Find the shortest distance between the following lines:
$\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} and \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}$

Solution

$\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}$

The vector form of this equation is:

$\vec{r} = 3 \hat{i} + 5 \hat{j} + 7 \hat{k} + λ \hat{i} - 2 \hat{j} + \hat{k} \vec{r} = {\vec{a}}_{1} + λ {\vec{b}}_{1} . . . . . . . . . . . . (1) \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}$

The vector form of this equation is:

$\vec{r} = - \hat{i} - \hat{j} - \hat{k} + λ 7 \hat{i} - 6 \hat{j} + \hat{k} Therefore, \vec{a_{1}} = 3 \hat{i} + 5 \hat{j} - 7 \hat{k} \vec{b_{1}} = \hat{i} - 2 \hat{j} + \hat{k} \vec{a_{2}} = - \hat{i} - \hat{j} - \hat{k} and \vec{b_{2}} = 7 \hat{i} - 6 \hat{j} + \hat{k}$

Now, the shortest distance between these two lines is given by:

$d = |\frac{\vec{b_{1}} x \vec{b_{2}} . \vec{a_{2}} - \vec{a_{1}}}{|\vec{b_{1}} x \vec{b_{2}}|}| \vec{b_{1}} x \vec{b_{2}} = [\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 1 \\ 7 & - 6 & 1 \end{matrix}] = \hat{i} (- 2 + 6) - \hat{j} (1 - 7) + \hat{k} (- 6 + 14) = 4 \hat{i} + 6 \hat{j} + 8 \hat{k} |\vec{b_{1}} x \vec{b_{2}}| = \sqrt{4^{2} + 6^{2} + 8^{2}} = \sqrt{116} \vec{a_{2}} - \vec{a_{1}} = - \hat{i} - \hat{j} - \hat{k} - 3 \hat{i} + 5 \hat{j} + 7 \hat{k} = - 4 \hat{i} - 6 \hat{j} - 8 \hat{k} ∴ d = |\frac{4 \hat{i} + 6 \hat{j} + 8 \hat{k} . - 4 \hat{i} - 6 \hat{j} - 8 \hat{k}}{\sqrt{116}}| = |\frac{- 16 - 36 - 64}{\sqrt{116}}| = |\frac{- 116}{\sqrt{116}}| = \sqrt{116}$

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Three Dimensional Geometry

Find the shortest distance between the following lines:
$\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} and \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}$

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.

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Three Dimensional Geometry

Find the shortest distance between the following lines:x - 31 = y - 5-2 = z - 71 and x + 17 = y + 1-6 = z + 11

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Find the shortest distance between the following lines:
$\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} and \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}$