Question

Find the distance between the two given lines. The line that joins (0,0,0) to \((1,2,-1),\) and the line that joins (1,1,1) to (2,3,4).

Short answer

The distance between the lines is \(\frac{\sqrt{5}}{5}\).

Step-by-step solution

Identify points and vectors

The first line passes through points (0,0,0) and \(1,2,-1\). The second line passes through points (1,1,1) and \(2,3,4\). Identify direction vectors for each line: \(\vec{d}_1 = (1-0, 2-0, -1-0) = (1,2,-1)\) and \(\vec{d}_2 = (2-1, 3-1, 4-1) = (1,2,3)\).

Form parametric equations of the lines

The parametric equations for the first line: \(L_1: (x,y,z) = (0,0,0) + t(1,2,-1) = (t,2t,-t)\) and for the second line: \(L_2: (x,y,z) = (1,1,1) + s(1,2,3) = (1+s,1+2s,1+3s)\).

Find the vector between points on the lines

Let \(\vec{r}(t) = (t,2t,-t)\) be a point on \(L_1\), and \(\vec{q}(s) = (1+s,1+2s,1+3s)\) be a point on \(L_2\). The vector joining these points is \(\vec{R}(t,s) = \vec{q}(s) - \vec{r}(t) = (1+s-t, 1+2s-2t, 1+3s+t)\).

Perpendicular distance and cross product

The shortest distance between two lines can be found using the formula: \[ D = \frac{|\vec{R} \cdot (\vec{d}_1 \times \vec{d}_2)|}{|\vec{d}_1 \times \vec{d}_2|} \]. Calculate \(\vec{d}_1 \times \vec{d}_2\): \(\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -1 \ 1 & 2 & 3 \end{vmatrix} = (8,-4,0)\).

Calculate magnitudes

Find magnitude of the cross product: \(|\vec{d}_1 \times \vec{d}_2| = \sqrt{8^2+(-4)^2+0^2} = \sqrt{80} = 4\sqrt{5}\). Now compute \(\vec{R}(t, s) \cdot (\vec{d}_1 \times \vec{d}_2)\): use any convenient \(t\) and \(s\), for instance, \(t=0\) and \(s=0\), then \(\vec{R}(0, 0) = (1,1,1)\), so \((1,1,1) \cdot (8,-4,0) = 8 - 4 = 4\).

Calculate distance

Finally, calculate the distance: \[ D = \frac{|4|}{4\sqrt{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}\].

Key concepts

parametric equations

When dealing with lines in three-dimensional space, parametric equations provide a powerful way to describe the position of points on those lines. A parametric equation expresses the coordinates of the points as functions of one or more parameters (usually represented by variables like t or s).

For instance, the line passing through the origin and the point \( (1,2,-1) \) can be represented using a single parameter t as follows: \((x,y,z) = (0,0,0) + t(1,2,-1) = (t, 2t, -t)\).

Similarly, the line passing through \( (1,1,1) \) and \( (2,3,4) \) can be expressed as: \((x, y, z) = (1, 1, 1) + s(1, 2, 3) = (1+s, 1+2s, 1+3s)\).

These equations allow us to compute the coordinates of any point on these lines by simply changing the value of the parameters.

vector operations

Vector operations are essential tools when working with lines in space. Understanding basic vector operations is key to solving many problems, like finding the distance between two parallel lines.

• Vector Addition: Adding two vectors combines their respective components. For example, \((a, b, c) + (d, e, f) = (a+d, b+e, c+f)\).
• Vector Subtraction: Subtracting one vector from another does the opposite: \((a, b, c) - (d, e, f) = (a-d, b-e, c-f)\).
• Scalar Multiplication: Multiplying a vector by a scalar changes its magnitude, but not its direction: \(k(a, b, c) = (ka, kb, kc)\).

These operations allow us to find vectors that connect points on different lines, as shown in the exercise where we formed the vector \(\vec{R}(t, s) = \vec{q}(s) - \vec{r}(t)\).

cross product

The cross product is a fundamental vector operation that takes two vectors and produces a third vector perpendicular to the plane of the initial vectors. Its magnitude is equal to the area of the parallelogram formed by the two input vectors.

For vectors \(\vec{a} = (a_1, a_2, a_3)\) and \(\vec{b} = (b_1, b_2, b_3)\), the cross product \(\vec{a} \times \vec{b} = (a_2 b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1)\)

In our exercise, we calculated the cross product of direction vectors. For \(\vec{d}_1 = (1, 2, -1)\) and \(\vec{d}_2 = (1, 2, 3)\), the cross product is:
\[ \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -1 \ 1 & 2 & 3 \end{vmatrix} = (8, -4, 0). \]
This vector is perpendicular to both initial direction vectors.

magnitude of a vector

The magnitude of a vector (also called its length or norm) is a measure of how long the vector is. It is computed using the Pythagorean theorem.

For a vector \(\vec{v} = (v_1, v_2, v_3)\), the magnitude is given by:
\[ |\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}. \]

In our solution, we calculated the magnitude of the cross product \(\vec{d}_1 \times \vec{d}_2 = (8, -4, 0)\) as:
\[ |\vec{d}_1 \times \vec{d}_2| = \sqrt{8^2 + (-4)^2 + 0^2} = \sqrt{80} = 4\sqrt{5}. \]
Understanding how to find the magnitude helps in computing distances and determining the spatial relationships between vectors.