Sum of a vector parallel and orthogonal to.

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The discussion focuses on expressing vector u as a sum of a vector parallel to v and a vector orthogonal to v. The projection of u onto v is calculated, resulting in [1/2](3i - j), which is then added to u to express it in the desired form. Clarifications are made regarding the notation and the relationship between the vectors, emphasizing that the orthogonal component w can be derived from u - Pu. The dot product of the orthogonal vector with v confirms its orthogonality. The conversation concludes with insights on the properties of linear operators and projections in vector spaces.
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Homework Statement


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.

Homework 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

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.
 
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Differentiate1 said:

Homework Statement


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.

Homework 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

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.
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).
 
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.
 
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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