Find perpendicular vectors in subspace of $V_{3}$

Question

Find all vectors of $V_{3}$ which are perpendicular to the vector $(7,0,-7)$ and belong to the subspace $L((0,-1,4), (6,-3,0)$.

As a note, this is an extra question of a long exercise, the vectors found above are replaced by results I found at the above questions.

Let $\overline{x}$ = {${ x_{1}, x_{2}, x_{3} } $} then since it's perpendicular to {${7,0, -7}$}

from there I get

$<\overline{x}, 2\overline{u_{2}} + \overline{u_{3}} > = 0$

$7x_{1} + 7x_{3} = 0$

$x_{1} = x_{3}$

So $\overline{x} = \left \{ \left. x_{1}, x_{2}, x_{1} \right \} \right.$ , $x_{i} \epsilon \mathbb{R}$, $i=1,2,3$

How do I proceed from here?

Answer 1

To belong to the subspace $L((0,−1,4),(6,−3,0)$ our vector must have the form as below (where $p,q\in R$) $$p(0,-1,4)+q(6,-3,0)=(6q,-p-3q,4p)$$

To be perpendicular to $(7,0,-7)$, we need to satisfy $$6q(7)+(-p-3q)(0)+(4p)(-7)=0 \Rightarrow 42q-28p=0\Rightarrow p=\frac{3}{2}q$$

We can choose an arbitrary $q$, and set $p=\frac{3}{2}q$, leading to our desired vector being of the form $$\frac{3q}{2}(0,-1,4)+q(6,-3,0)=(6q,-\frac{3q}{2},6q)$$

Answer 2

Well, you need to solve the system

$$\begin{pmatrix}x\\y\\x\end{pmatrix}=a\begin{pmatrix}\;0\\\!\!-1\\\;4\end{pmatrix}+b\begin{pmatrix}\;6\\\!\!-3\\\;0\end{pmatrix}=\begin{pmatrix}\;6b\\\!\!-a-3b\\\;4a\end{pmatrix}$$

So it must be

$$x=6b=4a\implies b=\frac23a\;,\;\;\text{and thus}\;\ldots$$