Question

Determine whether the lines $$\frac{x-1}{2}=\frac{y+3}{1}=\frac{z-4}{-3} \quad \text { and } \quad \frac{x+3}{4}=\frac{y+4}{1}=\frac{8-z}{4}$$ intersect. Two suggestions: (1) Can you find the intersection point, if any? (2) Consider the distance between the lines.

Short answer

Yes, the lines intersect at the point (1, -3, 4).

Step-by-step solution

- Parametrize the first line

The first line can be parametrized by setting each fraction equal to a parameter, say t. Hence, we get: \[ x = 2t + 1 \] \[ y = t - 3 \] \[ z = -3t + 4 \]

- Parametrize the second line

Similar to the first line, we parametrize the second line using a different parameter, say s: \[ x = 4s - 3 \] \[ y = s - 4 \] \[ z = -4s + 8 \]

- Set equations equal to find intersection

To find the intersection, set the parametrized coordinates equal to each other:\[ 2t + 1 = 4s - 3 \] \[ t - 3 = s - 4 \] \[ -3t + 4 = -4s + 8 \]

- Solve the system of equations for t and s

From the second equation: \[ t - s = -1 \] From the first equation: \[ 2t - 4s = -4 \] From the third equation: \[ -3t + 4s = 4 \] The first two equations can be solved for t and s: \[ t = s - 1 \] \[ 2(s - 1) + 1 = 4s - 3 \] \[ 2s - 1 = 4s - 3 \] \[ -2s = -2 \] \[ s = 1 \] \[ t = 1 - 1 \] \[ t = 0 \] Lastly check if these values satisfy the third equation:\[ -3(0) + 4 = -4(1) + 8 \] \[ 4 = 4 \] Thus, they satisfy.

- Verify intersection point

Substitute t and s back into the parametric equations to find the intersection point: For t=0, the parametric coordinates for the first line are: \[ x = 1 \] \[ y = -3 \] \[ z = 4 \] For s=1, the parametric coordinates for the second line are: \[ x = 1 \] \[ y = -3 \] \[ z = 4 \] Thus, the lines intersect at the point (1, -3, 4).

Key concepts

parametric equations

Parametric equations describe a way of expressing the coordinates of points on a line using a parameter. This is particularly useful in 3D geometry, where lines can be complex. For instance, in the solution provided, the first line is expressed as:

• \( x = 2t + 1 \)
• \( y = t - 3 \)
• \( z = -3t + 4 \)

Here, \( t \) is the parameter that varies, generating different points along the line as it changes. Similarly, the second line is expressed as:

• \( x = 4s - 3 \)
• \( y = s - 4 \)
• \( z = -4s + 8 \)

Using parametric equations is a common method in 3D geometry to create a clear and manageable form of lines for further analysis, such as finding intersections.

system of equations

To see if two lines intersect, we set up a system of equations. This involves equating the parametric expressions of each line's coordinates.
For example, set the coordinates derived from the parameters equal to each other:

• \( 2t + 1 = 4s - 3 \)
• \( t - 3 = s - 4 \)
• \( -3t + 4 = -4s + 8 \)

Solving these equations simultaneously provides the values of \( t \) and \( s \) that locate the point of intersection. The ability to solve such systems is crucial for finding intersections. Here's the process:

• From \( t - 3 = s - 4 \): \( t = s - 1 \)
• Substitute into another equation: \( 2(s - 1) + 1 = 4s - 3 \)
• Solve for \( s \): \( s = 1 \),\( t = 0 \)
Thus, we verify these values in the third equation to ensure consistency across all equations.

3D geometry

3D geometry involves shapes and figures in a three-dimensional space, described with coordinates \( x, y, \text{ and } z \). In the context of lines in 3D, it's crucial to understand that lines can be represented parametrically, and their relative positions can be analyzed through their equations. Finding intersections in 3D space helps determine if lines cross each other.
From the exercise, we can take the intersection point of two 3D lines, ascertained by solving the parametric equations. The lines intersect if there exists a common \( (x, y, z) \) point satisfying both line equations. In this example, both lines share the point \((1, -3, 4)\), confirming their intersection.

line parametrization

Line parametrization is a technique of expressing the coordinates of points on a line using a parameter. This parameter varies, allowing the representation of every point along the line. The general form for 3D line parametrization is:
\[ \mathbf{r} = \mathbf{r_0} + t\mathbf{d} \]Here, \( \mathbf{r_0} \) is a point on the line, \( \mathbf{d} \) is the direction vector of the line, and \( t \) is the parameter. For example, the first line in the solution is expressed as:

• \( x = 2t + 1 \)
• \( y = t - 3 \)
• \( z = -3t + 4 \)

This form clearly describes how the coordinates change with the parameter \( t \). Similarly, for the second line:

• \( x = 4s - 3 \)
• \( y = s - 4 \)
• \( z = -4s + 8 \)

By parametrizing lines, complex geometry problems become more manageable and understandable.