SOLUTION: Prove that two vectors must have equal magnitudes if their sum is perpendicular to their difference

Question 1206236: Prove that two vectors must have equal magnitudes if their sum is perpendicular to their difference
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

I'll use a period to indicate dot product.
A . B = A dot product B
where A and B are vectors.

Here are the dot product rules to memorize
A.B = B.A
A.(B+C) = A.B+A.C
0.A = 0
|A| = sqrt(A.A)
(k*A).B = k*(A.B) = A.(k*B) where k is a scalar and * means the usual multiplication
Let
S = A+B
represent the sum of the vectors

Recall that vectors are perpendicular if and only if their dot product is zero.
We'll start with the idea A+B is perpendicular to A-B.
That must mean (A + B) . (A - B) = 0
The goal is to end up with |A| = |B| to show they have the same magnitude.

(A + B) . (A - B) = 0
S . (A - B) = 0
S . A - S . B = 0
S . A = S . B
(A+B) . A = (A+B) . B
A . (A+B) = B . (A+B)
A.A + A.B = B.A + B.B
A.A + A.B = B.B + A.B
A.A = B.B
sqrt(A.A) = sqrt(B.B)
|A| = |B|

We have shown that (A + B) . (A - B) = 0 leads to |A| = |B|
Therefore, if A+B is perpendicular to A-B, then vectors A and B must have the same magnitude.

Answer by ikleyn(52800) (Show Source): You can put this solution on YOUR website!
.

Geometrically, the sum of two vectors is a DIAGONAL of the parallelogram, built on these vectors as on consecutive sides.

Then the difference of the vectors is the OTHER DIAGONAL of the same parallelogram, built on these vectors as on consecutive sides.

If the sum of two vectors is perpendicular to their difference, it means that
the diagonals of the parallelogram are perpendicular.

But if a parallelogram has perpendicular diagonals, then this parallelogram is a rhombus,
and, hence, all his sides have equal lengths.

It is a geometric proof of your statement.

At this point, this geometric proof is complete.

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