Are ||b||a + ||a||b and ||b||a - ||a||b always orthogonal?

[1MileCrash]

Homework Statement

Show that

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

Homework Equations

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.

 

[Mark44]

Homework Statement

Show that

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

Homework Equations

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?

 

[1MileCrash]

Nevermind, careless error.

 

[1MileCrash]

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]

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.

 

[1MileCrash]

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]

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.)