# Sum of a vector parallel and orthogonal to.

1. Sep 23, 2014

### Differentiate1

1. The problem statement, all variables and given/known data
v = 3i - j u = 2i + j - 3k
Express vector u as a sum of a vector parallel to v and a vector orthogonal to v.

2. Relevant equations
Proj of u onto v = [ (u • v) / |v|^2 ]v
Expressing vector u as sum of a vector parappel to v and a vector vector orthogonal to v
>> u = [Proj of u onto v] + u

3. The attempt at a solution
I found the projection of vector u onto v which is [1/2](3i - j) then added it to vector u = 2i + j - 3k to get
u = [7/2] i + [1/2] j - 3 k <<==>> Vector u expressed as the sum of a vector parallel to v and orthogonal to v.

I would like to receive assistance to assure that what I've done works.

2. Sep 23, 2014

### Fredrik

Staff Emeritus
I will use the notation Pu for the projection of u onto the 1-dimensional subspace spanned by v. You're saying that u=Pu+u. But that would mean that Pu=0 (since you can cancel u from both sides).

If u=Pu+w, then what is w? (I think you will find that the answer is automatically orthogonal to v).

3. Sep 23, 2014

### Differentiate1

Thanks for the reply. I believe I've found it and the check I used enhances the answer I got. Instead of just Pu + w, I did, w - Pu to get an orthogonal vector to v. The dot product of (w - Pu) with vector v came out with a scalar of 0, which proves it's orthogonal.

4. Sep 24, 2014

### Fredrik

Staff Emeritus
Right, if you solve u=Pu+w, you get w=u-Pu, and this vector is orthogonal to v.

If P is any linear operator, you can write u=Pu+(1-P)u. If P is the orthogonal projection onto some subspace M, then Pu will be an element of M, and (1-P)u will be orthogonal to every element of M.

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