JEE MAIN - Mathematics (2021 - 24th February Evening Shift - No. 14)

Let $$\lambda$$ be an integer. If the shortest distance between the lines

x $$-$$ $$\lambda$$ = 2y $$-$$ 1 = $$-$$2z and x = y + 2$$\lambda$$ = z $$-$$ $$\lambda$$ is $${{\sqrt 7 } \over {2\sqrt 2 }}$$, then the value of | $$\lambda$$ | is _________.
Answer
1

Explanation

$${{x - \lambda } \over 1} = {{y - {1 \over 2}} \over {{1 \over 2}}} = {z \over { - {1 \over 2}}}$$

$${{x - \lambda } \over 2} = {{y - {1 \over 2}} \over 1} = {2 \over { - 1}}$$ ....... (1)

Point on line = $$\left( {\lambda ,{1 \over 2},0} \right)$$

$${x \over 1} = {{y + 2\lambda } \over 1} = {{z - \lambda } \over 1}$$ ....... (2)

Point on line = $$(0, - 2\lambda ,\lambda )$$

Distance between skew lines $$ = {{\left[ {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}{{\overrightarrow b }_1}{{\overrightarrow b }_2}} \right]} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}}$$

$$\left| {\matrix{ \lambda & {{1 \over 2} + 2\lambda } & { - \lambda } \cr 2 & 1 & { - 1} \cr 1 & 1 & 1 \cr } } \right|$$

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

$$ = {{\left| { - 5\lambda - {3 \over 2}} \right|} \over {\sqrt {14} }} = {{\sqrt 7 } \over {2\sqrt 2 }}$$

$$ = |10\lambda + 3| = 7 \Rightarrow \lambda = - 1$$

$$ \Rightarrow |\lambda | = 1$$

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