JEE MAIN - Mathematics (2020 - 8th January Morning Slot - No. 3)
The shortest distance between the lines
$${{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}$$ and
$${{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}$$ is :
$${{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}$$ and
$${{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}$$ is :
3
$${7 \over 2}\sqrt {30} $$
$$3\sqrt {30} $$
$$2\sqrt {30} $$
Explanation
$$\overrightarrow a $$
= < 3, 8, 3 >
$$\overrightarrow b $$ = < – 3, – 7, 6 >
$$\overrightarrow p $$ = < 3, – 1, 1 >
$$\overrightarrow q $$ = < –3, 2, 4 >
$$\overrightarrow p \times \overrightarrow q = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & { - 1} & 1 \cr { - 3} & 2 & 4 \cr } } \right|$$ = < -6, -15, 3 >
Shortest distance = $$\left| {{{\left( {\overrightarrow b - \overrightarrow a } \right).\left( {\overrightarrow p \times \overrightarrow q } \right)} \over {\left| {\overrightarrow p \times \overrightarrow q } \right|}}} \right|$$
= $$\left| {{{\left( { - 6, - 15,3} \right).\left( { - 6, - 15,3} \right)} \over {\sqrt {36 + 225 + 9} }}} \right|$$
= $$\left| {{{36 + 225 + 9} \over {\sqrt {36 + 225 + 9} }}} \right|$$
= $$\sqrt {270} $$ = $$3\sqrt {30} $$
$$\overrightarrow b $$ = < – 3, – 7, 6 >
$$\overrightarrow p $$ = < 3, – 1, 1 >
$$\overrightarrow q $$ = < –3, 2, 4 >
$$\overrightarrow p \times \overrightarrow q = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & { - 1} & 1 \cr { - 3} & 2 & 4 \cr } } \right|$$ = < -6, -15, 3 >
Shortest distance = $$\left| {{{\left( {\overrightarrow b - \overrightarrow a } \right).\left( {\overrightarrow p \times \overrightarrow q } \right)} \over {\left| {\overrightarrow p \times \overrightarrow q } \right|}}} \right|$$
= $$\left| {{{\left( { - 6, - 15,3} \right).\left( { - 6, - 15,3} \right)} \over {\sqrt {36 + 225 + 9} }}} \right|$$
= $$\left| {{{36 + 225 + 9} \over {\sqrt {36 + 225 + 9} }}} \right|$$
= $$\sqrt {270} $$ = $$3\sqrt {30} $$
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