Question

The shortest distance between the lines \(x+1=2 y=-12 z\) and \(x=y+2=6 z-6\) is

• A. $\frac{3}{2}$
• B. 3
• C. 2
• D. $\frac{5}{2}$

Answer 1

The Correct Option is C

The given lines are:

\( x+1 = 2y = -12z \)

and

\( x = y+2 = 6z - 6 \).

The formula for the shortest distance between two skew lines is:

Shortest distance (S.D.) = \(\frac{|(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{p} \times \mathbf{q})|}{|\mathbf{p} \times \mathbf{q}|} \)

Here:

\(\mathbf{a} = (-1, 0, 0), \quad \mathbf{b} = (0, -2, -1), \quad \mathbf{p} = (1, \frac{1}{2}, -\frac{1}{12}), \quad \mathbf{q} = (1, 1, \frac{1}{6}).\)

Step 1: Compute \(\mathbf{p} \times \mathbf{q}\)

The cross product \(\mathbf{p} \times \mathbf{q}\) is given by:

\( \mathbf{p} \times \mathbf{q} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & \frac{1}{2} & -\frac{1}{12} \\ 1 & 1 & \frac{1}{6} \end{vmatrix}. \)

Expanding the determinant:

\( \mathbf{p} \times \mathbf{q} = \mathbf{i} \left( \frac{1}{12} - \left( -\frac{1}{12} \right) \right) - \mathbf{j} \left( \frac{1}{6} - \left( -\frac{1}{12} \right) \right) + \mathbf{k} \left( 1 - \frac{1}{2} \right). \)

Simplifying:

\( \mathbf{p} \times \mathbf{q} = \mathbf{i} \cdot \frac{1}{6} - \mathbf{j} \cdot \frac{1}{4} + \mathbf{k} \cdot 0. \)

So:

\( \mathbf{p} \times \mathbf{q} = \frac{1}{6}\mathbf{i} - \frac{1}{4}\mathbf{j}. \)

Step 2: Compute \(|\mathbf{p} \times \mathbf{q}|\)

The magnitude of \(\mathbf{p} \times \mathbf{q}\) is:

\( |\mathbf{p} \times \mathbf{q}| = \sqrt{\left( \frac{1}{6} \right)^2 + \left( -\frac{1}{4} \right)^2}. \)

Simplifying:

\( |\mathbf{p} \times \mathbf{q}| = \sqrt{\frac{1}{36} + \frac{1}{16}} = \sqrt{\frac{4}{144} + \frac{9}{144}} = \sqrt{\frac{13}{144}} = \frac{\sqrt{13}}{12}. \)

Step 3: Compute \(( \mathbf{b} - \mathbf{a}) \cdot (\mathbf{p} \times \mathbf{q} )\)

The vector \(\mathbf{b} - \mathbf{a}\) is:

\( \mathbf{b} - \mathbf{a} = (0 - (-1), -2 - 0, -1 - 0) = (1, -2, -1). \)

15

The dot product \(( \mathbf{b} - \mathbf{a}) \cdot (\mathbf{p} \times \mathbf{q} )\) is:

\( (1, -2, -1) \cdot \left( \frac{1}{6}, -\frac{1}{4}, 0 \right) = 1 \cdot \frac{1}{6} + (-2) \cdot \left( -\frac{1}{4} \right) + (-1) \cdot 0. \)

Simplifying:

\( (\mathbf{b} - \mathbf{a}) \cdot (\mathbf{p} \times \mathbf{q}) = \frac{1}{6} + \frac{2}{4} + \frac{1}{6} = \frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3}. \)

Step 4: Compute the shortest distance

Using the formula:

\( \text{S.D.} = \frac{|(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{p} \times \mathbf{q})|}{|\mathbf{p} \times \mathbf{q}|}. \)

Substituting the values:

\( \text{S.D.} = \frac{|\frac{2}{3}|}{\frac{\sqrt{13}}{12}} = \frac{2}{3} \cdot \frac{12}{\sqrt{13}} = \frac{8}{\sqrt{13}}. \)

Finally, rationalizing the denominator gives:

\( \text{S.D.} = \frac{8\sqrt{13}}{13}. \)

Correct Option: 3 (2)

Answer 2

The correct answer is (C) : 2
\(\frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{-1}{12}} \text{ and } \frac{x}{1}=\frac{y+2}{1}=\frac{z-6}{\frac{1}{6}}\)
\(⇒ \text{Shortest distance}= \frac{(\vec{b}−\vec{a})⋅(\vec{p}\times\vec{q})}{|\vec{p}\times\vec{q}|}\)
\(⇒ \text{S.D.}= (-\hat{i}+2\hat{j}-\hat{k})⋅\frac{(\vec{p}\times\vec{q})}{|\vec{p}\times\vec{q}|}\)
\(\Biggl\{\vec{p}\times\vec{q} \equiv \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&\frac{1}{2}&\frac{-1}{12}\\1&1&\frac{1}{6}\end{vmatrix}=\frac{1}{6}\hat{i}-\frac{1}{4}\hat{j}+\frac{1}{2}\hat{k} \text{ or } 2\hat{i}-3\hat{j}+6\hat{k} \Biggl\}\)
\(⇒ \text{S.D.}= \frac{(-\hat{i}+2\hat{j}-\hat{k}).(\hat{2i}-3\hat{j}+6\hat{k})}{\sqrt{2^2+3^2+6^2}}\)