# Thread: Resolving a vector into a sum of two vectors

1. ## Resolving a vector into a sum of two vectors

"Resolve the vector u = 5i + j + 6k into a sum of two vectors, one of which is parallel and the other perpendicular to v = 3i - 6j + 2k."

What I want to know with this question is how you are able to know how to find the sum of each vector. I know that you find the vector projection of u in the direction of v and the vector component u orthogonal to v for the sum of two vectors. I know how to solve and find the answers from this so, I just basically want to know why you have to use these two in particular to get the answer.

2. Originally Posted by cottontails
"Resolve the vector u = 5i + j + 6k into a sum of two vectors, one of which is parallel and the other perpendicular to v = 3i - 6j + 2k."

What I want to know with this question is how you are able to know how to find the sum of each vector. I know that you find the vector projection of u in the direction of v and the vector component u orthogonal to v for the sum of two vectors. I know how to solve and find the answers from this so, I just basically want to know why you have to use these two in particular to get the answer.
Maybe the picture will well. I can only draw in 2D.

First notice the right triangle in red. We need to find the adjacent and opposite sides of it.

Using trigonometry we know that the adjacent side of a triangle is

$\displaystyle \displaystyle \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenu se}}=\frac{A}{H} \iff A=H\cos(\theta)$

The hypotenuse is the magnitude of the vector
$\displaystyle \mathbf{u}$

This gives

$\displaystyle A=||\mathbf{u}||\cos(\theta)$

The object above is a scalar but is has the magnitude of the adjacent, now we just need to point it in the right direction without changing its length.

So we need a vector pointing in the direction of $\displaystyle \mathbf{v}$ with length $\displaystyle 1$.

So lets normalize $\displaystyle \mathbf{v}$

Let $\displaystyle \displaystyle \mathbf{n}=\frac{\mathbf{v}}{||\mathbf{v}||}$

Now lets combine these two to get

$\displaystyle \displaystyle A\mathbf{n}=(||\mathbf{u}||\cos(\theta))\left(\fra c{\mathbf{v}}{||\mathbf{v}||} \right)=\frac{||\mathbf{u}||||\mathbf{v}||\cos(\th eta)}{||\mathbf{v}||^2}\mathbf{v}=\frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{v}||^2}\mathbf{v}$

The last step comes from the geometric definition of the dot product.

To get the perpendicular $\displaystyle \mathbf{p}$ side

We use the geometric fact that vectors is Euclidean space add via the parallelogram rule. This gives

$\displaystyle A\mathbf{n}+\mathbf{p}=\mathbf{u}$ now we just solve for $\displaystyle \mathbf{p}, \mathbf{p}=A\mathbf{n}-\mathbf{u}$

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# vector as the sum of two vector

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