# Form the equation of a plane, given 2 points and a parallel straight line

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.

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: $\vec u=\vec r \times \vec v$

You are on the right track.

Note that those divisors will give you a vector parallel to the line. Once, you've computed a vector normal to the plane, you take a general point on the plane, say $P \equiv (x,y,z)$. This point lies on the plane. Note that the vector parallel to $PB$ will be perpendicular to the vector normal to the plane (Can you visualize it?)

Therefore, $\vec{PB}.\vec{n} = 0 \Rightarrow (x-3,y-4,z+2).\vec{n} = 0$, where $\vec{n}$ is the normal vector.

This will give you equation of the plane.

• Yes, you're right. But that's not the problem I'm having. I know how to form the plane equation from the point where I am using the same formula you gave me. The question was that why the normal vector is the cross product of the 2 vectors, about which I'm also writing an answer now because a little bit of googling actually helped me. :) Nov 23, 2016 at 10:22
• I'm sorry. That was not clear to me by your question. Nov 23, 2016 at 10:24
• To see if $a \times b$ gives you a vector normal to $a$ and $b$, try computing $a.(a \times b)$ and $b.(a \times b)$ using the definition of $a \times b$, both will come out to be zero, implying that $a \times b$ is perpendicular to $a$ and $b$. Nov 23, 2016 at 10:46

A little bit of googling about cross product could've help me before making this question.

Wikipedia: Given two linearly independent vectors a and b, the cross product, a × b, is a vector that is perpendicular to both a and b and therefore normal to the plane containing them.

As I got the vector from the line and the vector $BC$, the cross product will be perpendicular with both of them, means it will be the normal vector of the plane as well.