Write the vector equations of the following lines and hence determine the distance between them $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6} and \frac{x - 3}{4} = \frac{y - 3}{6} = \frac{z + 5}{12}$

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Solution

We have
$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6} \frac{x - 3}{4} = \frac{y - 3}{6} = \frac{z + 5}{12}$

Since the first line passes through the point (1, 2, $-$ 4) and has direction ratios proportional to 2, 3, 6, its vector equation is
$\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} . . . (1) \Rightarrow \vec{r} = \hat{i} + 2 \hat{j} - 4 \hat{k} + λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$

Also, the second line passes through the point (3, 3, $-$ 5) and has direction ratios proportional to 4, 6, 12.
Its vector equation is
$\vec{r} = \vec{a_{2}} + μ \vec{b_{2}} . . . (2) \Rightarrow \vec{r} = 3 \hat{i} + 3 \hat{j} - 5 \hat{k} + μ (4 \hat{i} + 6 \hat{j} + 12 \hat{k}) \Rightarrow \vec{r} = 3 \hat{i} + 3 \hat{j} - 5 \hat{k} + 2 μ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$

These two lines pass through the points having position vectors $\vec{a_{1}} = \hat{i} + 2 \hat{j} - 4 \hat{k} and \vec{a_{2}} = 3 \hat{i} + 3 \hat{j} - 5 \hat{k}$ and are parallel to the vector $\vec{b} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ .

Now,

$\vec{a_{2}} - \vec{a_{1}} = 2 \hat{i} + \hat{j} - \hat{k}$

and
$(\vec{a_{2}} - \vec{a_{1}}) \times \vec{b} = (2 \hat{i} + \hat{j} - \hat{k}) \times (2 \hat{i} + 3 \hat{j} + 6 \hat{k}) = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & - 1 \\ 2 & 3 & 6 \end{matrix}| = 9 \hat{i} - 14 \hat{j} + 4 \hat{k} \Rightarrow |(\vec{a_{2}} - \vec{a_{1}}) \times \vec{b}| = \sqrt{9^{2} + {(- 14)}^{2} + 4^{2}} = \sqrt{81 + 196 + 16} = \sqrt{293} and |\vec{b}| = \sqrt{2^{2} + 3^{2} + 6^{2}} = \sqrt{4 + 9 + 36} = 7$

The shortest distance between the two lines is given by

$\frac{|(\vec{a_{2}} - \vec{a_{1}}) \times \vec{b}|}{|\vec{b}|} = \frac{\sqrt{293}}{7} units$

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Similar questions

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}; \frac{x - 3}{4} = \frac{y - 3}{6} = \frac{z + 5}{12}

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

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

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

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

\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}{2} = \frac{y - 3}{4} = \frac{z + 2}{1}

3 x - y - 2 z + 4 = 0 = 2 x + y + z + 1

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

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3} and \frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}

\frac{5 - x}{- 2} = \frac{y + 3}{1} = \frac{1 - z}{3} and \frac{x}{3} = \frac{1 - y}{- 2} = \frac{z + 5}{- 1}

\frac{x - 2}{3} = \frac{y + 3}{- 2}, z = 5 and \frac{x + 1}{1} = \frac{2 y - 3}{3} = \frac{z - 5}{2}

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

\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} and \frac{x + 2}{- 1} = \frac{2 y - 8}{4} = \frac{z - 5}{4}

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