Consider the point $B(1,0,1)$ and the line $R=(1,1,1) + \alpha[1,1,-1]$. Find the equation of a plane that passes through $B$ and contains $R$.
What I tried doing was simply setting two different values for $\alpha$, since if the points are part of the line, then they are also part of the plane.
Doing that allowed me to find the points $P(1,1,1)$ when $\alpha=0$ and $Q(0,-1,0)$ when $\alpha=-1$.
With that, I found the vectors $\vec{PQ}=[-1,-1,1]$ and $\vec{PB}=[0,-1,0]$. Finding the cross product of these two vectors would then give me a vector normal to both, which I could use to find the cartesian equation of the plane.
Then the vector $n=[1,0,1]$ is the result of the cross product of $\vec{PQ}$ and $\vec{PB}$.
Now, after finding the $d$ term in $ax+by+cz=d$, I got the plane equation $x+z=2$.
However, I noticed that $\vec{PQ}$ and $\vec{PB}$ aren't linearly independent. Does this affect my result? If so, how do I choose the right points from the line? Is there a different, more effective way of solving this?
To construct a (general) equation of a plane you will typically need a normal vector as well as a specific point on the plane. You can obtain a normal vector by taking the cross product of two non-parallel (linearly independent) direction vectors that lie on the plane. Since the line R lies on the plane, you can simply use its direction vector $[1,1,-1]$ as one of the direction vectors that lie on the plane. To obtain another direction vector that lies on the plane, you can consider the vector $\vec{BR} = [0, 1, 0]$ which happens to be linearly independent to $[1,1,-1]$. Then cross these two vectors to obtain a normal vector. After that you can easily obtain the general equation of the plane.