Vector form of a plane

Question

Express the following plane in vector form:

$\mathcal P_1\subseteq \Bbb R^3$ with equation $4x-z=0$.

The answer is $t(1,0,4)+s(0,1,0)$. I don't understand how they got $(0,1,0)$ for the direction vector.

Answer 1

The cartesian equation is $z=4x$

The parametrization is not unique. Anyway You need two linearly independent points on the plane.

One point can be $A=(1,0,4)$. The second one cannot be $(2,0,8)$ or other points you can get from $A$ multiplying by a constant. Otherwise you get a line not a plane.

The second point cannot obviously be $(0,0,0$.

So we look for a point where the only nonzero coordinate is $y$, like $(0,1,0)$ or $(0,10,0)$

Finally the parametric equation of the plane is $$(1,0,4)t+(0,1,0)s=0$$

Note that $(2,0,8)t+(0,10,0)s=0$ would be right.

Hope this helps

Answer 2

Let $\pi\equiv 4x-z=0$ be your plane. We notice that the point $(0,0,0)\in\pi$ and the orthogonal vector to the plane is $\underline u=(4,0,-1)$.
The vector space $\text{dir}(\pi)$ has dimension $2$ and the vectors that span this space are orthogonal to the vector $\underline u$.
We can take for example the vectors $\underline v=(1,0,4)$ and $\underline w=(0,1,0)$, infact we have $$\langle\underline u,\underline v\rangle=1\cdot4+0+4\cdot(-1)=0\implies\text{ they are orthogonal}$$ $$\langle\underline u,\underline w\rangle=0\cdot4+1\cdot0+0\cdot(-1)=0\implies\text{ they are orthogonal}$$ Finally the equation of the plane $\pi$ is$$\begin{cases}x=0+\alpha\\y=0+\beta\\z=0+4\alpha \end{cases},\alpha,\beta\in\mathbb R.$$