Question

A pair of lines in \(\mathbb{R}^{3}\) are said to be skew if they are neither parallel nor intersecting. Determine whether the following pairs of lines are parallel, intersecting, or skew. If the lines intersect. determine the point(s) of intersection. $$\begin{aligned} &\mathbf{r}(t)=\langle 3+4 t, 1-6 t, 4 t\rangle;\\\ &\mathbf{R}(s)=\langle-2 s, 5+3 s, 4-2 s\rangle \end{aligned}$$

Short answer

Answer: The given pair of lines are parallel and intersecting, which means they are coplanar. The point of intersection is \(\langle 3, 1, 0 \rangle\).

Step-by-step solution

Check if the lines are parallel

The direction vector of the first line, \(\mathbf{r}(t)\), is given by \(\langle 4t,-6t,4t \rangle = t\langle 4,-6,4\rangle\). As for the second line, \(\mathbf{R}(s)\), the direction vector is given by \(\langle -2s,3s,-2s \rangle = s\langle -2,3,-2\rangle\). To check if the lines are parallel, we need to see if the direction vectors are scalar multiples of each other. Let's try to find a scalar \(k\) such that: $$k\langle -2,3,-2\rangle = \langle 4,-6,4\rangle .$$ Divide the first components of each vector to find \(k\): \(k = \frac{4}{-2} = -2.\) Then: $$-2\langle -2,3,-2\rangle = \langle 4,-6,4\rangle .$$ We can see that these two vectors are scalar multiples of each other. Therefore, the lines are parallel.

Determine if the lines intersect

Parallel lines either intersect or are skew, and we are not yet sure whether the lines in the given parametric equations intersect or not. To check whether the lines intersect, we need to find a point that exists on both lines. By equating the two equations, we will attempt to find a \(t\) and a \(s\) for which both represent \(\mathbf{a}\), a common point. $$\begin{aligned} &\mathbf{r}(t)=\langle 3+4 t, 1-6 t, 4 t\rangle ;\\ &\mathbf{R}(s)=\langle-2 s, 5+3 s, 4-2 s\rangle . \end{aligned}$$ Equating each component: $$ \begin{aligned} 3 + 4t &= -2s, \\ 1 - 6t &= 5 + 3s, \\ 4t &= 4 - 2s . \end{aligned} $$ Taking the third equation, we get: \(t = 1-s.\) Substitute \(t\) into the first equation: $$3+ 4(1-s) = -2s.$$ Solving for \(s\), we get: \(s=1.\) Now, substitute \(s=1\) back into the equation \(t = 1-s\) to find the value of \(t\): \(t= 1-1\), yielding \(t=0\). Now, plug the values of \(t\) into \(\mathbf{r}(t)\) to find the common point; in this case, the only point is when \(t=0\): $$\begin{aligned} \mathbf{r}(0)=\langle 3, 1, 0 \rangle. \end{aligned}$$ Similarly, we can plug the values of \(s\) into \(\mathbf{R}(s)\) to find \(\mathbf{a}\). In this case, the only point is when \(s=1\): $$\begin{aligned} \mathbf{R}(1)=\langle 3, 1, 0 \rangle. \end{aligned}$$ Since both lines have the common point \(\langle 3, 1, 0 \rangle\), they intersect at this point. The given pair of lines are parallel and intersecting, which means they are coplanar. The point of intersection is \(\langle 3, 1, 0 \rangle\).

Key concepts

Parallel Lines

Parallel lines are fascinating because they follow the same path without ever meeting. Essentially, two lines are parallel if they travel in the same direction and have no tendency to converge or diverge. In mathematical terms, this means their direction vectors are scalar multiples of each other.

For example, in three-dimensional space, consider the direction vector of the line \( \mathbf{r}(t) \) given by \( \langle 4, -6, 4 \rangle \) and the direction vector of the line \( \mathbf{R}(s) \) given by \( \langle -2, 3, -2 \rangle \). To check if these lines are parallel, you would multiply the direction vector of one line by a scalar, \( k \), to see whether it matches the other direction vector.

For these lines, we found \( k = -2 \), making the vectors \( \langle 4, -6, 4 \rangle \) and \( -2 \times \langle -2, 3, -2 \rangle \) equal. This confirms that the lines are indeed parallel. Remember, parallel lines run side by side, but that doesn’t mean they can't meet if they are coplanar, like running on train tracks.

Intersection Point

When two lines share a point, they intersect. To find the intersection point of two lines, you need to find a common set of parameter values where both vectors meet. For lines in three dimensions, this involves solving a set of equations derived from their parametric forms.

Consider the parametric lines:

• \( \mathbf{r}(t) = \langle 3 + 4t, 1 - 6t, 4t \rangle \)
• \( \mathbf{R}(s) = \langle -2s, 5 + 3s, 4 - 2s \rangle \)

To find an intersection, set the individual components equal to each other and solve for \( t \) and \( s \):

• \( 3 + 4t = -2s \)
• \( 1 - 6t = 5 + 3s \)
• \( 4t = 4 - 2s \)

Solve these equations to find \( t = 0 \) and \( s = 1 \), giving a common point \( \langle 3, 1, 0 \rangle \). The presence of this intersection means that the lines intersect and are not skew.

Direction Vectors

Direction vectors serve as the roadmap for lines in space. They point the way a line travels and determine whether lines are parallel, intersecting, or neither (skew).

For a line defined by the vector function \( \mathbf{r}(t) = \langle x_0 + at, y_0 + bt, z_0 + ct \rangle \), the direction vector is \( \langle a, b, c \rangle \). This simple vector denotes the line's inclination and stretch in space. Comparison of these vectors across different lines provides insights into their relationships:

• If direction vectors are scalar multiples, the lines are parallel.
• If they share a common point, they intersect.
• If neither, the lines are skew.

Checking direction vectors is straightforward, involving arithmetic checks for equality using scalar multiplication. In our example, comparing \( \langle 4, -6, 4 \rangle \) to \( \langle -2, 3, -2 \rangle \) involves simply determining if one can be transformed into the other via a consistent scalar, revealing parallelism. These vectors are vital tools in spatial geometry.