Write the vector equations of the following lines & have determine the distance between then: $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}; \frac{x - 3}{4} = \frac{y - 3}{6} = \frac{z + 5}{12}$ .

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Solution

$\begin{matrix} L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6} L_{2} : \frac{x - 3}{4} = \frac{y - 3}{6} = \frac{z + 5}{12} \to r_{1} = ˆ i + 2 ˆ j - 4 ˆ k + λ (2 ˆ i + 3 ˆ j + 6 ˆ k) \to r_{2} = 3 ˆ i + 3 ˆ j - 5 ˆ k + λ (4 ˆ i + 6 ˆ j + 12 ˆ k) \end{matrix}$
Lines are parallel.
$\begin{matrix} d = \frac{∣ ∣ (- 2 ˆ i - ˆ j + ˆ k) \times (2 ˆ i + 3 ˆ j + 6 ˆ k) ∣ ∣}{∣ ∣ 2 ˆ i + 3 ˆ j + 6 ˆ k ∣ ∣} = \frac{∣ ∣ - 9 ˆ i + 14 ˆ j - 4 ˆ k ∣ ∣}{∣ ∣ 2 ˆ i + 3 ˆ j + 6 ˆ k ∣ ∣} = \frac{\sqrt{81 + 196 + 16}}{\sqrt{4 + 9 + 36}} \end{matrix}$
Distance b/w lines $= \frac{\sqrt{293}}{7}$

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\frac{x + 1}{2} = \frac{y + 1}{- 6} = \frac{z + 1}{1} a n d \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}

\frac{x + 1}{4} = \frac{y - 3}{- 6} = \frac{z + 1}{1}

\frac{x + 3}{3} = \frac{y - 5}{2} = \frac{z - 7}{6}

\frac{x - 1}{1} = \frac{y - 2}{3} = \frac{z - 3}{2}

\frac{x - 4}{2} = \frac{y - 5}{3} = \frac{z - 6}{1}

\frac{x + 5}{4} = \frac{y + 3}{- 18} = \frac{z - 6}{- 4}

\frac{x - 3}{2} = \frac{y + 10}{- 9} = \frac{z - 1}{- 2}

\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2}

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