Science:Math Exam Resources/Courses/MATH105/April 2015/Question 01 (a)/Solution 1

For a plane parallel to $x-3y+2z=1$, we can take its normal vector to be $\langle 1,-3,2\rangle$ (two parallel planes have normal vectors that are scalar multiples of each other, so without loss of generality we may assume their normal vectors are identical).

Thus, the plane we seek has equation $1(x-x_{0})+(-3)(y-y_{0})+2(z-z_{0})=0,$ where $(x_{0},y_{0},z_{0})$ is any point on the plane. We are told that the plane passes through $(1,0,-1)$ so we take $x_{0}=1$, $y_{0}=0$, and $z_{0}=-1$ to obtain:

$(x-1)-3(y-0)+2(z-(-1))=0$, or equivalently, $\color {blue}x-3y+2z=-1$.