Question

A question on vectors

Answer 1

If A,B, A+B are three non-zero vectors. Such that A+B is perpendicular to B then which one is correct
\[(A):A \ge B\]
\[(B):A \ge \frac{ B }{ \sqrt{2} }\]
\[(C):A>B\]
\[(D):A > \frac{ B}{ \sqrt{2} }\]

Answer 2

@LolWolf

Answer 3

I would imagine this is by magnitude? We can show the inequality by using the dot product. See what you can do with the following:
\[
A\cdot (A+B)=0
\]Since \(A\perp A+B\).

Answer 4

To me, I choose C)

Answer 5

... try to not just give the answer, mate.

Answer 6

Also, I would disagree.

Answer 7

MY intuition was telling me (C) , yeah that's the answer , but why

Answer 8

ok, I am sorry Lolwolf. I know No.name is not the person looks for the answer only.

Answer 9

Hm. I thought it'd be A.

Answer 10

intuitively, I got it, lol.

Answer 11

Yeah, no, you're right. I forgot to exclude that \(A\not \perp B\).

Answer 12

For the more analytical way:
\[
\begin{align}
(A+B)\cdot B&=
\\A\cdot B+B^2&=0\implies\\
|A||B|\cos \theta &=-B^2\implies\\
|A|=-\frac{|B|}{\cos \theta}
\end{align}
\]Since \(|\cos\theta| <1\), then:
\[
|A|\ge |B|
\] But, since \(\theta <\frac{\pi}{2}\), then:
\[
|A|>|B|
\]

Answer 13

Aligned wrong:
\[
\begin{align}
(A+B)\cdot B&=
\\A\cdot B+B^2&=0\implies\\
|A||B|\cos \theta &=-B^2\implies\\
|A|&=-\frac{|B|}{\cos \theta}
\end{align}
\]

Answer 14

Can you explain me what you did pleasee?

Answer 15

Sure. So, since \(A+B\) is perpendicular to \(B\), like I explained in the previous question, we have that their dot product is 0. Hence, we dot them together to receive the above expression.

We expand on \(B\), giving us \(A\cdot B+B\cdot B=0\). Using the definition of the dot product, we receive:
\[
|A||B|\cos \theta =-|B|^2
\]Dividing through by \(|B|\) gives:
\[
|A|\cos \theta =-|B|
\]And, since we know that \(\cos\theta \) is always \(\le 1\), then the conclusion follows.

Answer 16

I got everything except the last line