Question

Determine the equation of the plane in the form

z = f(x,y) that is parallel to the vectors <9,6,2> and <−8,−4,−5,>,
and passes through the point (5,−5,8)

Answer 1

To determine the equation of a plane parallel to given vectors and passing through a point, we need to find the normal vector, take the cross product of the given vectors, use the point-normal form of the equation of a plane, and substitute the values to find the equation.

Step-by-step explanation:

To determine the equation of a plane, we need to find the normal vector to the plane. Since the given plane is parallel to the vectors <9,6,2> and <-8,-4,-5>, the normal vector will be perpendicular to both of these vectors. We can find the normal vector by taking the cross product of the two given vectors.

So, the cross product of <9,6,2> and <-8,-4,-5> is:

<6,-13,-60>

Since the plane is parallel to the normal vector <6,-13,-60> and passes through the point (5,-5,8), we can use the point-normal form of the equation of a plane to find the equation. The point-normal form is given by:

ax + by + cz = d

Substituting the values of the normal vector and the point, we get:

6(x-5) - 13(y+5) - 60(z-8) = 0

Therefore, the equation of the plane is 6x - 13y - 60z = 76.