JEE Advance - Mathematics (2008 - Paper 2 Offline - No. 4)

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 unit vector perpendicular to both $${L_1}$$ and $${L_2}$$ is :
$${{ - \widehat i + 7\widehat j + 7\widehat k} \over {\sqrt {99} }}$$
$${{ - \widehat i - 7\widehat j + 5\widehat k} \over {5\sqrt 3 }}$$
$${{ - \widehat i + 7\widehat j + 5\widehat k} \over {5\sqrt 3 }}$$
$${{7\widehat i - 7\widehat j - \widehat k} \over {\sqrt {99} }}$$

Explanation

Vector in the direction of $${L_1} = \overrightarrow {{n_1}} = 3\widehat i + \widehat j + 2\widehat k$$

Vector in the direction of $${L_2} = \overrightarrow {{n_2}} = \widehat i + 2\widehat j + 3\widehat k$$

$$\therefore$$ Vector perpendicular to both L$$_1$$ and L$$_2$$

$$\overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $$

$$ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & 1 & 2 \cr 1 & 2 & 3 \cr } } \right|$$

$$ = - \widehat i - 7\widehat j + 5\widehat k$$

$$\therefore$$ Required unit vector

$$ = \widehat n = {{ - \widehat i - 7\widehat j + 5\widehat k} \over {\sqrt {1 + 49 + 25} }} = {{ - \widehat i - 7\widehat j + 5\widehat k} \over {5\sqrt 3 }}$$

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