Equation of plane parallel vectors

AI Thread Summary
The discussion focuses on finding the Cartesian equation of a plane defined by two parallel vectors, a and b, and a point R. The user correctly identifies that the cross product of the vectors yields a normal vector to the plane, which is then normalized. The resulting equation of the plane is derived as 3x - 14y + 18z = 92. Additionally, the distance from the plane to the origin is calculated to be 4 units. The solution appears to be accurate based on the provided calculations.
Jbreezy
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Homework Statement


The vectors a= <-4,3,3> and b = <2,-6,-5> are parallel to a plane PI and R is a point on
with position vector <104,8,-6> . Find the Cartesian equation of the plane. What is the
distance of the plane from the origin?


Homework Equations





The Attempt at a Solution



This is my thinking.

Since we are told the vectors a and b are parallel to the plane if we cross them we can get a vector perpendicular to the plane. We can turn that into a unit vector and dot it with the point r.

So,

a cross b = <3,-14,18> : call this vector v
v = √(3^2) + (-14^2)+(18)^2 = 23
So v(hat) = 1/23<3,-14,18>
Then to get the equation of the plane we can do

x dot v(hat) = r dot v(hat)

<x,y,z> dot 1/23<3,-14,18> = <104,8,-6> dot <1/23<3,-14,18>

So I got
3/23x - 14/23y + 18/23z = 92/23
Then just multiply through to clear out the fraction

3x - 14y +18z = 92

and the distance from the plane to the origin is 4 units because r dot v hat is 4
Did I do this correct?
 
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