# What is the formula for a 3D line?

Just like we have the formula $y=mx+b$ for $\mathbb{R}^{2}$, what would be a formula for $\mathbb{R}^{3}$? Thanks.

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$$(a_0 + a_1 t, b_0 + b_1 t,c_0 + c_1 t)$$ where $a_1, b_1, c_1$ are not all $0.$ –  Will Jagy May 28 '13 at 3:39
Was there something in particular that was not clear from the many resources available upon Googling "3d line equation"? –  Zev Chonoles May 28 '13 at 3:39

You can describe a line in space as the intersection of two planes. Thus, $$\{(x,y,z)\in{\mathbb R}^3: a_1x+b_1y+c_1z=d_1 \text{ and } a_2x+b_2y+c_2z=d_2\}.$$ Alternatively, you can use vector notation to describe it as $$\vec{p}(t) = \vec{p}_0 + \vec{d}t.$$

I used this relationship to generate this picture:

This is largely a topic that you will learn about in a third semester calculus course, at least in the states.

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Here are three ways to describe the formula of a line in $3$ dimensions. Let's assume the line $L$ passes through the point $(x_0,y_0,z_0)$ and is traveling in the direction $(a,b,c)$.

Vector Form

$$(x,y,z)=(x_0,y_0,z_0)+t(a,b,c)$$

Here $t$ is a parameter describing a particular point on the line $L$.

Parametric Form

$$x=x_0+ta\\y=y_0+tb\\z=z_0+tc$$

These are basically the equations that result from the three components of vector form.

Symmetric Form

$$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$$

Here we assume $a,b,$ and $c$ are all nonzero. All we've done is solve the parametric equations for $t$ and set them all equal.

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In my opinion, the symmetric form is the most useless one. –  sidht May 28 '13 at 4:05
I am giving you an example. Let $A(-2,0,1),~~B(4,5,3)$ be two points in $\mathbb R^3$. And let $C$ be the end point for the vector which is drawn from the orrigin. In addition, we assume that this vector has the same direction as the vector $AB$. So we have its coordinates is $(4,5,3)-(-2,0,1)=(6,5,2)$. Therefore the equation of the line passing through $A$ and $B$ is $$L_{AB}: x=(-2,0,1)+t(6,5,2)$$