Vector problem involving orthogonal

• Mar 5th 2011, 05:41 PM
cb220
Vector problem involving orthogonal
Hi, I'm struggling with this problem:

For $\displaystyle \vec{u} =$[-4, 1, 10]^T and $\displaystyle \vec{v} =$ [−12, −6, 8]^T find the vectors $\displaystyle \vec{u1}$ and $\displaystyle \vec{u2}$ such that:

(i) $\displaystyle \vec{u1}$ is parallel to $\displaystyle \vec{v}$
(ii) $\displaystyle \vec{u2}$ is orthogonal to $\displaystyle \vec{v}$
(iii) $\displaystyle \vec{u} = \vec{u1} + \vec{u2}$

I figured I should firstly try and find u2 by (ii), and then after I found that I would be able to use (iii) to get u1. This approach didn't work out too well for me, heh. Basically I tried to set it up with dot product, and solve for u2:

$\displaystyle \vec{v} * \vec{u2} = 0$

Didn't work out, just ended up with something like: -12a - 6b + 8c = 0
Where a, b, c are the numbers in u2.

From there I couldn't see what more I could do to find a, b, c. Basically, I don't really know how to approach this problem :(.

Anyone mind helping out a math newbie? Thanks in advance!
• Mar 5th 2011, 07:13 PM
TheEmptySet
Quote:

Originally Posted by cb220
Hi, I'm struggling with this problem:

For $\displaystyle \vec{u} =$[-4, 1, 10]^T and $\displaystyle \vec{v} =$ [−12, −6, 8]^T find the vectors $\displaystyle \vec{u1}$ and $\displaystyle \vec{u2}$ such that:

(i) $\displaystyle \vec{u1}$ is parallel to $\displaystyle \vec{v}$
(ii) $\displaystyle \vec{u2}$ is orthogonal to $\displaystyle \vec{v}$
(iii) $\displaystyle \vec{u} = \vec{u1} + \vec{u2}$

I figured I should firstly try and find u2 by (ii), and then after I found that I would be able to use (iii) to get u1. This approach didn't work out too well for me, heh. Basically I tried to set it up with dot product, and solve for u2:

$\displaystyle \vec{v} * \vec{u2} = 0$

Didn't work out, just ended up with something like: -12a - 6b + 8c = 0
Where a, b, c are the numbers in u2.

From there I couldn't see what more I could do to find a, b, c. Basically, I don't really know how to approach this problem :(.

Anyone mind helping out a math newbie? Thanks in advance!

Let

$\displaystyle \vec{u}_1=\text{Proj}_{\vec{v}}\vec{u}=\frac{\vec{ u}\cdot \vec{v}}{||v||^2}\vec{v}$

Then

$\displaystyle \vec{u}_2=\vec{u}-\vec{u}_1$
• Mar 6th 2011, 04:37 AM
HallsofIvy
Quote:

Originally Posted by cb220
Hi, I'm struggling with this problem:

For $\displaystyle \vec{u} =$[-4, 1, 10]^T and $\displaystyle \vec{v} =$ [−12, −6, 8]^T find the vectors $\displaystyle \vec{u1}$ and $\displaystyle \vec{u2}$ such that:

(i) $\displaystyle \vec{u1}$ is parallel to $\displaystyle \vec{v}$
(ii) $\displaystyle \vec{u2}$ is orthogonal to $\displaystyle \vec{v}$
(iii) $\displaystyle \vec{u} = \vec{u1} + \vec{u2}$

I figured I should firstly try and find u2 by (ii), and then after I found that I would be able to use (iii) to get u1. This approach didn't work out too well for me, heh. Basically I tried to set it up with dot product, and solve for u2:

$\displaystyle \vec{v} * \vec{u2} = 0$

Didn't work out, just ended up with something like: -12a - 6b + 8c = 0
Where a, b, c are the numbers in u2.

From there I couldn't see what more I could do to find a, b, c. Basically, I don't really know how to approach this problem :(.

Anyone mind helping out a math newbie? Thanks in advance!

That's not a bad start! Yes, to satisfy (ii) you want -12a- 6b+ 8c= 0.
And to satisfy (iii) you want u1= <-4- a, 1- b, 10- c>.
And, then, to satisfy (i) you want u1 to be a multiple of v: u1= <-4-a, 1- b, 10- c>= d<12, 6, 8>. That is you have four equations, -12a- 6b+ 8c= 0, -4-a= 12d, 1- b= 6d, and 10- c= 8d, for the four numbers, a, b, c, and d.