I am given 2 points:
$B(3; 4; -2)$ and $C(1; 5; 3)$.
I am also given a straight line which is parallel to the plane:
$$
\frac{x+6}{-3}=\frac{y-7}{7}=z
$$
The task is to form an equation of a plane with the given information.
I know how to solve it, the problem is that I don't know how to describe it.
So what I did, was write out that the normal vector of the plane is perpendicular to the vector BC, and is also perpendicular to the vector $(-3, 7, 1)$, which I got from the divisors of the straight line. Don't know how they're called in English. From that point forward, I found the cross product of vectors BC and s (s being the divisors of the straight line), which gave me the normal vector and allowed me to form the plane equation.
Could someone describe, why it is done like this so I'd understand it more clearly for future assignments as well. Thank you.
Best Answer
It seems that your problem is to understand where comes from the vector $(-3,7,1)^T$.
If it is so, note that the equation of the line is: $$ \frac{x+6}{-3}=\frac{y-7}{7}=z=t $$ write this equation as: $$ \begin{pmatrix} \frac{x+6}{-3}\\ \frac{y-7}{7}\\ z \end{pmatrix} =t\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix} $$ that is $$ \begin{pmatrix} x+6\\ y-7\\ z \end{pmatrix} =t\begin{pmatrix} -3\\ 7\\ 1 \end{pmatrix} $$ so the equation of the line is: $$ \begin{pmatrix} x\\ y\\ z \end{pmatrix} =t\begin{pmatrix} -3\\ 7\\ 1 \end{pmatrix} +\begin{pmatrix} -6\\ 7\\ 0 \end{pmatrix} $$
and $\vec r=(-3,7,1)^T$ is the vector that gives the orientation of the line.
And, as you have noted, the orthogonal vector to the plane $\vec u$ must be orthogonal to this vector and to the vector $BC=(2,-1,-5)^T=\vec v$, so we can find it as the cross product: $\vec u=\vec r \times \vec v $