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

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.

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:

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

• insufficent data – Bak1139 Apr 14 '14 at 9:55
• @Bak1139 What is it that you don't understand? More than this would be a fully written out solution to your homework problem, which is not commonly done on this site. Especially since you haven't shown any effort yourself. – Matt L. Apr 14 '14 at 9:57
• I see...I have no basic knowledge in this subject, and no idea as for solving this, so pure theroy is not really enough for me to comprihand this.. – Bak1139 Apr 14 '14 at 9:59
• @Bak1139 Do you know what a dot product between two vectors is? As I said in my answer, you get the equation of the plane by computing $n_xx+n_yy+n_zz=const$. If you evaluate this equation with the coordinates of the given point $A$ you can determine the constant. This is the complete equation of the plane in the form $ax+by+cz=d$. – Matt L. Apr 14 '14 at 10:01
• @Bak1139 OK, I added (a lot of) details to my previous answer. I hope it's clear now. – Matt L. Apr 14 '14 at 10:18

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:

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

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

where $$\pi_1$$ is the set of points in the plane.

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:

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

• Shouldn't it be $$\vec{n}=(n_x,n_y,n_z)$$ ? – Sen Jacob Aug 22 '19 at 8:51