Find all unit vectors in the plane determined by vectors u and v that are perpendicular to the vector w. Find all unit vectors in the plane determined by vectors $u=(0,1,1)$ and $ v=(2,-1,3)$ that are perpendicular to the vector $w=(5,7,-4)$.
This is the question. I found the plane that determined by $u$ and $v$, its equation is $4x+2y-2z=0$ I think. What should I do next, how can I find a relation between my plane and w?
 A: The vector must be in the plane determined by $\bf u$ and $\bf v$. You've already got that, check. It also must be in the plane orthogonal to $\bf w$, which is $5x+7y-4z = 0$ (why?). It also must have length $1$. You can make that "length squared $1$" (why?). If $(x,y,z)$ is the vector you're looking for, you can go for: $$\begin{cases} 2x+y-z = 0 \\ 5x+7y-4z = 0 \\ x^2+y^2+ z^2 = 1\end{cases}.$$
It is possible that there is more than one vector that satisfies these relations.
A: There are many ways to solve this problem: I will illustrate a couple.


*

*Vectorial approach
Take a linear combination of $\mathbf u$ and $\mathbf v$
and impose that it be normal to $\mathbf w$
$$
\left( {\lambda \,\mathbf{u} + \mu \,\mathbf{v}} \right) \cdot \mathbf{w} = 0
$$
i.e.:
$$
\lambda \,\mathbf{u} \cdot \mathbf{w} + \mu \,\mathbf{v} \cdot \mathbf{w} = 0
$$
note that the dot products are scalars, so the above is a simple equation in $\lambda$ and $\mu$,
that can be solved expressing one in terms of the other , or both in terms of a third parameter.
So the required vectors $\mathbf x$ will be those expressed as a combination of $\mathbf u$ and $\mathbf v$ depending on one parameter.
$$
\mathbf{x} = \lambda \,\mathbf{u} + \mu \left( \lambda  \right)\,\mathbf{v} = \lambda \left( t \right)\,\mathbf{u} + \mu \left( t \right)\,\mathbf{v}
$$
In your case, we have
$$
\lambda \,\mathbf{u} \cdot \mathbf{w} + \mu \,\mathbf{v} \cdot \mathbf{w} = \lambda \,3 + \mu \,\left( { - 9} \right) = 0\quad  \Rightarrow \quad \lambda  = 3\mu 
$$
and thus
$$
\mathbf{x} = 3\,\mu \,\mathbf{u} + \mu \,\mathbf{v} = \mu \left( {3\,\,\mathbf{u} + \,\mathbf{v}} \right) = \mu \left( {2,2,6} \right) = \nu \left( {1,1,3} \right)
$$


*

*Geometric approach
As you already found $\mathbf x =(x,y,z)$ shall lie in the $\mathbf {u,v}$ plane (we can choose that passing through the origin)
$$
\left| {\,\begin{array}{*{20}c}
   x & y & z  \\
   {u_{\,x} } & {u_{\,y} } & {u_{\,z} }  \\
   {v_{\,x} } & {v_{\,y} } & {v_{\,z} }  \\
 \end{array} \,} \right| = 4x + 2y - 2z = 2x + y - z = 0
$$
and it shall be normal to $\mathbf w$
$$
\mathbf{x} \cdot \mathbf{w} = 5x + 7y - 4z = 0
$$
i.e., it shall be on the line given by
$$
\left\{ \begin{gathered}
  2x + y - z = 0 \hfill \\
  5x + 7y - 4z = 0 \hfill \\ 
\end{gathered}  \right.
$$
So find the parametric equation of that line, with a method you know, for example solving in terms of $z$ as parameter
$$
\left\{ \begin{gathered}
  x = z/3 \hfill \\
  y = z/3 \hfill \\ 
\end{gathered}  \right.\quad  \Rightarrow \quad \mathbf{x} = \left( {z/3,z/3,z} \right) = z/3\left( {1,1,3} \right)
$$
