Question

Find the shortest distance between the lines:
\( \frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1} \) and
\( \frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1} \).

Hint:

To convert a line from Cartesian form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \) to vector form, identify the point \( (x_1, y_1, z_1) \) to get \( \vec{a} \) and the direction ratios \( \langle a, b, c \rangle \) to get \( \vec{b} \). Be careful with signs, for example, \( x+1 \) means \( x_1 = -1 \).

Answer 1

Step 1: Understanding the Concept:
The given lines are skew (non-parallel and non-intersecting). The shortest distance between two skew lines is found using a standard vector formula involving the position vectors of a point on each line and the direction vectors of the lines.
Step 2: Key Formula or Approach:
The shortest distance \( d \) between two lines \( \vec{r} = \vec{a_1} + \lambda\vec{b_1} \) and \( \vec{r} = \vec{a_2} + \mu\vec{b_2} \) is given by:
\[ d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right| \] Step 3: Detailed Explanation or Calculation:
First, identify the vectors from the Cartesian equations:
Line 1: Passes through \( P_1(-1, -1, -1) \) and has direction vector \( \vec{b_1} = 7\hat{i} - 6\hat{j} + \hat{k} \). So, \( \vec{a_1} = -\hat{i} - \hat{j} - \hat{k} \).
Line 2: Passes through \( P_2(3, 5, 7) \) and has direction vector \( \vec{b_2} = \hat{i} - 2\hat{j} + \hat{k} \). So, \( \vec{a_2} = 3\hat{i} + 5\hat{j} + 7\hat{k} \).
1. Calculate \( \vec{a_2} - \vec{a_1} \):
\[ \vec{a_2} - \vec{a_1} = (3 - (-1))\hat{i} + (5 - (-1))\hat{j} + (7 - (-1))\hat{k} = 4\hat{i} + 6\hat{j} + 8\hat{k} \] 2. Calculate the cross product \( \vec{b_1} \times \vec{b_2} \):
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
7 & -6 & 1
1 & -2 & 1 \end{vmatrix} = \hat{i}(-6 - (-2)) - \hat{j}(7 - 1) + \hat{k}(-14 - (-6)) \] \[ = -4\hat{i} - 6\hat{j} - 8\hat{k} \] 3. Calculate the dot product in the numerator:
\[ (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (4)(-4) + (6)(-6) + (8)(-8) = -16 - 36 - 64 = -116 \] 4. Calculate the magnitude of the cross product:
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{(-4)^2 + (-6)^2 + (-8)^2} = \sqrt{16 + 36 + 64} = \sqrt{116} \] 5. Calculate the shortest distance d:
\[ d = \left| \frac{-116}{\sqrt{116}} \right| = \sqrt{116} = \sqrt{4 \times 29} = 2\sqrt{29} \] Step 4: Final Answer:
The shortest distance between the lines is \( 2\sqrt{29} \) units.