Question

Find the minimum distance between the lines having parametric equations \(x=t-1, y=2 t, z=t+3\) and \(x=3 s\), \(y=s+2, z=2 s-1\)

Short answer

The minimum distance is \(\frac{3\sqrt{35}}{5}\).

Step-by-step solution

Identify direction vectors

The first line has parametric equations: \(x=t-1, y=2t, z=t+3\). The direction vector of this line, obtained from the coefficients of \(t\), is \(\mathbf{a} = \langle 1, 2, 1 \rangle\).The second line has parametric equations: \(x=3s, y=s+2, z=2s-1\). The direction vector of this line, obtained from the coefficients of \(s\), is \(\mathbf{b} = \langle 3, 1, 2 \rangle\).

Check if the lines are parallel

Lines are parallel if their direction vectors are scalar multiples of each other. To check this for lines \(\mathbf{a} = \langle 1, 2, 1 \rangle\) and \(\mathbf{b} = \langle 3, 1, 2 \rangle\), compare the components: \(1/3 eq 2/1\) and \(2/1 eq 1/2\), so they are not parallel.

Find a point on each line

Choose specific points from the lines to compute the shortest distance. For line 1, let \(t = 0\), giving point \(A = (-1, 0, 3)\). For line 2, let \(s = 0\), giving point \(B = (0, 2, -1)\).

Compute vector between points

Find the vector \(\mathbf{AB}\) between points \(A = (-1, 0, 3)\) and \(B = (0, 2, -1)\):\[\mathbf{AB} = \langle 0 - (-1), 2 - 0, -1 - 3 \rangle = \langle 1, 2, -4 \rangle \]

Calculate the cross product of direction vectors

Calculate the cross product \(\mathbf{a} \times \mathbf{b}\):\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 2 & 1 \ 3 & 1 & 2 \end{vmatrix} \] - Expanding the determinant, we get:\(\mathbf{i}(2\cdot2 - 1\cdot1) - \mathbf{j}(1\cdot2 - 3\cdot1) + \mathbf{k}(1\cdot1 - 3\cdot2)\)- Result: \(\mathbf{a} \times \mathbf{b} = \langle 3, -1, -5 \rangle\).

Use the cross product to find the distance

The distance \(d\) between skew lines is given by:\[ d = \frac{|\mathbf{AB} \cdot (\mathbf{a} \times \mathbf{b})|}{|\mathbf{a} \times \mathbf{b}|} \]- First, compute \(\mathbf{AB} \cdot (\mathbf{a} \times \mathbf{b})\): \(\langle 1, 2, -4 \rangle \cdot \langle 3, -1, -5 \rangle = 1\cdot3 + 2\cdot(-1) + (-4)\cdot(-5) = 3 - 2 + 20 = 21\)- Magnitude of \(\mathbf{a} \times \mathbf{b}\): \(|\mathbf{a} \times \mathbf{b}| = \sqrt{3^2 + (-1)^2 + (-5)^2} = \sqrt{35}\)- Thus, the distance is: \(d = \frac{21}{\sqrt{35}}\).

Simplify the distance expression

To simplify, you can rationalize the denominator if necessary, or leave it as is depending on the expected form.- The simplified form of the distance \(d = \frac{21}{\sqrt{35}}\) can also be expressed as:\[d = \frac{21\sqrt{35}}{35}\] which simplifies further to \[d = \frac{3\sqrt{35}}{5}\].

Key concepts

Parametric Equations

Parametric equations describe a way to represent the coordinates of the points that make up a geometric object, such as a line, in terms of one or more parameters. For instance, consider the line with the parametric equations \( x = t - 1 \), \( y = 2t \), and \( z = t + 3 \). Here, \( t \) is the parameter that varies over all real numbers, generating all points on the line.
This representation is convenient for 3D geometry as it clearly separates the variable (the parameter) from the geometry itself (the points that satisfy the conditions described by the equations).
Similarly, the second line is given by \( x = 3s \), \( y = s+ 2 \), \( z = 2s - 1 \). When solving for the minimum distance between such lines, their parametric forms make it easier to identify key elements like points and direction vectors.

Direction Vectors

A direction vector gives a line its direction. It is obtained by examining the coefficients of the parameter in the parametric equations of a line. For the line with equations \( x = t - 1 \), \( y = 2t \), \( z = t + 3 \), the direction vector is \( \mathbf{a} = \langle 1, 2, 1 \rangle \).
For the second line \( x = 3s \), \( y = s + 2 \), \( z = 2s - 1 \), the direction vector is \( \mathbf{b} = \langle 3, 1, 2 \rangle \).
To determine if two lines are parallel, their direction vectors are compared. If one vector is a scalar multiple of the other, the lines are parallel. For our vectors \( \mathbf{a} \) and \( \mathbf{b} \), they are not parallel since their components are not scalar multiples of each other.
This fact confirms that the lines are skew and can be analyzed further to find the minimum distance.

Cross Product

The cross product is a vector operation that outputs a vector perpendicular to two input vectors. This operation is instrumental in determining the perpendicularity of lines in space.
The cross product of direction vectors \( \mathbf{a} = \langle 1, 2, 1 \rangle \) and \( \mathbf{b} = \langle 3, 1, 2 \rangle \) is calculated as follows:

• The resulting vector \( \mathbf{a} \times \mathbf{b} = \langle 3, -1, -5 \rangle \)

This vector is perpendicular to both direction vectors and is critical in computing the minimum distance between skew lines.
The magnitude of this vector is needed to finalize the distance calculation.

Dot Product

The dot product of two vectors \( \mathbf{u} \cdot \mathbf{v} \) is a scalar value that gives information about the angle between them. It is calculated by summing the products of their corresponding components.
For the vectors \( \mathbf{AB} = \langle 1, 2, -4 \rangle \) and the cross product \( \mathbf{a} \times \mathbf{b} = \langle 3, -1, -5 \rangle \), the dot product is computed as:

• \( \mathbf{AB} \cdot (\mathbf{a} \times \mathbf{b}) = 1\cdot3 + 2\cdot(-1) + (-4)\cdot(-5) = 21 \)

The dot product is essential in finding the numerator when calculating the minimum distance between two skew lines.

Magnitude of a Vector

The magnitude of a vector quantifies its length. For a vector \( \mathbf{v} = \langle x, y, z \rangle \), the magnitude is calculated with the formula \( |\mathbf{v}| = \sqrt{x^2 + y^2 + z^2} \).
In the context of finding the minimum distance between skew lines, compute the magnitude of \( \mathbf{a} \times \mathbf{b} \), which is the cross product of direction vectors:

• \( |\mathbf{a} \times \mathbf{b}| = \sqrt{3^2 + (-1)^2 + (-5)^2} = \sqrt{35} \)

This magnitude forms the denominator for the distance formula and helps visualize the spatial relationship between the two lines.