JEE Advance - Mathematics (2008 - Paper 2 Offline - No. 3)

Consider the lines,

$${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$$

$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$

Consider the lines,

$${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$$

$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$

Consider the lines,

$${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$$

$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$

The shortest distance between $${L_1}$$ and $${L_2}$$ is :
$$0$$
$${17 \over {\sqrt 3 }}$$
$${41 \over {5\sqrt 3 }}$$
$${17 \over {5\sqrt 3 }}$$

Explanation

The shortest distance between L$$_1$$ and L$$_2$$ is

$${{(\overrightarrow {{a_2}} - \overrightarrow {{a_1}} )(\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} )} \over {|\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} |}} = (\overrightarrow {{a_2}} - \overrightarrow {{a_1}} )\,.\,\widehat n$$

Where, $${a_1} = - \widehat i - 2\widehat j - \widehat k$$

$${a_2} = 2\widehat i - 2\widehat j + 3\widehat k$$

$$\therefore$$ $$\overrightarrow {{a_2}} - \overrightarrow {{a_1}} = 3\widehat i + 4\widehat k$$

$$\therefore$$ $$({\widehat a_2} - {\widehat a_1})\,.\,\widehat n$$

$$ = (3\widehat i + 4\widehat k)\,.\,\left( {{{ - \widehat i - 7\widehat j + 5\widehat k} \over {5\sqrt 3 }}} \right)$$

$$ = {{ - 3 + 20} \over {5\sqrt 3 }} = {{17} \over {5\sqrt 3 }}$$

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