Hello Drdumbom Originally Posted by

**Drdumbom** Hi

I really need some assistance with the following problem:

"Find all unit vectors in the plane determined by **u** = (3,0,1) and **v** = (1,-1,1) that are perpendicular to the vector **w** = (1,2,0)".

The problem for me is that I donīt really know how to approach planes in general. For example, how do I calculate what plane is determined by the two vectors **u** and **v**?

What would **u** and **v** look like if they were serparate planes?

If someone know a website that explains the basics of vector planes I would appreciate the adress.

Thanks in advance

It may help if you think first of the unit vectors $\displaystyle \textbf{i}$ and $\displaystyle \textbf{j}$ that are used to define the $\displaystyle x-y$ plane. Any vector in this plane can be written $\displaystyle p\textbf{i} +q\textbf{j}$, representing as it does, displacements $\displaystyle p$ and $\displaystyle q$ parallel to $\displaystyle Ox$ and $\displaystyle Oy$ respectively. ($\displaystyle p$ and $\displaystyle q$ are, of course, any scalar numbers.)

In the same way, any vector of the form $\displaystyle p\textbf{u}+q\textbf{v}$ will lie in the plane defined by $\displaystyle \textbf{u}$ and $\displaystyle \textbf{v}$. We may think of the grid of unit squares defined by $\displaystyle \textbf{i}$ and $\displaystyle \textbf{j}$ as having been replaced by a grid of parallelograms defined by $\displaystyle \textbf{u} $ and $\displaystyle \textbf{v}$.

So if $\displaystyle \textbf{u} = \begin{pmatrix}3\\0\\1\end{pmatrix}$ and $\displaystyle \textbf{v} = \begin{pmatrix}1\\-1\\1\end{pmatrix}$, any vector lying in the plane determined by $\displaystyle \textbf{u}$ and $\displaystyle \textbf{v}$ will be of the form $\displaystyle p\begin{pmatrix}3\\0\\1\end{pmatrix}+q\begin{pmatr ix}1\\-1\\1\end{pmatrix}=\begin{pmatrix}3p+q\\-q\\p+q\end{pmatrix}$.

Now two vectors are perpendicular if their dot (scalar) product is zero. So this vector is perpendicular to $\displaystyle \textbf{w}=\begin{pmatrix}1\\2\\0\end{pmatrix}$, if:$\displaystyle \begin{pmatrix}3p+q\\-q\\p+q\end{pmatrix}\cdot\begin{pmatrix}1\\2\\0\end {pmatrix}=0$

$\displaystyle \Rightarrow (3p+q).(1) +(-q).(2)+(p+q).(0) = 0$

$\displaystyle \Rightarrow 3p-q=0$

$\displaystyle \Rightarrow q = 3p$

So the vector will be of the form $\displaystyle \begin{pmatrix}3p+3p\\-3p\\p+3p\end{pmatrix}=\begin{pmatrix}6p\\-3p\\4p\end{pmatrix}=p\begin{pmatrix}6\\-3\\4\end{pmatrix}$

All that remains is to find unit vectors in this form. So that will be where$\displaystyle p^2(6^2+(-3)^2+4^2) = 1$

$\displaystyle \Rightarrow 61p^2=1$

$\displaystyle \Rightarrow p = \pm\frac{1}{\sqrt{61}}$

So I reckon there's our answer: the unit vectors are $\displaystyle \pm\frac{1}{\sqrt{61}}\begin{pmatrix}6\\-3\\4\end{pmatrix}$

Grandad