Question

Show that if the magnitude of A-B = the magnitude of A+B then A and B are perpindicular

Answer 1

\[\left| A-B \right| = \left| A+B \right|\]

That is what the equation should actually look like. I am unsure of how to start. New to vector algebra

Answer 2

Now square both sides and remember that
\[ |v|^2 = v \cdot v \]

Answer 3

So I get

\[\left| A \right|^2-\left| B \right|^2 = \left| A \right|^2+\left| 2 AB \right| +\left| B \right|^2\] correct?

Answer 4

No.

It is not the case that
\[ ******WRONG \ \ \ \ \ |A+B|^2 = |A|^2 +|2AB| + |B|^2 \]

What is the case is that
\[ |A+B|^2 = (A+B).(A+B) = A.A + A.B + B.A + B.B \]
where . is the dot product

Now expand out the left and side.

Answer 5

which should become \[2ABcos\] sorry if that did not come through so it becomes

\[\left| A \right|^2 + 2ABcos \theta+ \left| B \right|^2\]

Answer 6

Ok. That will also work but is a bit more subtle. In any case, rewrite the left hand side.

Answer 7

\[\left| A \right|^2 - \left| B \right|^2 = \left| A \right|^2 +2ABcos \theta + \left| B \right|^2\]

Answer 8

No
\[ |A-B|^2 = (A-B).(A-B) = |A|^2 - 2|A||B|\cos\theta + |B|^2 \]

Answer 9

Sorry I just reworked it out and got that my bad

Answer 10

Hence if
\[ |A-B| = |A+B| \]
then
\[ |A|^2 - 2|A||B|\cos\theta + |B|^2 = |A|^2 + 2|A||B|\cos\theta + |B|^2 \]
hence ... you take it from here

Answer 11

It ultimately becomes \[\left| A \right|\left| B\right|\cos \theta = 0\]

which means the angle between them is 90 or 180 degrees and therefore they are perpindicular Thanks!

Answer 12

Almost, certainly it must be that \( \cos\theta = 0 \) which means they are perpendicular (but not 180 degrees, yes?)

Answer 13

Sorry 270

Answer 14

Right

Answer 15

Do you think you could help me with a similar problem?

Answer 16

Post it and if I can help I will.

Answer 17

Prove the law of sines using the cross product. Here is what I have done so far...



From there I am stuck

Answer 18

Post as a new question if you don't mind.