# IB Class Notes

## Intersection of Lines in 3-D

Aims: By the end of this note, you will be able to

1. classify different sort of lines both in 2-D and in 3-D, and
2. use vector representation of line to solve the following cases: intersections, shortest distance, parallel and skew lines.

### Intoduction

Given two lines in 3-D space, they may satisfy one of the following conditions:

1. intersect at a point
2. be parallel to each other and do not intersect
3. be parallel and coincident (i.e. two lines are the same, all points satisfy both linear equations)
4. be neither parallel nor intersect (skew lines).

To intersect at a point or coincident, the two lines must be on the same plane.

Lines that are lying in the same plane are called coplanar lines. Cases 1, 2 and 3 above are coplanar or on the same plane. Two lines that are parallel are also coplanar lines.

Skew lines on the other hand do not share the same plane. You can think of one line in the first floor and the other in the 25th floor so that they will never meet. Thus, two lines in 2-D cannot be skews.

Questions in this section may involve you to test whether two lines are parallel or perpendicular to each other. Recall that given two non-zero vectors $\vec{a}$ and $\vec{b}$ then they are parallel to each other if $\vec{a} \cdot\vec{b} = \pm |\vec{a}||\vec{b}|$ perpendicular to each other if $\vec{a} \cdot \vec{b} = 0$

Example 1:

Find the point of intersection between line L1 and L2 whose vector equations are
$\vec{r_1} = \left(\begin{matrix} 2 \\ 1 \\ -2 \end{matrix}\right) + \lambda \left(\begin{matrix} 6 \\ -8 \\ 10 \end{matrix}\right)$ and
$\vec{r_2} =\left( \begin{matrix} -1 \\ -10 \\ 5 \end{matrix}\right) + \mu \left(\begin{matrix} 6 \\ 7 \\ -2 \end{matrix} \right)$ respectively.

Solution

For these two lines to intersect there must be a value of λ and μ that will provide a point that lies both on L1 and L2. The strategy is to solve for λ and μ. Next we have to make sure the point satisfies both $\vec{r_1}$ and $\vec{r_2}$.
Equating coordinates x, y, and z allow us to have the following system of equations:
2 + 6λ = -1 + 6μ -------(1)
1 - 8λ = -10 + 7μ ------(2)
-2 + 10λ = 5 - 2μ ------(3)
Solving this system using (1) and (3) simultaneously, for example, allow us to obtain
λ = 1/2 and μ = 1.
Substituting λ = 1/2 and μ = 1 into (1), (2) and (3) gives us the point of intersection at (5, -3, 3).

 Remember. Once you found λ and μ then make sure you that x-coordinates, y-coordinates, and z-coordinates of both lines are equal. If they are all equal then you have at least one intersection. If at least one of the coordinates be it x, y or z are different between the two lines then they have no intersection. Study example 4 below.

Example 2

Find the point of intersection between line L1 and L2 whose Cartesian equations are
L1: $\large \frac{x-3}{2} = \frac{y+2}{-1} = \frac{z}{3}$ and
L2: $\large \frac{x+1}{1} = \frac{y-2}{1} = \frac{z-4}{9}$ respectively.

Solution.

From L1, we obtain the following:
x = 3 + 2λ -------(1)
y = -2 - λ -------(2)
z = 3λ ------(3)
Similarly from L2 we have
x = -1+ μ -------(4)
y = 2 + μ -------(5)
z = 4 + 9μ ------(6)

Equate (1) to (4) and (2) to (5) and solve this system simultaneously to obtain
λ = -8/3 and μ = -4/3.
Substituting λ = -8/3 into (1), (2) and (3) and μ = -4/3 into (4), (5) and (6) gives us the point of intersection at (-7/3, 2/3, -8).

### Shortest distance from a point to a line (2-D or 3-D) Let A be a point in 3-D space and $d$ be the shortest distance from the point A to a line represented by vector $\vec{b}$. If we know vector $\vec{a}$ and angle θ as in the diagram then $d = |\vec{a}| sin \theta$ . However, most questions may not come with as much information as the diagram above. Usually, the angle is unknown. Let $A(x_1,y_1,z_1)$ be a known point that is fixed and image P to be a point moving in the direction of vector $\vec{b}$ as shown in this diagram. Vector $\vec{b}$ is $\left( \begin{matrix} b_1 \\ b_2 \\b_3 \end{matrix} \right)$ Image P is moving on the line represented by vector equation $\left( \begin{matrix} x \\ y \\ z \end{matrix} \right) + t\left( \begin{matrix} b_1 \\ b_2 \\ b_3 \end{matrix} \right)$ . In fact, any image P is represented by $(x+tb_1, y+tb_2, z+tb_3)$ and the position vector $\vec{OP}$ is thus $\left( \begin{matrix} x+tb_1 \\ y+tb_2 \\ z+tb_3 \end{matrix} \right)$ . We can represent vector $\vec{AP}$ using the parameter $t$ where \begin{align} \vec{AP} & = \vec{AO} + \vec{OP} \\ &= \left( \begin{matrix} -x_1 \\ -y_1 \\ -z_1 \end{matrix} \right) + \left( \begin{matrix} x+tb_1 \\ y+tb_2 \\ z+tb_3 \end{matrix} \right) \\ &= \left( \begin{matrix} x-x_1+tb_1 \\ y-y_1+tb_2 \\ z-z_1+tb_3 \end{matrix} \right) \end{align} The shortest distance $d$ between A and $\vec{b}$ occurs when P is moved into a position such that vector $\vec{AP}$ is perpendicular to $\vec{b}$. Thus, the strategy is to find the value $t$ where vector $\vec{AP} \bullet \vec{b} = 0$ . If $t*$ is the solution then the shortest distance $d = | \vec{AP} \text{with t*} |$

Example 3

Find the shortest distance from A(-5,-2,2) to the line $\large \frac{x+5}{1} = \frac{y+7}{2} = \frac{z-14}{-3}$ .

Solution
Any point P on the line has coordinates
x = -5 + λ
y = -7 + 2λ
z = 14 - 3λ.
Moreover, the direction vector of the line, say $\vec{b} = \vec{i}+2\vec{j}-3\vec{k}$ Then
\begin{align} \vec{AP} & = \vec{AO} + \vec{OP} \\ & = \left(\begin{matrix} 5 -5 + \lambda \\ 2 -7 + 2 \lambda \\ -2 + 14 - 3\lambda \end{matrix} \right) \\ & = \left(\begin{matrix} \lambda \\ -5 + 2 \lambda \\ 12 - 3\lambda \end{matrix} \right) \\ \end{align}
The shortest distance occurs at a λ such that the scalar product between $\vec{AP}$ and $\vec{b}$ is zero. Thus, we set
\begin{align} \vec{AP} \bullet \vec{b} & = 0 \\ & = \left(\begin{matrix} \lambda \\ -5 + 2 \lambda \\ 12 - 3\lambda \end{matrix} \right) \bullet \left(\begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right) \\ \implies 14\lambda - 46 & = 0 \\ \implies \lambda & = \frac{23}{7} \end{align}

The vector $\vec{AP}$ associated with the shortest distance is
\begin{align} \vec{AP*} & = \left( \begin{matrix} \frac{23}{7} \\ \frac{11}{7} \\ \frac{15}{7} \end{matrix} \right) \\ |\vec{AP*}| &= \sqrt{(\frac{23}{7})^2+(\frac{11}{7})^2+(\frac{15}{7})^2} \\ &= \frac{\sqrt{125}}{7} \end{align}
Thus, the shortest distance from A(-5,-2,2) to the line $\frac{x+5}{1} = \frac{y+7}{2} = \frac{z-14}{-3}$ is $\frac{\sqrt{125}}{7}$.

### Example 4.

Let two straight lines be

L1 : $\left( \vec{i} + 2 \vec{j} - \vec{k} \right) + \lambda \left( 5 \vec{i} - 3 \vec{j} + 2\vec{k} \right)$ and
L2 : $\left( 10 \vec{i} -5 \vec{j} +3\vec{k} \right) + \mu \left( 7\vec{i} + 8\vec{j} -3\vec{k} \right)$
respectively.
Show that L1 and L2 are skew lines.

Solution.
We will need to show that (1) L1 and L2 are not parallel to each other. That is, the set of coefficients of the direction vector in L1 and L2 are not proportional. (2) Next we need to show that L1 and L2 do not intersect at any point. With these two conditions, we can establish that L1 and L2 are skew lines.

Let us first consider the set of directional vectors of L1 and L2. The ratio of the coefficients of the $\vec{i}$ basis is 5/7. Similarly, those for $\vec{j}$ basis and $\vec{k}$ basis are -3/8 and 2/(-3) respectively. However, these ratios do not equate to each other. Thus, we can conclude that L1 and L2 are not parallel.

The next step is to show that these lines have no intersection. We will start by assuming that these two lines actually intersect at a point and show that this lead to a contradiction. From L1 we obtain
x = 1 + 5λ -------(1)
y = 2 - 3λ -------(2)
z = -1 + 2λ ------(3)
Similarly from L2 we have
x = 10 + 7μ -------(4)
y = -5 + 8μ-------(5)
z = 3 - 3μ ------(6)

Equate (1) to (4) and (2) to (5) and solve this system simultaneously to obtain
λ = 121/61 and μ = 8/61
[Note the system above can be easily solved using Polysmlt apps in the Ti-calculator and remember to change your answer into fraction..]
Substituting λ = 121/61 into (3) and μ = 8/61 into (6) gives 181/61 and 159/61 respectively.
Since these x-coordinates do not equal, these lines do not intersect.
[Notice that (3) and (6) were subjected to the test because they were not used to solve for λ and μ. ]
Since L1 and L2 are not parallel and do not intersect then L1 and L2 must be skew lines.

Exercise

Email KokMing Lee
Li Po Chun United World College of Hong Kong,
10 Lok Wo Sha Lane, Sai Sha Road,
Shatin, New Territories, Hong Kong SAR.