Freddy Friedman

2022-07-18

Express the following plane in vector form:
${\mathcal{P}}_{1}\subseteq {\mathbb{R}}^{3}$ with equation $4x-z=0$
The answer is $t\left(1,0,4\right)+s\left(0,1,0\right)$. I don't understand how they got (0,1,0) for the direction vector.

Brienueentismvh

Expert

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
$\left(1,0,4\right)t+\left(0,1,0\right)s=0$
Note that $\left(2,0,8\right)t+\left(0,10,0\right)s=0$ would be right.
Hope this helps

Lexi Mcneil

Expert

Let $\pi \equiv 4x-z=0$ be your plane. We notice that the point $\left(0,0,0\right)\in \pi$ and the orthogonal vector to the plane is $\underset{_}{u}=\left(4,0,-1\right)$
The vector space $\text{dir}\left(\pi \right)$ has dimension 2 and the vectors that span this space are orthogonal to the vector $\underset{_}{u}$
We can take for example the vectors $\underset{_}{v}=\left(1,0,4\right)$ and $\underset{_}{w}=\left(0,1,0\right)$, infact we have

Finally the equation of the plane $\pi$ is
$\left\{\begin{array}{l}x=0+\alpha \\ y=0+\beta \\ z=0+4\alpha \end{array},\alpha ,\beta \in \mathbb{R}.$

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