Question

The shortest distance between the lines \[ x = y + 2 = 6z - 6 \] and \[ x + 1 = 2y = -12z \] is:

• A. \( \dfrac{1}{2} \)
• B. \( 2 \)
• C. \( 1 \)
• D. \( \dfrac{3}{2} \)

Hint:

For shortest distance problems involving skew lines: \[ d= \frac{|(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)|} {|\vec{b}_1\times\vec{b}_2|} \] Always convert symmetric equations carefully before extracting direction ratios.

Answer 1

The Correct Option is B

Concept: The shortest distance between two skew lines \[ \vec{r} = \vec{a}_1 + \lambda \vec{b}_1 \] and \[ \vec{r} = \vec{a}_2 + \mu \vec{b}_2 \] is given by: \[ d = \frac{ \left| (\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2) \right| }{ |\vec{b}_1\times\vec{b}_2| } \] Step 1: Convert the first line into symmetric form.
Given: \[ x = y+2 = 6z-6 \] Let the common value be \(t\). Then: \[ x=t,\qquad y=t-2,\qquad z=\frac{t+6}{6} \] Hence: \[ \frac{x-0}{1} = \frac{y+2}{1} = \frac{z-1}{1/6} \] Multiplying direction ratios by \(6\): \[ \frac{x}{6} = \frac{y+2}{6} = \frac{z-1}{1} \] Therefore, \[ \vec{a}_1=(0,-2,1) \] and \[ \vec{b}_1=(6,6,1) \]

Step 2: Convert the second line into symmetric form.
Given: \[ x+1 = 2y = -12z \] Let the common value be \(s\). Then: \[ x=s-1,\qquad y=\frac{s}{2},\qquad z=-\frac{s}{12} \] Thus, \[ \frac{x+1}{1} = \frac{y}{1/2} = \frac{z}{-1/12} \] Multiplying direction ratios by \(12\): \[ \frac{x+1}{12} = \frac{y}{6} = \frac{z}{-1} \] Therefore, \[ \vec{a}_2=(-1,0,0) \] and \[ \vec{b}_2=(12,6,-1) \]

Step 3: Find the required vector quantities.
First, \[ \vec{a}_2-\vec{a}_1 = (-1-0,\ 0-(-2),\ 0-1) \] \[ = (-1,2,-1) \] Now compute: \[ \vec{b}_1\times\vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 6 & 6 & 1 12 & 6 & -1 \end{vmatrix} \] \[ = \hat{i}(6(-1)-1(6)) -\hat{j}(6(-1)-1(12)) +\hat{k}(6\cdot6-6\cdot12) \] \[ = \hat{i}(-12) -\hat{j}(-18) +\hat{k}(-36) \] \[ = -12\hat{i}+18\hat{j}-36\hat{k} \] Its magnitude is: \[ |\vec{b}_1\times\vec{b}_2| = \sqrt{(-12)^2+18^2+(-36)^2} \] \[ = \sqrt{144+324+1296} \] \[ = \sqrt{1764} = 42 \]

Step 4: Compute the scalar triple product.
\[ (\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2) \] \[ = (-1)(-12)+(2)(18)+(-1)(-36) \] \[ = 12+36+36 \] \[ = 84 \] Therefore, \[ d = \frac{84}{42} = 2 \] Hence, the shortest distance is: \[ \boxed{2} \] Therefore, the correct answer is: \[ \boxed{\text{(B)}} \]