When calculating the distance between two skew lines, remember to use the formula that involves both the cross product and dot product. The key steps are to compute the cross product of the direction vectors, the difference between points on the lines, and then normalize by the magnitude of the cross product. This method works for lines that do not intersect and are not parallel, which are defined as skew lines.
The problem requires calculating the distance between two skew lines given in vector form. The lines are defined as:
Line 1: \(\vec{r} = \hat{i} - 2\hat{j} + 3\hat{k} + \lambda (2\hat{i} + 3\hat{j} + 6\hat{k})\)
Line 2: \(\vec{r} = 3\hat{i} - 2\hat{j} + \hat{k} + \mu (4\hat{i} + 6\hat{j} + 12\hat{k})\)
First, identify the direction vectors of each line:
\(\vec{d_1} = 2\hat{i} + 3\hat{j} + 6\hat{k}\)
\(\vec{d_2} = 4\hat{i} + 6\hat{j} + 12\hat{k}\)
The cross product \(\vec{d_1} \times \vec{d_2}\) is used to find the direction perpendicular to both lines, which is essential for calculating the shortest distance. The initial calculation of the cross product yielded zero:
\(\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 4 & 6 & 12 \end{vmatrix}\)
\(= \hat{i}(3 \cdot 12 - 6 \cdot 6) - \hat{j}(2 \cdot 12 - 6 \cdot 4) + \hat{k}(2 \cdot 6 - 3 \cdot 4)\)
\(= \hat{i}(0) - \hat{j}(0) + \hat{k}(0) = \vec{0}\)
A zero cross product indicates that the direction vectors are parallel. This means the lines are either parallel or coincident. Since the direction vectors are scalar multiples of each other (\(\vec{d_2} = 2\vec{d_1}\)), the lines are parallel. The problem then shifts to finding the distance between these parallel lines.
To find the distance between parallel lines, we select a point from each line and calculate the vector connecting them. Let's use the point \((1, -2, 3)\) from Line 1 and \((3, -2, 1)\) from Line 2.
\(\vec{C} = \vec{r_2} - \vec{r_1}\)
\(\vec{C} = (3\hat{i} - 2\hat{j} + \hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k})\)
\(\vec{C} = (3-1)\hat{i} + (-2-(-2))\hat{j} + (1-3)\hat{k}\)
\(\vec{C} = 2\hat{i} + 0\hat{j} - 2\hat{k}\)
The distance between two parallel lines is given by the formula: \(d = \frac{|\vec{C} \times \vec{d}|}{|\vec{d}|}\), where \(\vec{d}\) is the common direction vector of the lines.
Let's use \(\vec{d_1} = 2\hat{i} + 3\hat{j} + 6\hat{k}\) as the common direction vector.
\(\vec{C} \times \vec{d_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & -2 \\ 2 & 3 & 6 \end{vmatrix}\)
\(= \hat{i}(0 \cdot 6 - (-2) \cdot 3) - \hat{j}(2 \cdot 6 - (-2) \cdot 2) + \hat{k}(2 \cdot 3 - 0 \cdot 2)\)
\(= \hat{i}(0 + 6) - \hat{j}(12 + 4) + \hat{k}(6 - 0)\)
\(= 6\hat{i} - 16\hat{j} + 6\hat{k}\)
Now, calculate the magnitude of \(\vec{C} \times \vec{d_1}\):
\(|\vec{C} \times \vec{d_1}| = \sqrt{6^2 + (-16)^2 + 6^2} = \sqrt{36 + 256 + 36} = \sqrt{328}\)
Next, calculate the magnitude of the direction vector \(\vec{d_1}\):
\(|\vec{d_1}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\)
Finally, the distance between the parallel lines is:
\(d = \frac{|\vec{C} \times \vec{d_1}|}{|\vec{d_1}|} = \frac{\sqrt{328}}{7}\)
Thus, the distance between the two lines is \(\frac{\sqrt{328}}{7}\).