Question

The shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{4} = \frac{y-4}{6} = \frac{z-5}{8}$ is equal to

• A. 0
• B. $\frac{29}{\sqrt{5}}$
• C. $\sqrt{\frac{5}{29}}$
• D. $\sqrt{5}$

Hint:

When finding the distance between lines, always check if they are parallel first by comparing their direction ratios. If they are, use the formula for parallel lines. If not, use the formula for skew lines. If the skew line distance formula gives zero, the lines are intersecting. In an exam, if your calculated answer is not in the options, re-check your calculations, then re-read the question for typos. If it still doesn't match, the question may be flawed.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We are asked to find the shortest distance between two lines given in Cartesian form. First, we need to determine if the lines are parallel, intersecting, or skew.
Step 2: Key Formula or Approach:
The Cartesian form of a line is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. The vector form is $\vec{r} = (x_1\hat{i} + y_1\hat{j} + z_1\hat{k}) + \lambda(a\hat{i} + b\hat{j} + c\hat{k})$. The direction ratios of the first line are $(a_1, b_1, c_1)$ and for the second line are $(a_2, b_2, c_2)$. Two lines are parallel if their direction ratios are proportional, i.e., $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. If lines are parallel, the shortest distance is the perpendicular distance from a point on one line to the other line. If the lines are not parallel, they are either intersecting or skew.
Step 3: Detailed Explanation:
Let the first line be L1 and the second line be L2.
L1: $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$. The direction ratios are $\vec{d_1} = (2, 3, 4)$. L2: $\frac{x-2}{4} = \frac{y-4}{6} = \frac{z-5}{8}$. The direction ratios are $\vec{d_2} = (4, 6, 8)$.
First, check if the lines are parallel by comparing their direction ratios.
\[ \frac{4}{2} = 2 \] \[ \frac{6}{3} = 2 \] \[ \frac{8}{4} = 2 \] Since the ratios are equal, the direction vectors are proportional ($\vec{d_2} = 2\vec{d_1}$). Therefore, the lines are parallel.
Now, to find the distance between them, we need to check if they are the same line (coincident) or distinct parallel lines. We can do this by taking a point from one line and checking if it lies on the other.
A point on L1 is $P_1 = (1, 2, 3)$. Let's check if this point satisfies the equation of L2.
Substitute the coordinates of $P_1$ into the equation for L2:
\[ \frac{1-2}{4} = \frac{-1}{4} \] \[ \frac{2-4}{6} = \frac{-2}{6} = \frac{-1}{3} \] \[ \frac{3-5}{8} = \frac{-2}{8} = \frac{-1}{4} \] Since $\frac{-1}{4} \neq \frac{-1}{3}$, the point (1, 2, 3) does not lie on the second line. Therefore, the lines are distinct and parallel.
The shortest distance between two parallel lines $\vec{r} = \vec{a_1} + \lambda\vec{d}$ and $\vec{r} = \vec{a_2} + \mu\vec{d}$ is given by the formula:
\[ D = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{d}|}{|\vec{d}|} \] From the equations:
$\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$
$\vec{a_2} = 2\hat{i} + 4\hat{j} + 5\hat{k}$
$\vec{d} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ (we can use the simpler direction vector)
$\vec{a_2} - \vec{a_1} = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$
Now calculate the cross product:
\[ (\vec{a_2} - \vec{a_1}) \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 2 & 2
2 & 3 & 4 \end{vmatrix} = \hat{i}(8-6) - \hat{j}(4-4) + \hat{k}(3-4) = 2\hat{i} - \hat{k} \] Magnitude of the cross product: $|2\hat{i} - \hat{k}| = \sqrt{2^2 + (-1)^2} = \sqrt{5}$.
Magnitude of the direction vector: $|\vec{d}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4+9+16} = \sqrt{29}$.
Distance: $D = \frac{\sqrt{5}}{\sqrt{29}} = \sqrt{\frac{5}{29}}$.