\|u\|^2-2\langle u,v\rangle+\|v\|^2 =\|u\|^2+2\langle u,v\rangle+\|v\|^2$$ iff $\langle u,v\rangle=0$. Avoid coordinate whenever possible.

Michael Hoppe
- 18,103
- 3
- 32
- 49