# Equation of plane from normal

A line is given by $x = 2 + 5t , y = 1 + 2t , z = 3 − 3t$

Define an equation in normal form for the plane which is perpendicular to the line and which intersects the point $(2, −3, 4)$.

When I look at the answer it says that the normal to the plane is

$n = [5; 2; -3]$

which is also the direction vector of the line, which I understand. Then we get to the part that confuses me. Here they claim that

$5(x − 2) + 2(y + 3) − 3(z − 4) = 0$

or

$5x + 2y − 3z + 8 = 0$

is the equation of the plane we seek, but I don't follow. What's a more detailed way of describing this approach?

In the answer you’ve cited, they use the normal form of the the equation of a plane in $\mathbb R^3$: $\mathbf n\cdot(\mathbf x-\mathbf p)=0$, where $\mathbf n$ is a vector normal to the plane and $\mathbf p$ is some vector in the plane. (By the way, this same equation works for a hyperplane of dimension $n-1$ in $\mathbb R^n$). We can rewrite this as $\mathbf n\cdot\mathbf x=d$, which tells us that the plane consists of all vectors whose dot product with a fixed normal vector to the plane is equal to some constant. This dot product is $\|\mathbf n\|$ times the projection of $\mathbf x$ onto $\mathbf n$, so geometrically, this says that the component in a direction normal to the plane of all vectors in the plane has the same length. This length is, of course, just the (orthogonal) distance of the plane from the origin. Since this is true for all vectors in the plane, it’s true for a particular one, so given a normal vector $\mathbf n$ and any element $\mathbf p$ of the plane, you can get a cartesian equation for the plane by expanding $\mathbf n\cdot\mathbf x-\mathbf n\cdot\mathbf p=0$.

In this particular case, the plane is perpendicular to the given line, so we can use its direction vector for $\mathbf n$. We’re given a point on the plane, so using the above we can write the equation down immediately: $(5,2,-3)\cdot(x,y,z)-(5,2,-3)\cdot(2,-3,4)=0$ or $5x+2y-3z+8=0$.

hint

If $(a,b,c)$ is a normal vector to a plane, then the cartesian equation of this plane will be of the form

$$ax+by+cz=d.$$ if the plane contains the point $(x_0,y_0,z_0)$, the equation becomes $$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$$

with $d=ax_0+by_0+cz_0$.

parmetric equations of the plane

the vectors director of the plane can be taken as $u=(-b,a,0)\perp(a,b,c)$ and

$v=(-c,0,a)\perp(a,b,c).$

• Ah, I understand, took way too long for me to see the cartesian equation in this! How come we can use the normal vector like this though? When you just try to imagine it it's hard to see how it can play a role in the equation of the actual plane. Is there a way to visualize it better? Thanks! Commented Jan 10, 2017 at 19:18
• @André I just edited some lines. is it clearer now. Commented Jan 10, 2017 at 19:36