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 (integer)
1
Solution
$${{x - \lambda } \over 1} = {{y - {1 \over 2}} \over {{1 \over 2}}} = {z \over { - {1 \over 2}}}$$<br><br>${{x - \lambda } \over 2} = {{y - {1 \over 2}} \over 1} = {2 \over { - 1}}$ ....... (1)<br><br>Point on line = $\left( {\lambda ,{1 \over 2},0} \right)$<br><br>${x \over 1} = {{y + 2\lambda } \over 1} = {{z - \lambda } \over 1}$ ....... (2)<br><br>Point on line = $(0, - 2\lambda ,\lambda )$<br><br>Distance between skew lines $$ = {{\left[ {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}{{\overrightarrow b }_1}{{\overrightarrow b }_2}} \right]} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}}$$<br><br>$$\left| {\matrix{
\lambda & {{1 \over 2} + 2\lambda } & { - \lambda } \cr
2 & 1 & { - 1} \cr
1 & 1 & 1 \cr
} } \right|$$<br><br>$$\overline {\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
2 & 1 & { - 1} \cr
1 & 1 & 1 \cr
} } \right|} $$<br><br>$$ = {{\left| { - 5\lambda - {3 \over 2}} \right|} \over {\sqrt {14} }} = {{\sqrt 7 } \over {2\sqrt 2 }}$$<br><br>$= |10\lambda + 3| = 7 \Rightarrow \lambda = - 1$<br><br>$\Rightarrow |\lambda | = 1$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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