Vector question

[zzinfinity]

Hi, I'm a little stuck on this vector problem I'd appreciate any thoughts.

Suppose the vectors (A+B) and (A?B) are orthogonal. What, if anything, can you conclude about A and B?

I have a feeling that you can say A and B are also orthogonal. Is this right, and if so how could I prove it?

Thanks!

 

[daon]

If A=[ai], B=[bi] and (A-B) dot (A+B) = 0, then

SUM_i (ai-bi)(ai+bi)=0

Do a bit of algebra and see what happens.

 

[Deleted member 4993]

Hi, I'm a little stuck on this vector problem I'd appreciate any thoughts.

Suppose the vectors (A+B) and (A?B) are orthogonal. What, if anything, can you conclude about A and B?

I have a feeling that you can say A and B are also orthogonal. Is this right, and if so how could I prove it?

Thanks!

I would make it slightly more general by defining:

A = x[sub:1o8et2zq]a[/sub:1o8et2zq]i + y[sub:1o8et2zq]a[/sub:1o8et2zq]j+ z[sub:1o8et2zq]a[/sub:1o8et2zq]k


and

B = x[sub:1o8et2zq]b[/sub:1o8et2zq]i + y[sub:1o8et2zq]b[/sub:1o8et2zq]j + z[sub:1o8et2zq]b[/sub:1o8et2zq]k


then find

(A+B).(A-B)

 

[soroban]

Hello, zzinfinity!

I made a sketch and the answer smacked me upside the head.

\(\displaystyle \text{Suppose the vectors }(\vec A+\vec B)\text{ and }(\vec A-\vec B)\text{ are orthogonal.}\)

\(\displaystyle \text{What, if anything, can you conclude about }\vec A\text{ and }\vec B\,?\)

          S * - - - - - - - - * R
           /  *           *  /
          /     *     *     /
       B /        *        /
        /     *     *     /
       /  *           *  /
    P * - - - - - - - - * Q
               A

\(\displaystyle \text{We have parallelogram }PQRS.\)

\(\displaystyle \text{We have: }\:\begin{Bmatrix}\vec A &=& \overrightarrow{PQ} \\ \\[-3mm] \vec B &=& \overrightarrow{PS}\end{array}\)

\(\displaystyle \text{Also: }\:\begin{Bmatrix}\text{diagonal }\overrightarrow{PR} &=& \vec A + \vec B \\ \\[-3mm] \text{diagonal }\overrightarrow{SQ} &=& \vec A - \vec B \end{array}\)

\(\displaystyle \text{We are told that: }\:\overrightarrow{PR} \perp \overrightarrow{SQ}.\)


\(\displaystyle \text{We have parallelogram }PQRS\text{ with perpendicular diagonals.}\)
. . \(\displaystyle \text{Hence, }PQRS\text{ is a }rhombus.\)

\(\displaystyle \text{Therefore: }\:\left|\vec A\right| \:=\: \left|\vec B\right| \;\;\hdots\;\text{ Their magnitudes are equal.}\)