Question

The shortest distance between the lines \[\overrightarrow{r}\text{ }=\text{ }\left( 3\widehat{i}-2\widehat{j}-2\widehat{k} \right)+\widehat{i}t\]and \[\overrightarrow{r}\text{ }=\text{ }\widehat{i}-\widehat{j}+2\widehat{k}+\widehat{j}s\] (t and s being parameters) is
A \[\sqrt{21}\]
B \[\sqrt{102}\]
C 4
D 3

Answer 1

Hint: In this question, we need to find the shortest distance between the two lines. The general form of line is given as \[\overrightarrow{r}\text{ }=\text{ }\overrightarrow{a}\text{ }+\alpha \overrightarrow{b}\], where \[\overrightarrow{r}\] is the position vector of the arbitrary point lie on line, \[\overrightarrow{a}\]is any point on that line, \[\alpha \] is any real number and \[\overrightarrow{b}\]is the direction ratio. To find distance we need to calculate the difference between Position vector such as \[\overrightarrow{{{a}_{2}}}\] (position vector of point, C lie on second line) - \[\overrightarrow{{{a}_{1}}}\](position vector of point, A lie on first line). Also, we need the cross-product of the direction ratio of two lines such as\[\overrightarrow{{{b}_{1}}}\times \overrightarrow{{{b}_{2}}}\].

Formula Used: Formula used to determine the distance between two lines is given as
\[\left\{ \left( \overrightarrow{{{a}_{2}}}\text{ }\text{ }\overrightarrow{{{a}_{1}}} \right)\left( \overrightarrow{{{b}_{1}}}\text{ }\times \text{ }\overrightarrow{{{b}_{2}}} \right) \right\}/\left| \overrightarrow{{{b}_{1}}}\text{ }\times \text{ }\overrightarrow{{{b}_{2}}} \right|\], where \[\overrightarrow{{{a}_{2}}}-\overrightarrow{{{a}_{1}}}\] is the difference between position vector of points, A and C on two lines, (\[\overrightarrow{{{b}_{1}}}\times \overrightarrow{{{b}_{2}}}\]) is the cross product between the direction ratio of two lines, and |\[\overrightarrow{{{b}_{1}}}\times \overrightarrow{{{b}_{2}}}\]| is the mode of the cross product of direction ratio of two lines.

Complete step by step solution: In the given question, the equation of two lines is given as
\[\overrightarrow{{{r}_{1}}}\text{ }=\text{ }\left( 3\widehat{i}-2\widehat{j}-2\widehat{k} \right)+\widehat{i}t\](Let’s say line 1)
And
\[\overrightarrow{{{r}_{2}}}\text{ }=\text{ }\widehat{i}-\widehat{j}+2\widehat{k}+\widehat{j}s\](Let’s say line 2)
Comparing both the equation of lines with the general equation of line such as
\[\overrightarrow{r}\text{ }=\text{ }\overrightarrow{a}\text{ }+\alpha \overrightarrow{b}\], where α is any real number
Similarly,
\[\overrightarrow{{{r}_{1}}}\text{ }=\text{ }\overrightarrow{{{a}_{1}}}\text{ }+\alpha \overrightarrow{{{b}_{1}}}\](General equation for line 1), where\[\overrightarrow{{{a}_{1}}}\text{ }=\text{ }\left( 3\widehat{i}-2\widehat{j}-2\widehat{k} \right)\], \[\overrightarrow{{{b}_{1}}}\text{ }=\text{ }\widehat{i}\], and α = t

 And
\[\overrightarrow{{{r}_{2}}}\text{ }=\text{ }\overrightarrow{{{a}_{2}}}\text{ }+\alpha \overrightarrow{{{b}_{2}}}\](General equation for line 2), where\[\overrightarrow{{{a}_{2}}}\text{ }=\text{ }\left( \widehat{i}-\widehat{j}+2\widehat{k} \right)\], \[\overrightarrow{{{b}_{2}}}\text{ }=\text{ }\widehat{j}\], and α = s.


Now the formula to find the shortest distance between the two lines is given as


\[\left\{ \left( \overrightarrow{{{a}_{2}}}\text{ }\text{ }\overrightarrow{{{a}_{1}}} \right)\left( \overrightarrow{{{b}_{1}}}\text{ }\times \text{ }\overrightarrow{{{b}_{2}}} \right) \right\}/\left| \overrightarrow{{{b}_{1}}}\text{ }\times \text{ }\overrightarrow{{{b}_{2}}} \right|\]
Finding
(\[\overrightarrow{{{a}_{2}}}-\overrightarrow{{{a}_{1}}}\]) = (\[\widehat{i}-\widehat{j}+2\widehat{k}\]) - (\[3\widehat{i}-2\widehat{j}-2\widehat{k}\])
                 = \[-2\widehat{i}\text{ }+\text{ }\widehat{j}\text{ }+\text{ }4\widehat{k}\]
(\[\overrightarrow{{{b}_{1}}}\times \overrightarrow{{{b}_{2}}}\]) = \[\widehat{i}\times \widehat{j}=\widehat{k}\]
|\[\overrightarrow{{{b}_{1}}}\times \overrightarrow{{{b}_{2}}}\]| = \[\left| \widehat{k} \right|=1\]
Putting all these values in a formula such as
\[\left\{ \left( \overrightarrow{{{a}_{2}}}\text{ }\text{ }\overrightarrow{{{a}_{1}}} \right)\left( \overrightarrow{{{b}_{1}}}\text{ }\times \text{ }\overrightarrow{{{b}_{2}}} \right) \right\}/\left| \overrightarrow{{{b}_{1}}}\text{ }\times \text{ }\overrightarrow{{{b}_{2}}} \right|=\left\{ \left( -2\widehat{i}\text{ }+\text{ }\widehat{j}\text{ }+\text{ }4\widehat{k} \right)\left( \widehat{k} \right) \right\}/1\]
= 4

Option ‘C’ is correct

Note: It is important to note that the always difference is calculated by subtracting the first component from the second component such as \[\overrightarrow{{{a}_{2}}}-\overrightarrow{{{a}_{1}}}\] (difference between points of two lines). Whenever we find a cross product we need to notice the unit vector such that when we find a cross product between \[\widehat{i}\times \widehat{j}\] it will give \[\widehat{k}\]unit vector, \[\widehat{j}\times \widehat{k}\]give \[\widehat{i}\]unit vector, \[\widehat{k}\times \widehat{i}\]= \[\widehat{j}\]unit vector, \[\widehat{j}\times \widehat{i}\]= -\[\widehat{k}\]unit vector, \[\widehat{k}\times \widehat{j}\]= -\[\widehat{i}\] and \[\widehat{i}\times \widehat{k}\] = -\[\widehat{j}\].