This is my first time working on 3 dimensional vectors and it went fine until I had to do the following:
Given is the plane $V$ described by the equation $$-56x +7y+11z=-1.$$
And I have to find a vector equation for plane $V.$
I know that you will need two direction vectors and a normal vector. But I don't really know where to start. Could someone give me a push in the right direction?
My calculation by working with three points:
My calculation
Would this be correct?
Best Answer
“Vector equation for plane $V$” is ambiguous, but from other parts of your question I gather that you want to find a parameterization of the form $\vec p+s\vec u+t\vec v$. For this you need two linearly independent vectors $\vec u$ and $\vec v$ that are parallel to $V$ and a point $\vec p$ on the plane.
You can rewrite the equation that you have in the form $\vec n\cdot\vec x=d$, specifically, $(-56,7,11)\cdot(x,y,z)=-1$. The fixed vector $\vec n=(-56,7,11)$ is perpendicular to the plane—normal to it. Now, for any nonzero vector $(a,b,c)$, the vectors $(0,-c,b)$, $(c,0,-a)$ and $(-b,a,0)$ are all orthogonal to it, and at least two of them are nonzero and linearly independent (verify this!). So, you can find suitable vectors $\vec u$ and $\vec v$ from $\vec n$ with basically no work. The only other thing you need is a point on the plane, and you can find one easily by setting two of the variables in the equation to zero and solving for the remaining one.