Calculus Question

Question

Find the distance between the point P (−2,0,3) and the plane π: 2x − 3y + z − 6 = 0.

Answer 1

The thing about an equation like 2x-3y+z-6=0 is that if you look carefully, there's a dot product hiding in plain sight. The equation can also be written

2(x-0)-3(y-0)+1(z-6)=0

and also

〈2,-3,1〉•〈x-0,y-0,z-6〉= 0,

which means that if you take any point (x,y,z) in this plane, subtract the point (0,0,6), then the resulting vector is orthogonal to the vector〈2,-3,1〉. The vector 〈2,-3,1〉 called a normal of the plane. Conveniently, to find the minimum distance between a point and the plane, we need only subtract (0,0,6), dot with the plane normal, take the absolute value, and divide by the magnitude of the plane normal.

(-2,0,3) - (0,0,6) = 〈-2,0,-3〉

〈-2,0,-3〉 • 〈2,-3,1〉 = -7

||〈2,-3,1〉|| = ✓(14)

The distance is 7/✓(14) units.

Answer 2

I would take P as a position vector and find its projection onto the plane. This means finding two vectors on the plane that span the plane (find (x,y,z) that satisfy the plane equation above), finding the projection of P onto each vector and adding them up. Now you have the P vector and the projection of it onto the plane. To find the distance between these two just subtract it, so P - Projection Vector. The distance is the magnitude of that resulting distance vector.