The given vectors are $v = (3,0,1)$ and $u = (3,1,0)$.
I have used the following formula and plugged in what I have been given using vector $v$ as the variables for $i,j,k$ and vector $u$ for $x_0,y_0,z_0$ and have worked out the solution as follows: $$3(x-3) - 1(y-1) + 1(z-0) = 0$$ ultimately having $3x - y - 8 = 0$ as the equation of the plane.
Have I understood this correctly?
Any advice appreciated!
$\endgroup$ 03 Answers
$\begingroup$The plane perpendicular to $(a,b,c)$ and passing through $(x_0,y_0,z_0)$ is $$a(x-x_0)+b(y-y_0)+c(z-z_0)=0\ .$$ For your vectors, $$3(x-3)+0(y-1)+1(z-0)=0$$ which simplifies to $$3x+z-9=0\ .$$
$\endgroup$ 0 $\begingroup$Based on David's answer, the concept behind this formula is the following picture:
Since $\hat{n}$ is orthogonal to the plane, we have $$\begin{bmatrix}a\\b\\c \end{bmatrix}^T\begin{bmatrix}x-x_0\\y-y_0\\z-z_0 \end{bmatrix}=0,$$ then we obtain that formula.
Note that, if $(x_0,y_0,z_0)=(0,0,0)$, then this is a plane (a subspace of $\mathbb{R}^3$) passing through the origin; if not, this plane is an affine plane (not a subspace of $\mathbb{R}^3$ since the plane does not pass through the origin).
$\endgroup$ $\begingroup$$U=(3,1,0),\;V=(3,0,1)$. Let $P(x,y,z)$ be any point on the plane. Given the plane passes through $U=(3,1,0)$$P-U= (x-3,y-1,z-0)$. As $P-U$ is perpendicular to $V$,$$(P-U)\cdot V=0,$$ where the dot stands for the scalar product. $$(x-3,y-1,z)\cdot(3,0,1)=0\\ \Leftrightarrow 3(x-3)+0(y-1)+z=0\\ \Leftrightarrow 3x+z-9=0$$
$\endgroup$
