# Math Help - Perpendicular Vectors

1. ## Perpendicular Vectors

Problem:

Consider the vector

$\vec{v}=\begin{bmatrix}1\\ 2\\ 3\\ 4\end{bmatrix}$ in $\mathbb{R}^{4}$

Find a basis of the subspace of R^4 consisting of all vectors perpendicular to $\vec{v}$.

I'm not very sure about how to solve this problem. I assume that I need to find a set of vectors $\vec{u_{1}},...,\vec{u_{m}}$ for which the case $\vec{v} \cdot \vec{u_{i}} = 0$ holds.

I tried using the formula,

$\vec{v}^\perp=\vec{v}-\vec{v}^\parallel$

Since I already know $\vec{v}$, I would just need to figure out $\vec{v}^\parallel$. To do this, I used projection.

$proj_x(\vec{v}) = \sum_{i=1}^{m}(\vec{u}_{i} \cdot \vec{v})\vec{u}_{i}$

This is where I get stuck, because I'm LOOKING for the u vectors... correct?

Am I far off the path I should be on here or am I on the right track?

Any help is greatly appreciated. Thanks for reading.

2. $\begin{bmatrix}-2\\ 1\\ 0\\ 0\end{bmatrix}$ is one member of the basis. Can you find the rest?

3. I know, I saw that in the back of the book, but I can't figure out where that's coming from. Perhaps a hint to push me in the right direction would be good.

4. Originally Posted by tangibleLime
I know, I saw that in the back of the book, but I can't figure out where that's coming from. Perhaps a hint to push me in the right direction would be good.
Set two components of the basis vector to zero and a third component to 1. Solve for the fourth one.

This way, you should be able to find two more members of the basis.

5. here is one way the vector (-2,1,0,0) might have been found:

we want the dot product (1,2,3,4).(x,y,z,w) to be 0. for our first vector, we can choose any x,y,z and w that will work. so we can choose z = w = 0,

and find x,y so that x + 2y = 0. if y = 1, x will have to be -2.

now, if you choose x = y = 0, clearly whatever vector you wind up with, will be linearly independent of (-2,1,0,0) (unless the only vector that works is (0,0,0,0)).

that will give you 2 elements of the basis, a similar approach will yield the 3rd.

6. Thanks! I took an arbitrary column vector, dotted it with the given vector and solved for each variable which gave me the correct answer.