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I'm given that the plane $W$ in $\mathbb R^3$ can be written as

$$W: \mathbf{x} = (1, 0, 1) + s(1, 3, -1) + t(2, 2, 1)$$

where $s$ and $t$ are real numbers.

My task is to write $W$ as a general equation.

I can't seem to figure out how to do this. I've tried to find similar threads and working out a normal of two vectors in the plane, using the cross product. Can't get it to work. When I plot certain points for random values of s and t, the points don't actually lie on the plane.

Any help would be appreciated...

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    $\begingroup$ Can you explicitly show in your question, what you've tried so that we can see where you went wrong? $\endgroup$
    – Quadrics
    Commented Nov 25, 2023 at 9:33
  • $\begingroup$ The plane $W$ is in $\mathbb R^3$. $\endgroup$
    – gpassante
    Commented Nov 25, 2023 at 9:37
  • $\begingroup$ If $(x, y, z)$ is on the plane, what are x, y and z in terms of s and t? Can you now write s and t in terms of x and y? Put those values into your expression for z and you have the plane in scalar form. $\endgroup$
    – Paul
    Commented Nov 25, 2023 at 11:24

1 Answer 1

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First: Since $(1,3,-1)$ and $(2,2,1)$ are vectors parallel to the plane. We can use their cross product to find the components of the vector $\vec{n}=(a,b,c)$, which is perpendicular to the plane as follows. $$(a,b,c)=(1,3,-1)\times(2,2,1)=\text{?}$$

Having obtained $\vec{n}=(a,b,c)$, we can write the general form of the equation of a plane $W$ in $\mathbb{R}^3$ as $$W: ax+by+cz+d=0. \ \ \ \ \ \ (1)$$, where $a$, $b$, and $c$ are the components of $\vec{n}$ that you have found.

Second: Since $(1,0,1)$ is a point in $W$, it follows from Eq. (1) that $$a(1)+b(0)+c(1)+d=0\Rightarrow d=-a-c.$$ By substituting $a$ and $c$ that you have found before, we obtain $d$. Finally, substituting $a,b,c$, and $d$ into Eq. (1) you get the solution.

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    $\begingroup$ Though if you have 4 component vectors and 3 parameters there is no cross product to find the scalar form of the hyperplane. $\endgroup$
    – Paul
    Commented Nov 25, 2023 at 17:07
  • $\begingroup$ @Paul You are correct. However, the cross product is not necessary. You can find the vector $\vec{n}=(a,b,c)$, which is perpendicular to the plane, using the dot product as follows. Since $(1,3,−1)$ and $(2,2,1)$ are vectors parallel to the plane, then $(a,b,c)\cdot(1,3,-1)=0$ and $(a,b,c)\cdot(2,2,1)=0$. From this, we get the following system of equations: $$\begin{cases} a+3b-c=0&\\ 2a+2b+c=0& \end{cases}$$ The solution is $a=-\frac{5}{3}b$ and $c=\frac{4}{3}b$, with $b\in\mathbb{R}$. Pick $b=-3$, it follows that $a=5$ and $c=-4$. Therefore, $\vec{n}=(5,-3,-4)$. $\endgroup$
    – QuantumSuperfield
    Commented Nov 25, 2023 at 23:51

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