Find plane by normal and instance point + distance between origin and plane

Question

I have a question regarding Vectors; Find the equation of the plane perpendicular to the vector $\vec{n}\space=(2,3,6)$ and which goes through the point $ A(1,5,3)$. (A cartesian and parametric equation). Also find the distance between the beginning of axis and this plane.

I'm not really sure where to start. Any help would be appreciated.

Answer 1

Note that for any point $\vec{x}$ on the plane the dot product $\vec{n}\cdot\vec{x}$ must be constant:

$$2x+3y+6z=const$$

You can determine this constant using the given point $A$:

$$2\cdot1+3\cdot5+6\cdot3=35$$

So we get for the equation of the plane

$$2x+3y+6z=35\tag{1}$$

The distance from the origin is the length of the projection of a vector from the origin to any point on the plane onto the normal vector. This is simply the constant on the right hand side of (1) divided by the length of the normal vector:

$$\textrm{distance}=\frac{35}{|\vec{n}|}=\frac{35}{\sqrt{2^2+3^2+6^2}}=\frac{35}{7}=5$$

Answer 2

If a plane is perpendicular to a vector then this vector represents its normal. If you have a normal vector and a point in the plane you can determine equation of the plane like this:

$$\vec{n}=(n_x,n_x,n_z)$$

$$T(x_T,y_T,z_T)\in \pi_1$$

Then you have:

$$ \pi_1\dots n_x(x-x_T)+n_y(y-y_T)+n_z(z-z_T)=0$$

And you have the plane equation.

If you want to find distance from (0,0,0) consider the plane equation in the following form:

$$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$$