# I'm having trouble with these types of questions. I have the following

I'm having trouble with these types of questions. I have the following vector $$\displaystyle{u}={\left({4},{7},-{9}\right)}$$ and it wants me to find 2 vectors that are perpendicular to this one.
I know that $$\displaystyle{\left({4},{7},-{9}\right)}\cdot{\left({x},{y},{z}\right)}={0}$$

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Opeance1951
Take vector $$\displaystyle{n}={\left({x},{y},{z}\right)}$$. As you said by yourself, dot product should vanish. So
$$\displaystyle{\left({4},{7},-{9}\right)}\cdot{\left({x},{y},{z}\right)}={4}{x}+{7}{y}-{9}{z}={0}$$
As you might see, all points that lie on the plane $$\displaystyle{4}{x}+{7}{y}-{9}{z}={0}$$ satisfy the condition of perpendicularity. If you want two linear independent vectors, just pick two different points. So, pick any two different triples $$\displaystyle{n}_{{1}}={\left({x}_{{1}},{y}_{{1}},\frac{{{4}{x}_{{1}}+{7}{y}_{{1}}}}{{9}}\right)}$$ and $$\displaystyle{n}_{{2}}={\left({x}_{{2}},{y}_{{2}},\frac{{{4}{x}_{{2}}+{7}{y}_{{2}}}}{{9}}\right)}$$ where $$\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}\ne{\left({x}_{{2}},{y}_{{2}}\right)}$$, and you're done.