Find the cartesian equation of the plane passing through p(8, -2,0) and perpendicular to a- 5i+3j-k what is the distance of this plane to the point 0(2,2, 2)? (a) (b)

equation of plane, 5x+3y-z-36=0

Distance of point (2,2,2) from plane = 4.05 units

Step-by-step explanation:

Given,

Plane passing through the point = (8, -2, 0)

Let's say,

Plane perpendicular to the vector, a= 5i + 3j- k

Since, the vector is perpendicular to the plane, hence the equation of plane can be given by

Hence, the equation of plane can be given by, 5x+3y-z-36=0

Now, we have to calculate the distance of the point O(2,2,2) from the plane 5x+3y-z-36=0

Let's say,

a= 5, b= 3, c= -1, d=-36

So, distance of a point from the plane can be given by,

  = 4.05 units

So, the distance of the point O(2,2,2) from the given plane will be 4.05 units.

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step-by-step explanation:

this is a parabola.