JEE MAIN - Mathematics (2021 - 20th July Morning Shift - No. 19)
If the shortest distance between the lines $$\overrightarrow {{r_1}} = \alpha \widehat i + 2\widehat j + 2\widehat k + \lambda (\widehat i - 2\widehat j + 2\widehat k)$$, $$\lambda$$ $$\in$$ R, $$\alpha$$ > 0 and $$\overrightarrow {{r_2}} = - 4\widehat i - \widehat k + \mu (3\widehat i - 2\widehat j - 2\widehat k)$$, $$\mu$$ $$\in$$ R is 9, then $$\alpha$$ is equal to ____________.
Answer
6
Explanation
If $$\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b $$ and $$\overrightarrow r = \overrightarrow c + \lambda \overrightarrow d $$ then shortest distance between two lines is
$$L = {{(\overrightarrow a - \overrightarrow c ).(\overrightarrow b \times \overrightarrow d )} \over {|b \times d|}}$$
$$\therefore$$ $$\overrightarrow a - \overrightarrow c = ((\alpha + 4)\widehat i + 2\widehat j + 3\widehat k)$$
$${{\overrightarrow b \times \overrightarrow d } \over {|b \times d|}} = {{(2\widehat i + 2\widehat j + \widehat k)} \over 3}$$
$$\therefore$$ $$((\alpha + 4)\widehat i + 2\widehat j + 3\widehat k).{{(2\widehat i + 2\widehat j + \widehat k)} \over 3} = 9$$
or $$\alpha$$ = 6
$$L = {{(\overrightarrow a - \overrightarrow c ).(\overrightarrow b \times \overrightarrow d )} \over {|b \times d|}}$$
$$\therefore$$ $$\overrightarrow a - \overrightarrow c = ((\alpha + 4)\widehat i + 2\widehat j + 3\widehat k)$$
$${{\overrightarrow b \times \overrightarrow d } \over {|b \times d|}} = {{(2\widehat i + 2\widehat j + \widehat k)} \over 3}$$
$$\therefore$$ $$((\alpha + 4)\widehat i + 2\widehat j + 3\widehat k).{{(2\widehat i + 2\widehat j + \widehat k)} \over 3} = 9$$
or $$\alpha$$ = 6
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