Question

Find the shortest distance between the two skew lines whose vector equations are given by: \[ \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3hat{j} + 2\hat{k}) \] \[ \vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k}) \]

• A. \frac{3}{\sqrt{19}} \]
• B. \( 0 \)
• C. \( 9 \)
• D. \( \sqrt{151} \)

Hint:

To perform the vector cross product quickly and accurately during exams, always write down your grid layout explicitly. Missing a single negative sign during the middle \( -\hat{j} \) row expansion is the most common cause of calculation errors.

Answer 1

The Correct Option is A

Concept: The shortest distance (\( d \)) between two skew lines represented by the vector equations \( \vec{r} = \vec{a}_1 + \lambda\vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \mu\vec{b}_2 \) is determined by calculating the projection of the position difference vector along the common perpendicular direction vector: \[ d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| \]

Step 1: Extract the position and direction coordinates from the equations.
From the provided line equations, isolate the independent parameters:

• Line 1: \( \vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \vec{b}_1 = \hat{i} - 3\hat{j} + 2\hat{k} \)
• Line 2: \( \vec{a}_2 = 4\hat{i} + 5\hat{j} + 6\hat{k} \) and \( \vec{b}_2 = 2\hat{i} + 3\hat{j} + \hat{k} \)

Now, calculate the displacement vector connecting the two starting points: \[ \vec{a}_2 - \vec{a}_1 = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k} \]

Step 2: Compute the cross product vector \( \vec{b}_1 \times \vec{b}_2 \).
Set up the component matrix determinant using standard base unit vectors: \[ \vec{b}_1 \times \vec{b}_2 = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} 1 & -3 & 2 2 & 3 & 1 \end{matrix} \right| \] Expanding along the top row: \[ = \hat{i}(-3 - 6) - \hat{j}(1 - 4) + \hat{k}(3 - (-6)) = -9\hat{i} + 3\hat{j} + 9\hat{k} \] Calculate its magnitude: \[ |\vec{b}_1 \times \vec{b}_2| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171} = 3\sqrt{19} \]

Step 3: Evaluate the scalar dot product and solve for distance \( d \).
Multiply the vector terms calculated in our earlier steps: \[ (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (-9\hat{i} + 3\hat{j} + 9\hat{k}) \] \[ = (3)(-9) + (3)(3) + (3)(9) = -27 + 9 + 27 = 9 \] Plug these completed values back into the shortest distance projection formula: \[ d = \left| \frac{9}{3\sqrt{19}} \right| = \frac{3}{\sqrt{19}} \]