# Lines in Space

## Key Questions

• The symmetric equation of the line with the direction vector $\vec{v} = \left(a , b , c\right)$ passing through the point $\left({x}_{0} , {y}_{0} , {z}_{0}\right)$ is:

$\frac{x - {x}_{0}}{a} = \frac{y - {y}_{0}}{b} = \frac{z - {z}_{0}}{c}$,

where none of $a , b$ and $c$ are zero.

If one of $a , b$, and $c$ is zero; for example, $c = 0$, then we can write:

$\frac{x - {x}_{0}}{a} = \frac{y - {y}_{0}}{b}$ and $z = {z}_{0}$.

If two pf $a , b$, and $c$ are zero; for example, $b = c = 0$, then we can write:

$y = {y}_{0}$ and $z = {z}_{0}$

(There is no restriction on $x$, it can be any real number. )

I hope that this was helpful.

• We want to find the parametric equations of the line L passing through the point P and parallel to a vector A.

Let us take a 3-dimensional point in ${R}^{3}$, call it $P = \left({x}_{0} , {y}_{0} , {z}_{0}\right)$.

A line L is drawn such that it passes through P and is parallel to the vector $A = \left(u , v , w\right)$.

3 parametric equations can be written which express the components:

$x = {x}_{0} + t u$
$y = {y}_{0} + t v$ $\left(- \infty < t < \infty\right)$
$z = {z}_{0} + t w$

As an example, with a point $P = \left(2 , 4 , 6\right)$ and vector $A = \left(1 , 3 , 5\right)$, we have the following parametric equations:

$x = 2 + t u$
$y = 4 + 3 v$ $\left(- \infty < t < \infty\right)$
$z = 6 + 5 w$

For plane:

$a x + b x + c x + d = 0$

and line:

$\left(x , y , z\right) = \left({x}_{0} , {y}_{0} , {z}_{0}\right) + t \left(u , v , w\right)$

these are suitable conditions:

{ (ax_0+by_0+cz_0+d = 0), (au+bv+cw = 0) :}

#### Explanation:

The most general equation of a plane in three dimensional space is:

$a x + b y + c z + d = 0$

A line in three dimensional space can be represented in a variety of ways, but one representation that will work for any line is the parametric form:

$\left(x , y , z\right) = \left({x}_{0} , {y}_{0} , {z}_{0}\right) + t \left(u , v , w\right)$

This line will lie in the plane if and only if two distinct points of it both satisfy the equation of the plane.

Using $t = 0$ and $t = 1$, we obtain the conditions:

{ (ax_0+by_0+cz_0+d = 0), (a(x_0+u) + b(y_0 + v) + c(z_0 + w) + d = 0) :}

Subtracting the first of these from the second, we obtain the condition:

$a u + b v + c w = 0$

So a necessary and sufficient set of conditions is:

{ (ax_0+by_0+cz_0+d = 0), (au+bv+cw = 0) :}

• I'll show how to solve this problem using an example using the vector based form of a two straight lines. Bear in mind that there will be one of the following outcomes:

• a single unique point.
• no solution (if the lines do not intersect).
• infinitely many solutions (if the lines coincide).

Suppose we have;

${L}_{1} : \setminus \setminus \setminus \setminus \vec{{r}_{1}} = \left(\begin{matrix}5 \\ 2 \\ - 1\end{matrix}\right) + l a m \mathrm{da} \left(\begin{matrix}1 \\ - 2 \\ - 3\end{matrix}\right)$

${L}_{2} : \setminus \setminus \setminus \setminus \vec{{r}_{2}} = \left(\begin{matrix}2 \\ 0 \\ 4\end{matrix}\right) + \mu \left(\begin{matrix}1 \\ 2 \\ - 1\end{matrix}\right)$

It is vital that the variable parameters ($l a m \mathrm{da}$ and $\mu$) have different symbols, as they are different parameters relative to each line.

If the lines do meet then for some specific values of $l a m \mathrm{da}$ and $\mu$ then:

$\vec{{r}_{1}} = \vec{{r}_{2}}$

In which case we would have:

$\left(\begin{matrix}5 \\ 2 \\ - 1\end{matrix}\right) + l a m \mathrm{da} \left(\begin{matrix}1 \\ - 2 \\ - 3\end{matrix}\right) = \left(\begin{matrix}2 \\ 0 \\ 4\end{matrix}\right) + \mu \left(\begin{matrix}1 \\ 2 \\ - 1\end{matrix}\right)$
$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \left(\begin{matrix}5 + l a m \mathrm{da} \\ 2 - 2 l a m \mathrm{da} \\ - 1 - 3 l a m \mathrm{da}\end{matrix}\right) = \left(\begin{matrix}2 + \mu \\ 2 \mu \\ 4 - \mu\end{matrix}\right)$

By comparing the coefficients of $\hat{i}$, $\hat{j}$ and $\hat{k}$ we have three equations in two unknowns, so we can solve for $l a m \mathrm{da}$ and $\mu$ using the first two equations and check if the third is satisfied:

$\hat{i} : \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus 5 + l a m \mathrm{da} = 2 + \mu \text{ } \ldots \ldots . \left[1\right]$
$\hat{j} : \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus 2 - 2 l a m \mathrm{da} = 2 \mu \text{ } \ldots \ldots . \left[2\right]$
$\hat{k} : \setminus \setminus \setminus \setminus - 1 - 3 l a m \mathrm{da} = 4 - \mu \setminus \text{ } \ldots \ldots . \left[3\right]$

$E q \left[1\right] + \frac{1}{2} E q \left[2\right]$ gives $\implies 6 = 2 + 2 \mu \implies \mu = 2$
Subs $\mu = 2$ into $E q \left[1\right] \implies 5 + l a m \mathrm{da} = 4 \implies l a m \mathrm{da} = - 1$
Subs $\mu = 2$ and $l a m \mathrm{da} = - 1$ into $E q \left[3\right] \implies - 1 + 3 = 2 = 4 - 2$

So we have established that if we choose $l a m \mathrm{da} = - 1$ and $\mu = 2$ then we get a unique solution simultaneously satisfying all three equations.

We can then substitute $l a m \mathrm{da} = - 1$ into ${L}_{1}$, (equally $\mu = 2$ into ${L}_{2}$ would work) to determine the actual coordinate of intersection.

$\vec{{r}_{1}} = \left(\begin{matrix}5 \\ 2 \\ - 1\end{matrix}\right) - \left(\begin{matrix}1 \\ - 2 \\ - 3\end{matrix}\right)$
$\setminus \setminus \setminus = \left(\begin{matrix}5 - 1 \\ 2 + 2 \\ - 1 + 3\end{matrix}\right)$
$\setminus \setminus \setminus = \left(\begin{matrix}4 \\ 4 \\ 2\end{matrix}\right)$

so in this example, the lines ${L}_{1}$ and ${L}_{2}$ intersect at the coordinate $\left(4 , 4 , 2\right)$

• Since we are talking about a line in 3-D, it is more appropriate to talk about its direction vector than its slope. To find the direction vector $\vec{v}$ of the line passing through two points $\left({x}_{1} , {y}_{1} {z}_{1}\right)$ and $\left({x}_{2} , {y}_{2} , {z}_{2}\right)$ can be found by

$\vec{v} = \left({x}_{2} - {x}_{1} , {y}_{2} - {y}_{1} , {z}_{2} - {z}_{1}\right)$.