Vector Product

[Sweatyapples]

A plane is perpendicular to the vector n and contains the point P
a) show that the shortest distance from the plane to the origin is given by (n (dot product) OP)/magnitude n
b) Calculate the shortest distance from the plane 2x + 3y - z = 12 to the origin.

I have no idea about a but I'm pretty sure I've solved b correctly.
2x + 3y - z = 12
n1(x-1) + n2(y-b) + n3(z-c) = 0
2(x-0) + 3(y-0) + -(z+12) = 0

therefore, perpendicular vector (n) = 2i + 3j - k
and a point on the plane is P(0,0,12)
((2i + 3j - k) dot product 12k)/sqrt 14 = -3.2 = 3.2
any help with a or b would be appreciated. )

 

[pka]

A plane is perpendicular to the vector n and contains the point P
a) show that the shortest distance from the plane to the origin is given by (n (dot product) OP)/magnitude n
b) Calculate the shortest distance from the plane 2x + 3y - z = 12 to the origin.

@Sweatyapples, it greatly disturbs me by how little you seem to understand about planes.
If \(\displaystyle \vec{~R~}=<x.y,z>~\&~\vec{~n~}=<2,3,-1>\) then your plane is \(\displaystyle \vec{~n~}\cdot\vec{~R~}=12\).

Suppose that \(\displaystyle \Pi:\vec{~N~}\cdot\vec{~R~}=C\) is any plane and \(\displaystyle P: (a,b,c)\) is a point on \(\displaystyle \Pi\) then \(\displaystyle \vec{OP}=<a-0,b-0,c-0>\) is the vector from \(\displaystyle 0\to P\).

Now the distance from \(\displaystyle O\), the origin, to \(\displaystyle \Pi\) is \(\displaystyle \mathcal{D}(O,\Pi)=\left[\dfrac{|\vec{OP}\cdot N|}{\|N\|}\right].\) That is the answer for part a).
YOU should try to derive that formula.