Find the equation of the plane in $\mathbb R^3$ which passes through the point $(−10, 5, 4)$ and which is perpendicular to the line joining the points $(4, −1, 2)$ and $(−3, 2, 3)$.
I was trying this question many times. But I could not get the solution. I was trying to make two plane equations $4x-y+ 2z =0$ and $-3x + 2y+ 3z =0$ but I don't know what is other step to find the equation.
If anbody help me I would be very thankful to him.
Thank you.
If it is perpendicular to the line joining the points that you mentioned, then it is perpendicular to the vector $(7,-3,-1)$. So, its equation is $7x-3y-z=k$, for some $k$. Since you want it to pass through $(-10,5,4)$, what must be the value of $k$?