Question

Find the distance between the two given lines. $$\mathbf{r}=(4,3,-1)+(1,1,1) t \quad \text { and } \quad \mathbf{r}=(4,-1,1)+(1,-2,-1) t$$

Short answer

The distance between the lines is \( \sqrt{14} \).

Step-by-step solution

- Identify the direction vectors

The given lines are:\( \mathbf{r}_1 = (4,3,-1) + (1,1,1)t \)\( \mathbf{r}_2 = (4,-1,1) + (1,-2,-1)t \)Identify the direction vectors of both lines:\( \mathbf{d}_1 = (1,1,1) \) for the first line and \( \mathbf{d}_2 = (1,-2,-1) \) for the second line.

- Determine a vector connecting points from both lines

Select a point from each line by setting \( t = 0 \):Point on the first line: \( \mathbf{P}_1 = (4,3,-1) \)Point on the second line: \( \mathbf{P}_2 = (4,-1,1) \).Determine the vector between these points: \( \mathbf{P}_1\mathbf{P}_2 = \mathbf{P}_2 - \mathbf{P}_1 = (4,-1,1) - (4,3,-1) = (0,-4,2) \).

- Calculate the cross product of direction vectors

Calculate the cross product \( \mathbf{d}_1 \times \mathbf{d}_2 \):\( \mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \,\mathbf{i} & \,\mathbf{j} & \,\mathbf{k} \, \ 1 & 1 & 1 \ 1 & -2 & -1 \ \end{vmatrix} \ = (1*-1 - 1*-2)\mathbf{i} - (1*-1 - 1*1)\mathbf{j} + (1*-2 - 1*1)\mathbf{k} \ = (1)\mathbf{i} - (-2)\mathbf{j} + (-3)\mathbf{k} \ = (1,2,-3) \).

- Calculate the magnitude of the cross product

The magnitude of the cross product \( |\mathbf{d}_1 \times \mathbf{d}_2| \) is:\( \sqrt{(1)^2 + (2)^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \).

- Calculate the projection of the connecting vector onto the cross product

Find the projection of \(\mathbf{P}_1\mathbf{P}_2 \) onto \( \mathbf{d}_1 \times \mathbf{d}_2 \):\( \text{Proj}_{\mathbf{d}_1 \times \mathbf{d}_2}(\mathbf{P}_1\mathbf{P}_2) = \frac{|\mathbf{P}_1\mathbf{P}_2 \bullet (\mathbf{d}_1 \times \mathbf{d}_2)|}{|\mathbf{d}_1 \times \mathbf{d}_2|} \)Calculate the dot product: \( \mathbf{P}_1\mathbf{P}_2 \bullet (\mathbf{d}_1 \times \mathbf{d}_2) = (0)(1) + (-4)(2) + (2)(-3) = -8 - 6 = -14 \)So, the projection length is:\( \frac{|-14|}{\sqrt{14}} = \frac{14}{\sqrt{14}} = \sqrt{14} \).

Key concepts

Vector Calculus

Vector calculus is a branch of mathematics focused on the study of vector fields and differential operators. It's crucial in physics and engineering for understanding spatial relationships and motion. Vectors are quantities with both magnitude and direction.

In our problem, we use vector calculus to analyze skew lines. Skew lines are lines that do not intersect and are not parallel. Understanding the vector operations between these lines is key to calculating the distance between them.

Cross Product

The cross product, also known as the vector product, is an operation on two vectors in three-dimensional space. It results in a new vector that is perpendicular to the plane of the original vectors. The formula for the cross product of vectors \(\textbf{a} \) and \(\textbf{b} \) is:

\(\textbf{a} \times \textbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ \end{vmatrix} \).

In practical terms, we use the cross product to find a vector orthogonal to the direction vectors of our skew lines. This orthogonal vector helps in calculating the shortest distance between the lines, as it represents a normal vector to the plane containing both direction vectors.

Dot Product

The dot product, also known as the scalar product, is an algebraic operation that takes two equal-length sequences of numbers and returns a single number. The formula for the dot product of vectors \(\textbf{a} \) and \(\textbf{b} \) is:

\(\textbf{a} \cdot \textbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \).

In our problem, we use the dot product to find the projection of the vector connecting points on the two skew lines onto the orthogonal vector obtained from the cross product. This helps in determining the component of the connecting vector that is parallel to the orthogonal vector, which in turn aids in calculating the distance between the skew lines.

Vector Projection

Vector projection is the operation of projecting one vector onto another. The formula for projecting vector \(\textbf{a} \) onto vector \(\textbf{b} \) is:

\(\text{Proj}_{\textbf{b}}(\textbf{a}) = \frac{\textbf{a} \cdot \textbf{b}}{|\textbf{b}|^2} \textbf{b} \).

In the context of our exercise, we need to project the vector connecting the two points on the lines onto the vector orthogonal (from the cross product) to determine its magnitude. This magnitude represents the shortest distance between the skew lines, as it measures how far apart the points are in the direction perpendicular to both lines.