The given vectors are $v = (3,0,1)$ and $u = (3,1,0)$.
I have used the following formula and plugged in what I have been given using vector $v$ as the variables for $i,j,k$ and vector $u$ for $x_0,y_0,z_0$ and have worked out the solution as follows:
$$3(x-3) – 1(y-1) + 1(z-0) = 0$$
ultimately having $3x – y – 8 = 0$ as the equation of the plane.
Have I understood this correctly?
Any advice appreciated!
Best Answer
The plane perpendicular to $(a,b,c)$ and passing through $(x_0,y_0,z_0)$ is $$a(x-x_0)+b(y-y_0)+c(z-z_0)=0\ .$$ For your vectors, $$3(x-3)+0(y-1)+1(z-0)=0$$ which simplifies to $$3x+z-9=0\ .$$