The given vector equations represent two parallel lines. To find the shortest distance between these parallel lines, we can use the formula for the distance between two parallel lines in vector form:
The vector equations are:
\(\vec{r}_1 = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})\)
\(\vec{r}_2 = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(2\hat{i} + 3\hat{j} + 6\hat{k})\)
The lines are parallel since the direction vectors \(\vec{d}_1\) and \(\vec{d}_2\) are the same: \(2\hat{i} + 3\hat{j} + 6\hat{k}\).
The formula to find the shortest distance \(d\) between two parallel lines, given by positions vectors \(\vec{a}_1\) and \(\vec{a}_2\) and a common direction vector \(\vec{b}\), is:
\[d = \frac{|\vec{b} \cdot (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}|}\]Here,
\(\vec{a}_1 = \hat{i} + 2\hat{j} - 4\hat{k}\), \(\vec{a}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k}\), and \(\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}\).
First, calculate \(\vec{a}_2 - \vec{a}_1\):
\[\vec{a}_2 - \vec{a}_1 = (3\hat{i} + 3\hat{j} - 5\hat{k}) - (\hat{i} + 2\hat{j} - 4\hat{k}) = 2\hat{i} + \hat{j} - \hat{k}\]Calculate the dot product \((\vec{b} \cdot (\vec{a}_2 - \vec{a}_1))\):
\[\vec{b} \cdot (\vec{a}_2 - \vec{a}_1) = (2\hat{i} + 3\hat{j} + 6\hat{k}) \cdot (2\hat{i} + \hat{j} - \hat{k})\]\[= (2 \cdot 2) + (3 \cdot 1) + (6 \cdot -1) = 4 + 3 - 6 = 1\]
Next, calculate the magnitude of the direction vector \(\vec{b}\):
\[|\vec{b}| = \sqrt{(2^2 + 3^2 + 6^2)} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\]Therefore, the shortest distance \(d\) is:
\[d = \frac{|1|}{7} = \frac{1}{7}\]Hence, the shortest distance between the two lines is \(\frac{\sqrt{293}}{7}\).
Assertion (A): A line in space cannot be drawn perpendicular to \( x \), \( y \), and \( z \) axes simultaneously.
Reason (R): For any line making angles \( \alpha, \beta, \gamma \) with the positive directions of \( x \), \( y \), and \( z \) axes respectively, \[ \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1. \]