Answer to Question #113672 in Analytic Geometry for Rethabile

Question

Find the plane that passes through the point (2,5,3) and is perpendicular to the plane 2x-3y+4z+7=0

Answer 1

First we define the normal to the plane. In general for the plane defined by the equation A*x+B*y+C*z+D normal vector is n=(A,B,C). So in our case we have n=(2,-3,4).

Let's define equation for the plane that passes through the point (2,5,3) in general:

A₁*2+A₂*5+A₃*3 + A4 = 0.

For a plane passing through the origin and parallel to the plane above we have:

A₁*2+A₂*5+A₃*3 = 0. (1)

Normal vector for this plane is n1=(A₁,A₂,A₃).

For perpendicular planes, their normals are also perpendicular. Two nonzero vectors are perpendicular when their scalar product is zero. So we have 2*A₁-3*A₂+4*A₃ = 0. (2)

Considering (1) and (2) we obtain the following system of linear equations:

2*A₁+5*A₂+3*A₃=0

2*A₁-3*A₂+4*A₃=0

It can be seen that our underdetermined system of linear equations must have at least one solution because of the rank of its coefficient matrix is equal to the rank of its augmented matrix (Kronecker–Capelli theorem). Since in an underdetermined system this rank is necessarily less than the number of unknowns, there are indeed an infinitude of solutions.

Let's subtract the second equation from the first equation:

8*A₂-A₃=0 or A₃=8*A₂ (3)

Let's substitute A3 from (3) into our system of linear equations:

2*A₁+29*A₂=0

2*A₁+29*A₂=0

Assuming A₂ = 1 we get A₁=-29/2=-14.5.

Substituting A₂ =1 in (3) we have A₃ = 8.

Thus, the desired equation of the plane has the form:

-14.5*x+y+8*z+A₄=0, where A₄ is constant.

Any plane parallel to the specified will be perpendicular to the given plane.

Planes are parallel if their normal vectors are collinear. The normal vector is determined by the coefficients A₁, A₂, A₃. So for any A₄ we get plane perpendicular to the given plane.

For definiteness let's put A₄=0 (You can choose another constant number). So we have equation:

-14.5*x+y+8*z=0.