Question:medium

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

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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.
Updated On: Mar 27, 2026
  • 0
  • $\frac{29}{\sqrt{5}}$
  • $\sqrt{\frac{5}{29}}$
  • $\sqrt{5}$
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The Correct Option is C

Solution and Explanation

Step 1: Problem Definition:
The objective is to compute the minimum distance between two lines presented in Cartesian coordinates. The initial step involves classifying the relationship between the lines: are they parallel, intersecting, or skew?Step 2: Governing Principles:
A line in Cartesian form is expressed as $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. Its vector representation 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 for the first line are $(a_1, b_1, c_1)$, and for the second line, they are $(a_2, b_2, c_2)$.Parallelism is established when direction ratios are proportional, meaning $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.For parallel lines, the shortest distance is the perpendicular distance from a point on one line to the other.If the lines are not parallel, they must be either intersecting or skew.Step 3: Detailed Analysis:
Let the first line be L1 and the second line be L2.
L1: $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$. Direction ratios: $\vec{d_1} = (2, 3, 4)$.L2: $\frac{x-2}{4} = \frac{y-4}{6} = \frac{z-5}{8}$. Direction ratios: $\vec{d_2} = (4, 6, 8)$.
Parallelism Check: Compare direction ratios.
\[ \frac{4}{2} = 2 \]\[ \frac{6}{3} = 2 \]\[ \frac{8}{4} = 2 \]Proportional direction vectors ($\vec{d_2} = 2\vec{d_1}$) indicate that the lines are parallel.
To distinguish between coincident and distinct parallel lines, verify if a point from L1 lies on L2.
A point on L1 is $P_1 = (1, 2, 3)$. Test $P_1$ against L2's equation.
Substitute $P_1$'s coordinates into L2's equation:
\[ \frac{1-2}{4} = \frac{-1}{4} \]\[ \frac{2-4}{6} = \frac{-2}{6} = \frac{-1}{3} \]\[ \frac{3-5}{8} = \frac{-2}{8} = \frac{-1}{4} \]As $\frac{-1}{4} eq \frac{-1}{3}$, $P_1$ is not on L2. Thus, the lines are distinct and parallel.
The shortest distance (D) between two parallel lines, $\vec{r} = \vec{a_1} + \lambda\vec{d}$ and $\vec{r} = \vec{a_2} + \mu\vec{d}$, is computed using:
\[ D = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{d}|}{|\vec{d}|} \]From the line equations:
$\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$
$\vec{a_2} = 2\hat{i} + 4\hat{j} + 5\hat{k}$
$\vec{d} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ (using the simplified direction vector)
$\vec{a_2} - \vec{a_1} = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$
Cross product calculation:
\[ (\vec{a_2} - \vec{a_1}) \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 2 & 2
2 & 3 & 4 \end{vmatrix} = \hat{i}(8-6) - \hat{j}(4-4) + \hat{k}(3-4) = 2\hat{i} - \hat{k} \]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: $D = \frac{\sqrt{5}}{\sqrt{29}} = \sqrt{\frac{5}{29}}$.
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