Edit: the correct answer is $-4x +4y +z -31 =0$, but I don't know how I was so off.
Let $l$ be the line in $\mathbb{R}^3$ given by the Cartesian equation
$$x+8=\frac{y+1}3=\frac{3−z}8.$$
Let $P$ be the plane that contains $l$ and also contains the point $B(−4,7,−13)$
Find a Cartesian equation for P
I'm not sure where I went wrong with this question; I found the cartesian equation to be $4x +4y -z -25 =0$.
Working :
$x =t-8$,
$y =3t-1$,
$z =-8t +3$
$$A =(-8,-1,3)$$
$$AB = \begin{bmatrix}4\\8\\-16\end{bmatrix}$$
normal : $$n = AB \wedge \begin{bmatrix}-8\\-1\\3\end{bmatrix}$$
$$= \begin{bmatrix}16\\16\\-4\end{bmatrix}$$
Then I substituted the pt and normal vector into the point normal form
Best Answer
Let the sought equation of the plane be $\pi: ax+by+cz+d=0$ with normal vector $\overrightarrow{N}=(a,b,c)$.
Since we want the point $P(-4,7,-13)$ to belong to the plane, we make the point satisfy the plane equation: $$P\in \pi\iff-4a+7b-13c+d=0.\tag{1}$$
Notice that the given line equation can be written in standard form as: $$L:\quad \frac{x+8}{1}=\frac{y+1}{3}=\frac{z-3}{-8}$$ Hence, we identify the direction vector $\overrightarrow{S}=(1,3,-8)$ and an initial point $P_{0}(-8,-1,3)$.
Now, the line belongs to the plane if and only if the direction vector of the line and the normal vector of the plane are orthogonal, and furthermore, any point on the line satisfies the plane equation.
Thus, we find:
$$\overrightarrow{S}\cdot \overrightarrow{N}=0\iff a+3b-8c=0\tag{2}$$ and also, $$P_{0}\in\pi\iff-8a-b+3c+d=0\tag{3}$$ The system of equations $(1),(2)$, and $(3)$ has a solution: $$(a,b,c,d)=(a,-a,-a /4,31a/4),$$ with $a$ being a nonzero real number.
Hence, the sought plane is: $$\pi: ax+by+cz+d=0\iff ax-ay-\frac{a}{4}z+\frac{31a}{4}=0$$ Dividing everything by $a\neq 0$ and multiplying by $-4$, we get: $$\pi: -4x+4y+z-31=0$$ which coincides with the initial given answer.