# Resolving a vector into a sum of two vectors

• Apr 6th 2011, 06:25 AM
cottontails
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.
• Apr 6th 2011, 07:03 AM
TheEmptySet
Quote:

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.

Attachment 21375
http://www.mathhelpforum.com/math-he...5&d=1302097383

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

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

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

This gives

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

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 $\mathbf{v}$ with length $1$.

So lets normalize $\mathbf{v}$

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

Now lets combine these two to get

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

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

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

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

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