Question

Find the distance between the two given lines. $$\text{ The }x\text{ axis and }\mathbf{r}=\mathbf{j}-\mathbf{k}+(2\mathbf{i}-3\mathbf{j}+\mathbf{k})t$$.

Short answer

$\sqrt{2}$

Step-by-step solution

Identify the given lines

The problem gives the lines as the x-axis and $$\mathbf{\text{r}}=\mathbf{\text{j}}-\mathbf{\text{k}}+(2\mathbf{\text{i}}-3\mathbf{\text{j}}+\mathbf{\text{k}})t$$

Understand the x-axis

The x-axis can be represented by the vector equation $$\mathbf{\text{r}}=t\mathbf{\text{i}}$$, where t is any real number.

Vector form of the second line

The equation $$\mathbf{\text{r}}=\mathbf{\text{j}}-\mathbf{\text{k}}+(2\mathbf{\text{i}}-3\mathbf{\text{j}}+\mathbf{\text{k}})t$$ gives a direction vector $$\mathbf{\text{d}}=2\mathbf{\text{i}}-3\mathbf{\text{j}}+\mathbf{\text{k}}$$ and a point $${\mathbf{\text{p}}}_0=\mathbf{\text{j}}-\mathbf{\text{k}}.$$

Write the coordinates of the point on the x-axis and second line

Any point on the x-axis can be given by (t, 0, 0), and on the second line, a point can be given by (0, 1, -1).

Find the shortest distance from a point to a line

The shortest distance from the point (0, 1, -1) to the x-axis can be found using the formula for distance from a point to a line: $$\text{Distance}=\sqrt{1^2+(-1{)}^2}$$.

Calculate the distance

Let’s calculate: $$\text{Distance}=\sqrt{1+1}=\sqrt{2}.$$

Key concepts

Vector Equations

Vector equations provide a way to represent lines and curves in space using vectors. These equations are composed of a point vector and a direction vector. For example, in the line equation given by $\mathbf{\text{r}}=\mathbf{\text{j}}-\mathbf{\text{k}}+(2\mathbf{\text{i}}-3\mathbf{\text{j}}+\mathbf{\text{k}})t$, $\mathbf{\text{r}}$ is the position vector of any point on the line. The components of this equation can be broken down as follows:

• Initial point vector: $\mathbf{\text{j}}-\mathbf{\text{k}}$, which is the point from which the line starts.
• Direction vector: $2\mathbf{\text{i}}-3\mathbf{\text{j}}+\mathbf{\text{k}}$, which indicates the direction in which the line extends.

By varying the parameter \t\, we can find any point on the line. Understanding vector equations is crucial because it allows us to analyze the line's behavior and interaction with other elements in space, such as finding intersections or distances.

Direction Vector

The direction vector is a core component of the vector equation of a line. It tells us how the line moves in space. When the vector equation of the line is written as $\mathbf{\text{r}}=\mathbf{\text{p}}+t\mathbf{\text{d}}$, the vector $\mathbf{\text{d}}$ is the direction vector.

The direction vector in the given problem is: $\mathbf{\text{d}}=2\mathbf{\text{i}}-3\mathbf{\text{j}}+\mathbf{\text{k}}$. This means that as the parameter \t\ varies, the line moves 2 units in the x-direction, -3 units in the y-direction, and 1 unit in the z-direction. This vector defines the line's orientation and helps in computations like cross products, dot products, and finding distances.

To summarize, the direction vector is essential for:

• Defining the line's orientation in space
• Determining how to locate points along the line using the parameter \t\
• Facilitating the calculation of the shortest distance to other geometric entities

With direction vectors, we can understand the behavior of lines and solve complex geometrical problems.

Distance Formula

The distance formula is critical when we need to find the shortest distance between geometric entities. Specifically, in this exercise, we needed to find the shortest distance from a point to a line using:

• Convert both the line and point into a manageable form.
• Apply the distance formula correct to our context.

For a point (0, 1, -1) and the x-axis represented by $\mathbf{\text{r}}=t\mathbf{\text{i}}$, the distance formula tells us:

\text{Distance} = \sqrt{1^2 + (-1)^2}\This results from plugging in the coordinates into the formula. The calculation steps show:

• Square the differences in the y and z components:
• Square of 1 and square of -1.

Adding these gives $\text{Distance}=\sqrt{1+1}=\sqrt{2}$. This illustrates how we can break down and simplify complex distances using fundamental formulas.

Remember, the key to using the distance formula correctly is to:

• Identify the correct components of points and lines or planes.
• Apply the distance formula step-by-step.
• Ensure all units in the calculation match to avoid errors.

These steps facilitate solving distance-related problems accurately.