Straight Lines in Three Dimensions

Straight Lines in 3D

The equations of straight lines in 3D are not the same as those in 2D. So do not mix them up. In 2D, a line is given by a single equation in \(x\) and \(y\), but in 3D, one equation like that represents a plane, not a line.

In fact, the correct extension of a straight line from 2D to 3D is not another line, but a plane. This might seem surprising at first, but you will learn that the way planes are written in 3D is very similar to how straight lines are written in 2D.

This does not mean lines in 3D have nothing in common with lines in 2D. There are still some similarities, but their equations are quite different. You will understand this better when you study vector and parametric forms of lines in space.

Symmetric Form of Equation of a Straight Line

So how does one write the equation? Suppose you have a line in space. How do you correctly specify this line — that it is this particular line and no other? One point and a direction would be sufficient, if you think carefully.

A straight line in space is completely determined by:

1. A point through which it passes, and
2. A direction in which it extends.

Suppose a known point on the line is \((x_1, y_1, z_1)\), and a vector parallel to the line is

\[ \vec{d} = a\hat{i} + b\hat{j} + c\hat{k}. \]

Then \((a, b, c)\) are called the direction ratios of the line.

Let \((x, y, z)\) be any general point on the line. Since both \((x_1, y_1, z_1)\) and \((x, y, z)\) lie on the same line, the vector joining them is

\[ \vec{r} = (x - x_1)\hat{i} + (y - y_1)\hat{j} + (z - z_1)\hat{k}. \]

This vector must be parallel to \(\vec{d} = a\hat{i} + b\hat{j} + c\hat{k}\).

Since the two vectors are parallel, their corresponding components must be in the same ratio. Therefore,

\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}. \]

This is called the symmetric form of the equation of a line in 3D.

Any point \((x, y, z)\) that satisfies this equation lies on the line. Conversely, every point on the line must satisfy this relation.

For example:

A line passing through the point \((1, 2, -3)\) and having direction ratios \((-1, 2, -3)\) has the equation:

\[ \frac{x - 1}{-1} = \frac{y - 2}{2} = \frac{z + 3}{-3}. \]

Parametric Form of the line

Since all three ratios are equal to the same constant, we may assume this common value to be a parameter, say \(\lambda\). That is,

\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda. \]

From this, we get:

\[ \begin{aligned} x &= x_1 + \lambda a, \\ y &= y_1 + \lambda b, \\ z &= z_1 + \lambda c. \end{aligned} \]

This form is called the parametric form of the line, because the coordinates of every point on the line depend on a parameter \(\lambda\), which can take any real value.

Thus, any point on the line can be written as

\[ (x, y, z) = (x_1 + \lambda a,\ y_1 + \lambda b,\ z_1 + \lambda c) \quad \text{for some } \lambda \in \mathbb{R}. \]

Example:

Consider the line whose symmetric equation is

\[ \frac{x - 1}{2} = \frac{y + 3}{3} = \frac{z}{4}. \]

Let the common ratio be \(\lambda\). Then,

\[ \begin{aligned} x &= 1 + 2\lambda, \\ y &= -3 + 3\lambda, \\ z &= 0 + 4\lambda. \end{aligned} \]

So the parametric form of the line is:

\[ (x, y, z) = (1 + 2\lambda,\ -3 + 3\lambda,\ 4\lambda), \quad \lambda \in \mathbb{R}. \]

For example, taking \(\lambda = 1\), we get the point

\[ (1 + 2,\ -3 + 3,\ 4) = (3,\ 0,\ 4) \]

which lies on the line.

Every different value of \(\lambda\) gives a different point on the line. So two distinct points on the line always correspond to two distinct values of \(\lambda\).

Reading the Cartesian Form of a Line Correctly

The Cartesian (or symmetric) form of a line in 3D is usually written as:

\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}, \]

where \((x_1, y_1, z_1)\) is a point on the line and \((a, b, c)\) are the direction ratios of the line.

Let us now interpret each of the given cases:

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(a)

\[ \frac{x - 1}{2} = \frac{y}{3} = \frac{z + 1}{4} \]

Comparing with the standard form, we immediately see:

• Point on the line: \((1,\ 0,\ -1)\)
• Direction ratios: \((2,\ 3,\ 4)\)

Conclusion: This is the straight line passing through \((1,\ 0,\ -1)\) with direction ratios \(2,\ 3,\ 4\).

---

(b)

\[ \frac{2x - 1}{2} = \frac{y + 1}{3} = \frac{1 - z}{3} \]

This is not in standard form. We must first rewrite each term in the form \((\text{variable} - \text{constant}) / (\text{number})\).

Start with the first term:

\[ \frac{2x - 1}{2} = \frac{2(x - \frac{1}{2})}{2} = x - \frac{1}{2} \ \Rightarrow\ \frac{x - \frac{1}{2}}{1} \]

Third term:

\[ \frac{1 - z}{3} = \frac{-(z - 1)}{3} = \frac{z - 1}{-3} \]

Thus, the equation becomes:

\[ \frac{x - \frac{1}{2}}{1} = \frac{y + 1}{3} = \frac{z - 1}{-3} \]

So:

• Point on the line: \(\left(\frac{1}{2},\ -1,\ 1\right)\)
• Direction ratios: \((1,\ 3,\ -3)\)

Conclusion: This is the straight line passing through \(\left(\frac{1}{2},\ -1,\ 1\right)\) with direction ratios \(1,\ 3,\ -3\).

---

(c)

\[ \frac{x - 1}{0} = \frac{y + 1}{2} = \frac{z}{4} \]

At first glance, division by zero may seem undefined, but in this context, it carries meaning. Let:

\[ \frac{x - 1}{0} = \lambda \implies x - 1 = 0 \implies x = 1 \]

This means x is constant for all points on the line. The line is perpendicular to the x-axis.

For the other components:

\[ \frac{y + 1}{2} = \lambda \Rightarrow y = 2\lambda - 1,\quad \frac{z}{4} = \lambda \Rightarrow z = 4\lambda \]

Thus, the parametric equations are:

\[ x = 1,\quad y = 2\lambda - 1,\quad z = 4\lambda \]

Hence:

• Point on the line: when \(\lambda = 0\), we get \((1,\ -1,\ 0)\)
• Direction ratios: \((0,\ 2,\ 4)\), or simplified as \((0,\ 1,\ 2)\)

Conclusion: This is the straight line passing through \((1,\ -1,\ 0)\) and parallel to the vector \((0,\ 1,\ 2)\).

Vector Form of the Equation of a Straight Line

Let \(\vec{a}\) be the position vector of a fixed point \(A\) on the line, and let \(\vec{b}\) be a vector parallel to the line. Let \(\vec{r}\) be the position vector of a general point \(P\) on the line.

Then the vector \(\overrightarrow{AP}\) is given by:

\[ \overrightarrow{AP} = \vec{r} - \vec{a}. \]

Since the vector \(\overrightarrow{AP}\) lies along the line and \(\vec{b}\) is also parallel to the line, they must be collinear. Hence, we must have:

\[ \vec{r} - \vec{a} = \lambda \vec{b} \quad \text{for some } \lambda \in \mathbb{R}. \]

This gives the vector equation of the line:

\[ \vec{r} = \vec{a} + \lambda \vec{b}, \quad \lambda \in \mathbb{R}. \]

This is the parametric form of the line written in vector notation. The point \(\vec{a}\) fixes the location of the line in space, and the vector \(\vec{b}\) gives its direction.

To relate this to the Cartesian form, suppose:

\[ \vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}, \quad \vec{b} = a\hat{i} + b\hat{j} + c\hat{k}, \quad \vec{r} = x\hat{i} + y\hat{j} + z\hat{k}. \]

Substitute into the vector equation:

\[ x\hat{i} + y\hat{j} + z\hat{k} = (x_1 + \lambda a)\hat{i} + (y_1 + \lambda b)\hat{j} + (z_1 + \lambda c)\hat{k}. \]

Comparing components, we get:

\[ \begin{aligned} x &= x_1 + \lambda a, \\ y &= y_1 + \lambda b, \\ z &= z_1 + \lambda c. \end{aligned} \]

These are exactly the parametric equations of the line in Cartesian form. So the vector form \(\vec{r} = \vec{a} + \lambda \vec{b}\) is nothing but a compact and elegant way of writing the same information.

Alternative Vector Form of a Straight Line: Cross Product Form

Another way to express the equation of a straight line using vectors is through the idea of collinearity.

Let \(\vec{a}\) be the position vector of a fixed point \(A\) on the line, and let \(\vec{b}\) be a vector parallel to the line. Let \(\vec{r}\) be the position vector of a general point \(P\) on the line.

As before, the vector \(\vec{r} - \vec{a} = \overrightarrow{AP}\) joins the fixed point \(A\) to the variable point \(P\). Since both \(A\) and \(P\) lie on the line and \(\vec{b}\) is parallel to the line, it follows that \(\vec{r} - \vec{a}\) must be collinear with \(\vec{b}\).

Two vectors are collinear if and only if their cross product is zero. Hence, the condition for \(P\) to lie on the line is:

\[ (\vec{r} - \vec{a}) \times \vec{b} = \vec{0}. \]

This is an alternative form of the vector equation of a line. It does not involve any parameter and directly imposes a geometric condition: the vector from a fixed point on the line to any general point must be parallel to the given direction vector.

This form is especially useful when eliminating the parameter is desired or when checking whether a given point lies on a known line.

Example

Find the vector and Cartesian equations of the line passing through the points

\[ A(-1,\ 0,\ 1) \quad \text{and} \quad B(2,\ 3,\ 4). \]

Solution:
To determine the equation of a straight line in space, we require a fixed point on the line and a direction vector parallel to it. Here, both are naturally obtained from the given points.

We choose point \(A(-1, 0, 1)\) as the fixed point on the line. Its position vector is:

\[ \vec{a} = -\hat{i} + 0\hat{j} + \hat{k}. \]

To find a direction vector \(\vec{b}\) for the line, we consider the vector from point \(A\) to point \(B\), given by:

\[ \vec{b} = \overrightarrow{AB} = (2 - (-1))\hat{i} + (3 - 0)\hat{j} + (4 - 1)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}. \]

This direction vector may be simplified without changing the line’s direction, since any non-zero scalar multiple of a direction vector defines the same line. Dividing each component by 3, we obtain the simplified direction vector:

\[ \vec{b} = \hat{i} + \hat{j} + \hat{k}. \]

Therefore, the vector equation of the line is:

\[ \vec{r} = \vec{a} + \lambda \vec{b} = (-\hat{i} + \hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k}), \quad \lambda \in \mathbb{R}. \]

This expresses the position vector \(\vec{r}\) of any point on the line as a linear combination of the fixed vector \(\vec{a}\) and the direction vector \(\vec{b}\), controlled by the real parameter \(\lambda\).

To obtain the Cartesian form, we express the vector equation component-wise. Writing

\[ \vec{r} = x\hat{i} + y\hat{j} + z\hat{k}, \]

we equate:

\[ x\hat{i} + y\hat{j} + z\hat{k} = (-\hat{i} + \hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k}). \]

Comparing components, we obtain the parametric equations:

\[ \begin{aligned} x &= -1 + \lambda, \\ y &= 0 + \lambda = \lambda, \\ z &= 1 + \lambda. \end{aligned} \]

Eliminating the parameter \(\lambda\) from these expressions gives the symmetric (Cartesian) form:

\[ \frac{x + 1}{1} = \frac{y}{1} = \frac{z - 1}{1}, \quad \text{or simply} \quad x + 1 = y = z - 1. \]

Thus, the line passing through \(A(-1, 0, 1)\) and \(B(2, 3, 4)\) has the required vector and Cartesian equations as derived above.

Example

Find the coordinates of a point lying on the line

\[ \frac{x - 1}{2} = \frac{y + 1}{-2} = \frac{z}{1} \]

that is at a distance of 9 units from the point \((1,\ -1,\ 0)\).

Solution:
Note that the point \((1,\ -1,\ 0)\) lies on the line, as seen from the numerators of the symmetric form.

Let the common ratio be \(\lambda\). Then the coordinates of any point on the line are:

\[ x = 2\lambda + 1,\quad y = -2\lambda - 1,\quad z = \lambda. \]

We now compute the distance between this point and \((1,\ -1,\ 0)\). The distance is:

\[ \sqrt{(2\lambda + 1 - 1)^2 + (-2\lambda - 1 + 1)^2 + (\lambda - 0)^2} = \sqrt{(2\lambda)^2 + (-2\lambda)^2 + \lambda^2} = \sqrt{9\lambda^2} = 3|\lambda|. \]

Set this equal to 9:

\[ 3|\lambda| = 9 \implies |\lambda| = 3 \implies \lambda = \pm 3. \]

Substitute back to find the two points.

For \(\lambda = 3\):

\[ x = 7,\quad y = -7,\quad z = 3 \quad \Rightarrow \quad (7,\ -7,\ 3). \]

For \(\lambda = -3\):

\[ x = -5,\quad y = 5,\quad z = -3 \quad \Rightarrow \quad (-5,\ 5,\ -3). \]

Hence, the required points are

\[ \boxed{(7,\ -7,\ 3)\quad \text{and} \quad (-5,\ 5,\ -3)}. \]

Normalized Symmetric Form

When expressing a straight line in three-dimensional space, the symmetric form using direction ratios is commonly written as:

\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}, \]

where \((x_1, y_1, z_1)\) is a point on the line, and \((a, b, c)\) are direction ratios of a vector along the line.

This form, while valid, can be further refined by replacing the direction ratios with direction cosines, which are normalized components of a unit vector along the direction of the line. This leads to what is called the normalized symmetric form.

Let us define the direction cosines \((l, m, n)\) corresponding to the direction ratios \((a, b, c)\). These are given by:

\[ l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \quad m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \quad n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}. \]

These satisfy the identity:

\[ l^2 + m^2 + n^2 = 1. \]

Substituting \(l, m, n\) into the symmetric form yields:

\[ \frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}, \]

which is the normalized symmetric form of the line. It describes the same line, but now the direction vector has unit length.

Let the common value of the three ratios be \(\lambda\). Then we obtain the parametric form:

\[ x = x_1 + \lambda l, \quad y = y_1 + \lambda m, \quad z = z_1 + \lambda n. \]

This form has a key geometric feature: the parameter \(\lambda\) directly represents the signed distance between the general point \((x, y, z)\) and the fixed point \((x_1, y_1, z_1)\). To see this, compute the distance between \((x, y, z)\) and \((x_1, y_1, z_1)\):

\[ \sqrt{(x - x_1)^2 + (y - y_1)^2 + (z - z_1)^2} = \sqrt{(\lambda l)^2 + (\lambda m)^2 + (\lambda n)^2} = |\lambda| \cdot \sqrt{l^2 + m^2 + n^2}. \]

Since \(l^2 + m^2 + n^2 = 1\), this becomes:

\[ \text{Distance} = |\lambda|. \]

Thus, in the normalized symmetric form, the parameter \(\lambda\) is not arbitrary—it has direct geometric meaning: it's modulus is the exact distance from the point \((x_1, y_1, z_1)\) along the line to any other point \((x, y, z)\). This makes the normalized form both elegant and powerful, particularly when measuring distance along the line.

Example

Given the line

\[ \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 3}{-2}, \]

find the points on the line that lie at a distance of 3 units from the point \((1,\ 2,\ 3)\).

Solution:

This line is already in symmetric form with direction ratios

\[ (2,\ 1,\ -2). \]

We first normalize these direction ratios to obtain direction cosines. The magnitude of the direction vector is:

\[ \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3. \]

Hence, the direction cosines are:

\[ l = \frac{2}{3},\quad m = \frac{1}{3},\quad n = \frac{-2}{3}. \]

Substituting into the symmetric form, the line becomes:

\[ \frac{x - 1}{2/3} = \frac{y - 2}{1/3} = \frac{z - 3}{-2/3}. \]

Let this common ratio be \(\lambda\). Then:

\[ \frac{x - 1}{2/3} = \frac{y - 2}{1/3} = \frac{z - 3}{-2/3} = \lambda. \]

The parametric form is:

\[ x = 1 + \lambda \cdot \frac{2}{3}, \quad y = 2 + \lambda \cdot \frac{1}{3}, \quad z = 3 + \lambda \cdot \left(-\frac{2}{3}\right). \]

Since we are measuring distance from the point \((1, 2, 3)\), and this point appears directly in the equation, we know it lies on the line and corresponds to \(\lambda = 0\). As shown earlier, in the normalized form, the parameter \(\lambda\) directly measures the distance from \((1, 2, 3)\). Therefore, to find points at a distance of 3 units from \((1, 2, 3)\), we simply take:

\[ \lambda = \pm 3. \]

For \(\lambda = 3\):

[ x = 1 + \frac{6}{3} = 3,\quad y = 2 + \frac{3}{3} = 3,\quad z = 3 - \frac{6}{3} = 1, ]
so one point is \((3,\ 3,\ 1)\).

For \(\lambda = -3\):

[ x = 1 - \frac{6}{3} = -1,\quad y = 2 - \frac{3}{3} = 1,\quad z = 3 + \frac{6}{3} = 5, ]
so the other point is \((-1,\ 1,\ 5)\).

Hence, the required points are

\[ \boxed{(3,\ 3,\ 1) \quad \text{and} \quad (-1,\ 1,\ 5)}. \]

To write the equation of a straight line in three-dimensional space, either of the following data is sufficient:

1. One point and a direction vector

   If you are given a fixed point \((x_1, y_1, z_1)\) on the line and a non-zero vector \(\vec{d} = a\hat{i} + b\hat{j} + c\hat{k}\) parallel to the line, then the line is uniquely determined. The vector equation is:

   \[ \vec{r} = \vec{a} + \lambda \vec{d}, \]

   where \(\vec{a} = x_1 \hat{i} + y_1 \hat{j} + z_1 \hat{k}\) is the position vector of the fixed point, and \(\lambda \in \mathbb{R}\).

2. Two distinct points

   If you are given two distinct points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\), then you can construct the direction vector by taking their difference:

   \[ \vec{d} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}. \]

   Now, using this \(\vec{d}\) and the first point \((x_1, y_1, z_1)\), you reduce to the previous case. Hence, two points determine a line because they define both a location and a direction.

Conclusion:

To write the equation of a line, either:

• One point and a direction vector, or
• Two distinct points (from which you obtain both)
  is sufficient.

Unsymmetric Form of a Line

A straight line in space can also be described as the common solution of two linear equations in three variables. This representation is called the unsymmetric form of a line.

Consider the symmetric equation of a line:

\[ \frac{x - 1}{2} = \frac{y + 1}{5} = \frac{z + 1}{2}. \]

Let us isolate two of the expressions at a time to eliminate the parameter and obtain two separate equations.

First, from

\[ \frac{x - 1}{2} = \frac{y + 1}{5}, \]

cross-multiplying gives:

\[ 5(x - 1) = 2(y + 1) \implies 5x - 5 = 2y + 2 \implies 5x - 2y = 7. \tag{1} \]

Next, from

\[ \frac{y + 1}{5} = \frac{z + 1}{2}, \]

cross-multiplying gives:

\[ 2(y + 1) = 5(z + 1) \implies 2y + 2 = 5z + 5 \implies 2y - 5z = 3. \tag{2} \]

So, the line can also be described as the set of all points \((x, y, z)\) that satisfy both equations:

\[ \begin{cases} 5x - 2y = 7, \\ 2y - 5z = 3. \end{cases} \]

This is the unsymmetric form of the line.

In the next chapter, we will formally study planes in 3D space. Each linear equation in \(x, y, z\) like equation (1) or (2) represents a plane. The set of all points satisfying a single equation like \(5x - 2y = 7\) forms an infinite flat surface—a plane. Similarly, the second equation defines another plane.

Now, two planes in space may intersect in a line. This happens when they are not parallel and not coincident. The intersection of two non-parallel planes is always a straight line.

To visualize this, open your notebook and bend it in a V-shape. The two pages represent two planes. The crease where the two pages meet—the book binding—is exactly the line of intersection. That is the straight line common to both pages (planes), and this is how two linear equations can define a line.

Hence, in the unsymmetric form, we write a line as a pair of simultaneous linear equations:

\[ \begin{aligned} a_1x + b_1y + c_1z + d_1 &= 0, \\ a_2x + b_2y + c_2z + d_2 &= 0. \end{aligned} \]

Each equation represents a plane. The line is the set of points that lie on both planes simultaneously, i.e., their intersection.

Converting an Unsymmetric Form to Symmetric Form

While converting a symmetric form into unsymmetric form is theoretically valid, it is rarely useful in practice. On the other hand, converting an unsymmetric form—that is, a pair of linear equations in \(x, y, z\)—into symmetric form is extremely useful. This is because once we have the symmetric form, we can apply all standard tools and results about direction ratios, direction cosines, distance from a point to the line, angle between lines, and so on.

Let us understand the method with an example.

Example:
Convert the unsymmetric form

\[ x + y + z = 1, \quad 2x - y + z = 3 \]

into symmetric form.

Solution:

We are given two linear equations. The line is the common solution set of these equations.

To convert into symmetric form, we fix one variable arbitrarily as a parameter. Let us set:

\[ x = \lambda, \quad \lambda \in \mathbb{R}. \]

Substitute this into the two given equations to reduce the system to two equations in \(y\) and \(z\).

From the first equation:

\[ x + y + z = 1 \implies \lambda + y + z = 1 \implies y + z = 1 - \lambda \tag{1} \]

From the second equation:

\[ 2x - y + z = 3 \implies 2\lambda - y + z = 3 \implies -y + z = 3 - 2\lambda \tag{2} \]

Now, add (1) and (2):

\[ (y + z) + (-y + z) = (1 - \lambda) + (3 - 2\lambda) \Rightarrow 2z = 4 - 3\lambda \Rightarrow z = \frac{4 - 3\lambda}{2}. \]

Subtract (1) from (2):

\[ (-y + z) - (y + z) = (3 - 2\lambda) - (1 - \lambda) \Rightarrow -2y = 2 - \lambda \Rightarrow y = \frac{2 - \lambda}{2}. \]

Thus, the general point \((x, y, z)\) on the line is:

\[ x = \lambda, \quad y = \frac{2 - \lambda}{2}, \quad z = \frac{4 - 3\lambda}{2}. \]

This gives the parametric form. To convert to symmetric form, we eliminate \(\lambda\) between these equations. First, isolate \(\lambda\) in terms of \(y\) and \(z\).

From \(y = \frac{2 - \lambda}{2} \Rightarrow \lambda = 2 - 2y\),
and from \(z = \frac{4 - 3\lambda}{2} \Rightarrow \lambda = \frac{4 - 2z}{3}\).

Also, \(x = \lambda\), so we can now write:

\[ \lambda = x = 2 - 2y = \frac{4 - 2z}{3}. \]

On rearraging, the symmetric form is:

\[ \boxed{\frac{x}{1} = \frac{2y - 2}{-1} = \frac{2z - 4}{-3}}, \]

or equivalently, you may express it in standard symmetric form as:

\[ \frac{x - 0}{1} = \frac{y - 1}{-1/2} = \frac{z - 2}{-3/2}. \]

Direction ratios are \((1,\ -\tfrac{1}{2},\ -\tfrac{3}{2})\), and the line passes through the point \((0, 1, 2)\).

To simplify the direction ratios and avoid fractional components, we multiply all direction ratios by 2. This operation preserves the direction of the line, since multiplying by a non-zero constant does not change the direction.

Given the symmetric form:

\[ \frac{x - 0}{1} = \frac{y - 1}{-1/2} = \frac{z - 2}{-3/2}, \]

the direction ratios are:

\[ \left(1,\ -\frac{1}{2},\ -\frac{3}{2}\right). \]

Multiplying each by 2 gives:

\[ (2,\ -1,\ -3). \]

Hence, an equivalent symmetric form of the same line—now using integer direction ratios—is:

\[ \frac{x}{2} = \frac{y - 1}{-1} = \frac{z - 2}{-3}. \]

This form is cleaner to work with and makes the geometry easier to interpret, while still describing the same line through the point \((0,\ 1,\ 2)\).

Interaction of a Point with a Line

In Cartesain Forms

Let \(P(x_0,\ y_0,\ z_0)\) be a point in space, and let the line be given in symmetric form as:

\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}, \]

which passes through the point \((x_1,\ y_1,\ z_1)\) and has direction ratios \((a,\ b,\ c)\).

There are three important quantities we often wish to compute:

1. The perpendicular distance from the point \(P\) to the line.
2. The foot of the perpendicular, say \(N\), dropped from point \(P\) onto the line.
3. The image of the point \(P\) in the line, denoted by \(Q\), which lies on the line and satisfies \(N\) as the midpoint of segment \(PQ\).

---

Let us proceed to compute these quantities step by step.

Let \(N\) be the foot of the perpendicular from \(P\) to the line. Since \(N\) lies on the line, its coordinates can be written in parametric form:

\[ (x,\ y,\ z) = (a\lambda + x_1,\ b\lambda + y_1,\ c\lambda + z_1) \quad \text{for some } \lambda \in \mathbb{R}. \]

Now the vector \(\vec{PN}\), having direction ratios:

\[ (a\lambda + x_1 - x_0,\ b\lambda + y_1 - y_0,\ c\lambda + z_1 - z_0), \]

must be perpendicular to the line, whose direction ratios are \((a, b, c)\). Therefore, their dot product must be zero:

\[ (a\lambda + x_1 - x_0)\cdot a + (b\lambda + y_1 - y_0)\cdot b + (c\lambda + z_1 - z_0)\cdot c = 0. \]

Expanding:

\[ \lambda(a^2 + b^2 + c^2) + a(x_1 - x_0) + b(y_1 - y_0) + c(z_1 - z_0) = 0. \]

From this equation, \(\lambda\) is easily determined. Once \(\lambda\) is known, we substitute it back into:

\[ N = (a\lambda + x_1,\ b\lambda + y_1,\ c\lambda + z_1), \]

to obtain the coordinates of the foot \(N\).

The perpendicular distance from \(P\) to the line is then simply the Euclidean distance between the points \(P(x_0, y_0, z_0)\) and \(N\), using the distance formula:

\[ \text{Distance} = \sqrt{(x_0 - x_N)^2 + (y_0 - y_N)^2 + (z_0 - z_N)^2}. \]

We do not derive a general formula for this, as the expression becomes complex and is not easy to memorize or apply. It is far better to remember the method.

Finally, to find the image point \(Q\) of \(P\) in the line, we use the fact that the foot \(N\) is the midpoint of segment \(PQ\). So if \(Q = (x', y', z')\), then:

\[ \begin{aligned} x' &= 2x_N - x_0, \\ y' &= 2y_N - y_0, \\ z' &= 2z_N - z_0. \end{aligned} \]

This gives the coordinates of the image point.

In Vector forms

Let a line be given in vector form as:

\[ \vec{r} = \vec{a} + \lambda \vec{b}, \]

where \(\vec{a}\) is the position vector of a fixed point \(A\) on the line, and \(\vec{b}\) is a non-zero vector in the direction of the line.

Let \(\vec{p}\) be the position vector of a point \(P\) in space. Then the vector \(\vec{AP}\) from the point \(A\) on the line to the point \(P\) is:

\[ \vec{AP} = \vec{p} - \vec{a}. \]

In coordinate methods, we typically first compute the foot of the perpendicular from \(P\) to the line, and then use the distance formula. However, using vector methods, we can avoid finding the foot altogether.

---

The key idea is to compute the perpendicular component of \(\vec{AP}\) with respect to the direction \(\vec{b}\). This component is the part of \(\vec{AP}\) orthogonal to \(\vec{b}\), and its magnitude gives the perpendicular distance from the point to the line.

This is given by:

\[ \text{Distance} = |\vec{AP} \times \hat{b}| = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}. \]

In terms of \(\vec{p}\) and \(\vec{a}\), the formula becomes:

\[ \boxed{ \text{Perpendicular distance from point } P \text{ to the line} = \frac{|(\vec{p} - \vec{a}) \times \vec{b}|}{|\vec{b}|}. } \]

This direct formula eliminates the need to find the foot of the perpendicular and gives the required distance using only vector operations.

Interaction of Two Lines

There are many things we are interested in when two lines are given in space. Do they intersect? Are they parallel or perpendicular? What is the angle between them? What is the shortest distance between them? These questions form the basis of understanding how two lines relate to each other in three-dimensional geometry.

Angle Between Two Straight Lines

Let two lines be given in vector form as:

\[ \vec{r} = \vec{a}_1 + \lambda \vec{b}_1, \quad \vec{r} = \vec{a}_2 + \mu \vec{b}_2, \]

where \(\vec{b}_1\) and \(\vec{b}_2\) are non-zero direction vectors of the lines. The acute angle \(\theta\) between the two lines is defined as the angle between their direction vectors.

Hence, the angle \(\theta\) is given by:

\[ \cos\theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1||\vec{b}_2|}. \]

We take the modulus of the dot product to ensure that \(\theta\) is the acute angle between the lines.

Condition for Two Lines to Be Parallel

Two lines are said to be parallel if the angle between them is zero. This happens exactly when their direction vectors are collinear.

Let the lines be given in vector form:

\[ \vec{r} = \vec{a}_1 + \lambda \vec{b}_1, \quad \vec{r} = \vec{a}_2 + \mu \vec{b}_2. \]

Then the lines are parallel if and only if:

\[ \vec{b}_1 = k \vec{b}_2 \quad \text{for some scalar } k \in \mathbb{R}. \]

This means the two direction vectors are scalar multiples of each other, i.e., they have the same or exactly opposite direction.

If the lines are given in Cartesian symmetric form:

\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}, \quad \frac{x - x_2}{p} = \frac{y - y_2}{q} = \frac{z - z_2}{r}, \]

then the lines are parallel if and only if their direction ratios are proportional:

\[ \frac{a}{p} = \frac{b}{q} = \frac{c}{r}. \]

Condition for Two Lines to Be Perpendicular

Let two lines in space be given in vector form:

\[ \vec{r} = \vec{a}_1 + \lambda \vec{b}_1, \quad \vec{r} = \vec{a}_2 + \mu \vec{b}_2, \]

where \(\vec{b}_1\) and \(\vec{b}_2\) are non-zero direction vectors of the lines.

Then the lines are perpendicular if and only if their direction vectors are perpendicular, that is:

\[ \vec{b}_1 \cdot \vec{b}_2 = 0. \]

In terms of Cartesian symmetric form, suppose the lines are:

\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}, \quad \frac{x - x_2}{p} = \frac{y - y_2}{q} = \frac{z - z_2}{r}, \]

then the condition for the lines to be perpendicular is:

\[ ap + bq + cr = 0. \]

Distance Between Two Parallel Lines

In vector form, the distance between two parallel lines is very straightforward to compute.

Let the two lines be:

\[ L_1: \vec{r} = \vec{a}_1 + \lambda \vec{b}, \quad L_2: \vec{r} = \vec{a}_2 + \mu \vec{b}, \]

where \(\vec{b}\) is a common direction vector, indicating that the lines are parallel. Let \(A_1\) and \(A_2\) be points on \(L_1\) and \(L_2\) respectively, with position vectors \(\vec{a}_1\) and \(\vec{a}_2\).

To find the distance between these lines, we consider the vector \(\overrightarrow{A_2A_1} = \vec{a}_1 - \vec{a}_2\). The shortest distance between the lines is the component of this vector perpendicular to the direction \(\vec{b}\). That is, we project \(\vec{a}_1 - \vec{a}_2\) onto a direction perpendicular to \(\vec{b}\).

Let \(\hat{\vec{b}}\) be the unit vector along \(\vec{b}\), given by:

\[ \hat{b} = \frac{\vec{b}}{|\vec{b}|}. \]

Then the required perpendicular distance is the magnitude of the cross product:

\[ \text{Distance} = \left| \overrightarrow{A_2A_1} \times \hat{b}\right| = \left|(\vec{a}_1 - \vec{a}_2) \times \hat{b}\right| = \left|\frac{(\vec{a}_1 - \vec{a}_2) \times \vec{b}}{|\vec{b}|}\right|. \]

Hence, the distance between the two parallel lines is:

\[ \boxed{ \text{Distance} = \frac{|(\vec{a}_1 - \vec{a}_2) \times \vec{b}|}{|\vec{b}|}. } \]

Coincident Lines

Two lines are said to be coincident if they lie exactly on top of each other — that is, they share all points in common.

Let the two lines be given in vector form as:

\[ L_1: \vec{r} = \vec{a}_1 + \lambda \vec{b}_1, \quad L_2: \vec{r} = \vec{a}_2 + \mu \vec{b}_2. \]

Then the lines are coincident if and only if the following two conditions are satisfied:

1. Direction vectors are collinear:

   The direction vectors must be scalar multiples of each other:

   \[ \vec{b}_1 = k \vec{b}_2 \quad \text{for some } k \in \mathbb{R}. \]

2. A point on one line lies on the other:

   The point \(\vec{a}_1\) on \(L_1\) must lie on \(L_2\). That is, there exists some \(\mu \in \mathbb{R}\) such that:

   \[ \vec{a}_1 = \vec{a}_2 + \mu \vec{b}_2. \]

If both conditions hold, then the two lines coincide.

Check Whether Two Lines Are Coincident

Consider the two lines:

\[ L_1: \frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z + 2}{-3}, \quad L_2: \frac{x - 3}{-2} = \frac{y + 2}{1} = \frac{z + 5}{3}. \]

We proceed to check whether \(L_1\) and \(L_2\) are coincident.

Step 1: Check Direction Vectors

The direction ratios of \(L_1\) are:

\[ (2,\ -1,\ -3), \]

and of \(L_2\):

\[ (-2,\ 1,\ 3). \]

Clearly,

\[ \frac{2}{-2} = \frac{-1}{1} = \frac{-3}{3} = -1, \]

so the direction vectors are scalar multiples of each other. Hence, the lines are parallel.

Step 2: Check If a Point on One Line Lies on the Other

Take the point \((1,\ -1,\ -2)\) on line \(L_1\). We check whether this point lies on line \(L_2\):

\[ \frac{1 - 3}{-2} = \frac{-1 + 2}{1} = \frac{-2 + 5}{3} \Rightarrow \frac{-2}{-2} = \frac{1}{1} = \frac{3}{3} = 1. \]

All three expressions are equal, so the point lies on \(L_2\).

Conclusion:

The lines are parallel and share a common point. Therefore,

\[ \boxed{\text{The two lines are coincident.}} \]

Skew Lines

In two dimensions, any pair of straight lines falls into one of three possible configurations:
1. The lines intersect at a unique point.
2. The lines are parallel but not coincident, and hence do not intersect.
3. The lines are coincident, and thus intersect in infinitely many points.

However, in three-dimensional space, there is one more possible orientation for two lines, which does not arise in the plane.

To understand this, begin with two parallel lines \(L_1\) and \(L_2\) in space, separated by a fixed positive distance \(d\). Take a point \(P\) on \(L_1\) and a point \(Q\) on \(L_2\) such that the segment \(PQ\) is perpendicular to both lines.

Now imagine rotating line \(L_2\) slightly about the axis defined by \(PQ\). After rotation, line \(L_2\) is no longer parallel to \(L_1\), and since the direction has changed but not the position of \(PQ\), the two lines also do not intersect.

This new configuration describes a situation where the lines are neither intersecting nor parallel. Such lines are said to be skew.

Hence, two lines in three-dimensional space are called skew lines if they do not intersect and are not parallel. Equivalently, skew lines are lines that do not lie in the same plane.

There exists a unique plane through \(P\) that is perpendicular to the segment \(PQ\) and contains the line \(L_1\). Similarly, there is a unique plane through \(Q\), also perpendicular to \(PQ\), containing the line \(L_2\).

These two planes are parallel to each other, since they are both perpendicular to the same line segment \(PQ\) and contain lines (\(L_1\) and \(L_2\)) that are not parallel and do not intersect. Thus, skew lines can be thought of as lying in distinct parallel planes, each perpendicular to the common perpendicular segment joining them.

Shortest Distance Between Two Skew Lines

Let two straight lines be given in vector form:

\[ L_1: \vec{r} = \vec{a}_1 + \lambda \vec{b}_1, \quad L_2: \vec{r} = \vec{a}_2 + \mu \vec{b}_2, \]

where \(\vec{a}_1\) and \(\vec{a}_2\) are position vectors of points \(A_1\) and \(A_2\) on lines \(L_1\) and \(L_2\) respectively, and \(\vec{b}_1,\ \vec{b}_2\) are direction vectors of the lines.

Let \(P\) be a point on \(L_1\) and \(Q\) a point on \(L_2\). Then the vector \(\vec{PQ} = \vec{a}_2 + \mu \vec{b}_2 - (\vec{a}_1 + \lambda \vec{b}_1)\) joins a general point on one line to a general point on the other.

We are interested in the shortest possible value of \(|\vec{PQ}|\) — this occurs when the segment \(PQ\) is perpendicular to both lines. That is, the direction vector of \(PQ\) is perpendicular to both \(\vec{b}_1\) and \(\vec{b}_2\).

Thus, the direction of the segment \(PQ\) in this minimal configuration must be parallel to the vector:

\[ \vec{b}_1 \times \vec{b}_2, \]

which is perpendicular to both \(\vec{b}_1\) and \(\vec{b}_2\).

Let us now consider the vector \(\vec{A_1A_2} = \vec{a}_2 - \vec{a}_1\). The shortest distance between the two lines is the projection of this vector onto the direction \(\vec{b}_1 \times \vec{b}_2\).

Hence, the shortest distance \(D\) is:

\[ D = \left|\frac{(\vec{a}_1 - \vec{a}_2) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|}\right|. \]

Thus,

\[ \boxed{ \text{Shortest distance between the two skew lines} = \left| \frac{(\vec{a}_1 - \vec{a}_2) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|. } \]

This is the length of the line segment \(PQ\), which is perpendicular to both lines and joins them at the closest possible approach.

Example

Find the shortest distance between the skew lines:

\[ L_1: \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{2}, \quad L_2: \frac{x + 1}{-1} = \frac{y - 1}{2} = \frac{z - 3}{1}. \]

Solution:

From the given equations, we take:
Point on \(L_1\):

\[ \vec{a}_1 = \hat{i} - \hat{j} + 0\hat{k}. \]

Point on \(L_2\):

\[ \vec{a}_2 = -\hat{i} + \hat{j} + 3\hat{k}. \]

Direction vector of \(L_1\):

\[ \vec{b}_1 = 2\hat{i} + 3\hat{j} + 2\hat{k}, \quad \text{(from the denominators)} \]

Direction vector of \(L_2\):

\[ \vec{b}_2 = -\hat{i} + 2\hat{j} + \hat{k}. \]

Compute the vector \(\vec{a}_1 - \vec{a}_2\):

\[ \vec{a}_1 - \vec{a}_2 = (\hat{i} - (-\hat{i})) + (-\hat{j} - \hat{j}) + (0\hat{k} - 3\hat{k}) = 2\hat{i} - 2\hat{j} - 3\hat{k}. \]

Now compute \(\vec{b}_1 \times \vec{b}_2\):

\[ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 2 \\ -1 & 2 & 1 \end{vmatrix} = \hat{i}(3 \cdot 1 - 2 \cdot 2) - \hat{j}(2 \cdot 1 - 2 \cdot (-1)) + \hat{k}(2 \cdot 2 - 3 \cdot (-1)) \]

\[ = \hat{i}(3 - 4) - \hat{j}(2 + 2) + \hat{k}(4 + 3) = -\hat{i} - 4\hat{j} + 7\hat{k}. \]

So,

\[ \vec{b}_1 \times \vec{b}_2 = -\hat{i} - 4\hat{j} + 7\hat{k}. \]

Now compute the scalar triple product:

\[ (\vec{a}_1 - \vec{a}_2) \cdot (\vec{b}_1 \times \vec{b}_2) = (2\hat{i} - 2\hat{j} - 3\hat{k}) \cdot (-\hat{i} - 4\hat{j} + 7\hat{k}) \]

\[ = 2 \cdot (-1) + (-2) \cdot (-4) + (-3) \cdot 7 = -2 + 8 - 21 = -15. \]

Magnitude of the cross product:

\[ |\vec{b}_1 \times \vec{b}_2| = \sqrt{(-1)^2 + (-4)^2 + 7^2} = \sqrt{1 + 16 + 49} = \sqrt{66}. \]

Hence, the shortest distance between the lines is:

\[ \boxed{ \text{Distance} = \left| \frac{-15}{\sqrt{66}} \right| = \frac{15}{\sqrt{66}}. } \]

Shortest Distance betwee Skew Lines vs Parallel Lines

Before applying the formula for the shortest distance between two skew lines, it is essential to first verify that the lines are not parallel. The formula:

\[ \text{Distance} = \left| \frac{(\vec{a}_1 - \vec{a}_2) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| \]

is valid only when \(\vec{b}_1 \times \vec{b}_2 \ne \vec{0}\), i.e., the direction vectors are not collinear.

If the lines are parallel, that is, \(\vec{b}_1 = k\vec{b}_2\) for some scalar \(k\), then \(\vec{b}_1 \times \vec{b}_2 = \vec{0}\), and the above formula becomes undefined.

In that case, we must use the shortest distance formula for parallel lines, which is:

\[ \text{Distance} = \left| \frac{(\vec{a}_1 - \vec{a}_2) \times \vec{b}}{|\vec{b}|} \right|, \]

where \(\vec{b}\) is the common direction vector of the two lines.

Hence, always check for parallelism before applying the skew line distance formula.

Equation of the Line of Shortest Distance

Given two skew lines in space, such as:

\[ L_1: \vec{r} = \vec{a}_1 + \lambda \vec{b}_1, \quad L_2: \vec{r} = \vec{a}_2 + \mu \vec{b}_2, \]

the line of shortest distance is the line segment joining a point \(P\) on \(L_1\) to a point \(Q\) on \(L_2\), such that the segment \(PQ\) is perpendicular to both lines. This line is sometimes referred to as the common perpendicular.

The direction of this line is simple to obtain. Since \(PQ\) is perpendicular to both \(\vec{b}_1\) and \(\vec{b}_2\), a vector along the line of shortest distance is given by:

\[ \vec{d} = \vec{b}_1 \times \vec{b}_2. \]

However, to write the equation of a line, we need both a direction vector and a point on the line. The direction vector we already have. To find a point that lies on the line of shortest distance, we must determine the foot points \(P\) and \(Q\) on the two lines such that \(\vec{PQ}\) is perpendicular to both \(\vec{b}_1\) and \(\vec{b}_2\). This requires solving a system of equations.

Let us learn how to do this with an example.

Example: Find the Equation of the Line of Shortest Distance Between Two Skew Lines

Let the two skew lines be given in symmetric form:

\[ L_1: \frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z - 1}{1}, \quad L_2: \frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}. \]

The direction vector of line \(L_1\) is

\[ \vec{b}_1 = 2\hat{i} - \hat{j} + \hat{k}, \]

and of line \(L_2\) is

\[ \vec{b}_2 = \hat{i} + 2\hat{j} + 3\hat{k}. \]

The line of shortest distance is perpendicular to both \(L_1\) and \(L_2\), so its direction vector is the cross product \(\vec{b}_1 \times \vec{b}_2\):

\[ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}((-1)(3) - (1)(2)) - \hat{j}((2)(3) - (1)(1)) + \hat{k}((2)(2) - (-1)(1)) \]

\[ = \hat{i}(-3 - 2) - \hat{j}(6 - 1) + \hat{k}(4 + 1) = -5\hat{i} -5\hat{j} + 5\hat{k}. \]

Removing the common scalar multiple, we choose the direction vector:

\[ \vec{d} = \hat{i} + \hat{j} - \hat{k}. \]

Let the line of shortest distance intersect \(L_1\) at point \(P\) and \(L_2\) at point \(Q\). These can be written in terms of parameters \(\lambda\) and \(\mu\) as:

\[ P = (2\lambda + 1,\ -\lambda - 1,\ \lambda + 1), \quad Q = (\mu - 1,\ 2\mu + 2,\ 3\mu + 3). \]

The direction vector \(\vec{PQ} = \vec{P} - \vec{Q}\) has components:

\[ \begin{aligned} x\text{-component:} &\quad (2\lambda + 1) - (\mu - 1) = 2\lambda - \mu + 2, \\ y\text{-component:} &\quad (-\lambda - 1) - (2\mu + 2) = -\lambda - 2\mu - 3, \\ z\text{-component:} &\quad (\lambda + 1) - (3\mu + 3) = \lambda - 3\mu - 2. \end{aligned} \]

Since this direction vector is along the line of shortest distance, it must be proportional to \((1,\ 1,\ -1)\). Thus,

\[ \frac{2\lambda - \mu + 2}{1} = \frac{-\lambda - 2\mu - 3}{1} = \frac{\lambda - 3\mu - 2}{-1}. \]

Equating the first and second expressions:

\[ 2\lambda - \mu + 2 = -\lambda - 2\mu - 3 \implies 3\lambda + \mu + 5 = 0. \tag{1} \]

Equating the second and third expressions:

\[ -\lambda - 2\mu - 3 = -(\lambda - 3\mu - 2) \Rightarrow -\lambda - 2\mu - 3 = -\lambda + 3\mu + 2 \Rightarrow -5\mu = 5 \Rightarrow \mu = -1. \tag{2} \]

Substituting \(\mu = -1\) into equation (1):

\[ 3\lambda -1 + 5 = 0 \Rightarrow 3\lambda = -4 \Rightarrow \lambda = -\frac{4}{3}. \]

We now substitute \(\mu = -1\) into coordinates of point \(Q\):

\[ Q = (-1 - 1,\ 2(-1) + 2,\ 3(-1) + 3) = (-2,\ 0,\ 0). \]

Thus, the line of shortest distance passes through the point \((-2,\ 0,\ 0)\) and is directed along \(\vec{d} = \hat{i} + \hat{j} - \hat{k}\). Hence, its symmetric form is:

\[ \frac{x + 2}{1} = \frac{y}{1} = \frac{z}{-1}. \]

\(\blacksquare\)

Intersection of Two Lines in 3D

In two dimensions, any two non-parallel lines always intersect at a point. However, in three dimensions, the situation is more delicate. Even if two lines are not parallel, they may still not intersect. This possibility arises because lines in 3D space can miss each other while pointing in different directions and lying in different planes — such lines are called skew lines.

This raises a natural question: How can we determine whether two non-parallel lines in 3D intersect?

Suppose the lines are given in vector form as:

\[ L_1: \vec{r} = \vec{a}_1 + \lambda \vec{b}_1, \quad L_2: \vec{r} = \vec{a}_2 + \mu \vec{b}_2, \]

where \(\vec{b}_1\) and \(\vec{b}_2\) are not collinear, so the lines are not parallel.

We know that for skew lines, the shortest distance between them is given by:

\[ \text{Distance} = \left| \frac{(\vec{a}_1 - \vec{a}_2) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|. \]

Therefore, if the shortest distance between the two lines is zero, they must intersect. This happens exactly when:

\[ (\vec{a}_1 - \vec{a}_2) \cdot (\vec{b}_1 \times \vec{b}_2) = 0. \]

This condition simply says that the vector \(\vec{a}_1 - \vec{a}_2\) is perpendicular to the vector \(\vec{b}_1 \times \vec{b}_2\), or equivalently, that the three vectors \(\vec{a}_1 - \vec{a}_2\), \(\vec{b}_1\), and \(\vec{b}_2\) are coplanar.

This geometric interpretation can be restated using the scalar triple product:

\[ [\vec{a}_1 - \vec{a}_2,\ \vec{b}_1,\ \vec{b}_2] = 0. \]

Hence, two non-parallel lines intersect if and only if the scalar triple product of \(\vec{a}_1 - \vec{a}_2,\ \vec{b}_1,\ \vec{b}_2\) is zero. This expresses the condition that these three vectors lie in the same plane — that is, the lines are coplanar and hence intersect.

\[ \boxed{ \text{Two non-parallel lines intersect} \iff [\vec{a}_1 - \vec{a}_2,\ \vec{b}_1,\ \vec{b}_2] = 0. } \]

Coplanarity of Two Lines

Let two lines in space be given in vector form as:

\[ L_1: \vec{r} = \vec{a}_1 + \lambda \vec{b}_1, \quad L_2: \vec{r} = \vec{a}_2 + \mu \vec{b}_2. \]

We say that the lines \(L_1\) and \(L_2\) are coplanar if there exists a single plane in which both lines lie completely.

This occurs in exactly two situations:

1. The lines are parallel: that is, the direction vectors \(\vec{b}_1\) and \(\vec{b}_2\) are collinear. In this case, any plane containing one line and a point on the other line contains both lines.

2. The lines intersect: that is, they meet at a common point. In this case, both lines lie in the unique plane determined by the point of intersection and the two direction vectors.

Hence, two lines in space are coplanar if and only if they are either parallel or intersecting. If neither is true — that is, they are not parallel and do not meet — then they are skew and not coplanar.

\[ \boxed{ \text{Two lines are coplanar} \iff \text{they are parallel or intersecting}. } \]

Example

Determine Whether Two Lines in Space Intersect

Consider the lines:

\[ L_1: \frac{x - 1}{2} = \frac{y + 3}{1} = \frac{z - 1}{-2}, \quad L_2: \frac{x - 5}{-3} = \frac{y + 1}{2} = \frac{z + 3}{-1}. \]

To determine whether these lines intersect, we extract from the symmetric form:

A point on line \(L_1\) is

\[ \vec{a}_1 = \hat{i} - 3\hat{j} + \hat{k}, \]

and its direction vector is

\[ \vec{b}_1 = 2\hat{i} + \hat{j} - 2\hat{k}. \]

A point on line \(L_2\) is

\[ \vec{a}_2 = 5\hat{i} - \hat{j} - 3\hat{k}, \]

and its direction vector is

\[ \vec{b}_2 = -3\hat{i} + 2\hat{j} - \hat{k}. \]

Since the direction vectors \(\vec{b}_1\) and \(\vec{b}_2\) are not scalar multiples of each other, the lines are not parallel. Therefore, the lines intersect if and only if they are coplanar, i.e.,

\[ [\vec{a}_1 - \vec{a}_2,\ \vec{b}_1,\ \vec{b}_2] = 0. \]

We compute:

\[ \vec{a}_1 - \vec{a}_2 = (\hat{i} - 5\hat{i}) + (-3\hat{j} + \hat{j}) + (\hat{k} + 3\hat{k}) = -4\hat{i} - 2\hat{j} + 4\hat{k}. \]

Now compute the scalar triple product:

\[ [\vec{a}_1 - \vec{a}_2,\ \vec{b}_1,\ \vec{b}_2] = \begin{vmatrix} -4 & -2 & 4 \\ 2 & 1 & -2 \\ -3 & 2 & -1 \end{vmatrix}. \]

We expand:

\[ = -4 \begin{vmatrix} 1 & -2 \\ 2 & -1 \end{vmatrix} + 2 \begin{vmatrix} 2 & -2 \\ -3 & -1 \end{vmatrix} + 4 \begin{vmatrix} 2 & 1 \\ -3 & 2 \end{vmatrix}. \]

\[ = -4(1 \cdot (-1) - 2 \cdot (-2)) + 2(2 \cdot (-1) - (-3) \cdot (-2)) + 4(2 \cdot 2 - (-3) \cdot 1) \]

\[ = -4(-1 + 4) + 2(-2 - 6) + 4(4 + 3) = -4(3) + 2(-8) + 4(7) = -12 -16 + 28 = 0. \]

Since the scalar triple product is zero, the three vectors are coplanar. Therefore, the lines intersect.

\[ \boxed{\text{The given lines intersect.}} \]