$ \vec{x} = \vec{p} + t\vec{d}$ where $ t \in \mathbb{R} $
$ \vec{d} = (d_1, d_2, d_3)$ is a direction vector
$ \vec{p} = (p_1, p_2, p_3)$ is a speical point on $L$
Suppose that:
$\delta_1 = a_1x + b_1y + c_1z = d_1, a_1, b_1, c_1, d_1 \in \mathbb{R}$
let a plane in $\mathbb{R}^3$ with a normal vector: $ \vec{n}_1 = (a_1, b_1, c_1), \vec{n}_2 = (a_2, b_2, c_2) \in \mathbb{R}^2$
$\delta_2 = a_2x + b_2y + c_2z = d_2$
What is the equation of $L$. Let $\vec{d}$, the direction vector to $L$ then $d_1 \perp n_1$ and $d_1 \perp n_2$
$ \vec{d} = n_1 \times n_2 $
Example: Let $L$ be the line of intersection between the planes:
$\delta_1: 3x - 2y + 4z = 11$
$\delta_2: 2x - 5y + 3z = 3$ \
$\rightarrow$ normal vectors of the equation are: $n_1 = (3, -2, 4), n_2 = (2, -5, 3)$
$d = n_1 \times n_2 = (14, -1, -11)$
Then then find a point that satisfies both equation like $ (7, 1, -2) $
$ \vec{x} (7, 1, -2) + t (14, -1, -11) $
Note that the equation above is not unique. Any scaler multiple of $(\vec{n} \times \vec{n})$ and any point that satisfies the equations will make a valid equation. i
Example: Find the equation of the plane that contains (1, 3, 0) and $ \vec{x} = (1, 3, 0) + t (2, -4, 1)
let $ \vec{u} = (2, -4, -1)$
make a new vector between the point and the start of of the vector:
$ \vec{v} = (3, 0, 7) - (1, 3, 0) = (2, -3, 7)
$\therefore$ the equation of the plane is:
$ \vec{x} = (3, 0, 7) + s(2, -4, 1) + t(2, -3, 7) $
or it can be written in general form, in which case we will have to find the normal first
$ \vec{n} = (2, -4, -1) \times (2, -3, 7) = (-31, -16, 2)$
$ \vec{n} \circ ( \vec{x} - \vec{p}) = 0 $
$ \Rightarrow (-31, -16, 2) \circ (x - 2, y, z - 7) = 0 $
$ \Rightarrow 31x + 16y - 2z = 79 $
Example: let $B = (x_0, y_0, z_0) \in \mathbb{R}_3$ and not contained in $\delta$. Find the distance $ d(B, \delta)$
Let $\vec{v} = Q \vec{B} = \vec{B} - Q = \vec (x_0 - x, y_0 - y, z_0 - z)$
Let $\vec{n} = (a, b, c)$ be a normal vector of $\delta$
Therefore $d(B, \delta) = || proj_\vec{n} \vec{v} || = || \frac{\vec{v} \circ \vec{n}}{|| \vec{n} ||^2} \vec{n} ||$
$ = \frac{|\vec{v} \circ \vec{n}|}{|| \vec{n} ||} = \frac{|a( x_0 + x) + b(y_0 + y) + c(z_0 + z) |}{\sqrt{a^2 + b^2 + c^2}} = \frac{| ax_0 + by_0 + cz_0 + d |}{\sqrt{a^2 + b^2 + c^2}}$