# Vector Orthogonality proof

## Homework Statement

Show that

||b||a + ||a||b and ||b||a - ||a||b are orthogonal vectors.

## The Attempt at a Solution

After analyzing it and trying to prove it to no avail, I don't even think it's a true statement.

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Mark44
Mentor

## Homework Statement

Show that

||b||a + ||a||b and ||b||a - ||a||b are orthogonal vectors.

## The Attempt at a Solution

After analyzing it and trying to prove it to no avail, I don't even think it's a true statement.
If two vectors are othogonal, their dot product will be zero. What do you get if you dot the two vectors in this problem?

Nevermind, careless error.

Though I was able to prove it, I do find it difficult to wrap my head around the fact that what makes it work is the scalar multiplication..

Mark44
Mentor
That's probably the most important use of the dot product - determining whether two vectors are orthogonal. Less important, IMO, is the ability to find the angle between two vectors.

I mean the scalar multiplied by vector multiplication. The dot product is not always 0 for (a+b) * (a-b). I find it hard to grasp that the magnitudes are what made them orthogonal.

Mark44
Mentor
I mean the scalar multiplied by vector multiplication. The dot product is not always 0 for (a+b) * (a-b).
True. If you form a parallelogram with a and b as two adjacent sides, then a + b will be a diagonal, and a - b will be the other diagonal. Most of the time these diagonals won't be perpendicular, so the dot product (a + b)$\cdot$(a - b) won't be zero.
I find it hard to grasp that the magnitudes are what made them orthogonal.
That's because you're essentially working with unit vectors, and the parallelogram is actually a square. In that case, the diagonals are perpendicular.

|b|a + |a|b = |a| * |b| * (a/|a| + b/|b|)

|b|a - |a|b = |a| * |b| * (a/|a| - b/|b|)

The vectors in the parentheses on the right sides are unit vectors, with a/|a| + b/|b| being one diagonal of a unit square, and a/|a| - b/|b| being the other diagonal.

(Here a/|a| means 1/|a| * a, of course. I'm also using just a single pair of | vertical bars to denote vector magnitude.)