QJEEM23Mafce7: JEE Main 2023 Mathematics - Class 12 JEEM23Mafce7 Solution: The shortest distance between the lines and is equal to ________

For ${L_1}:$ ${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 6} \over 2}$

$\therefore$ Point on line is $A(2,-1,6)$

Parallel vector to this line is,

$\overrightarrow p = 3\widehat i + 2\widehat j + 2\widehat k$

For ${L_2}:$ ${{x - 6} \over 3} = {{y - 1} \over -2} = {{z + 8} \over 0}$

$\therefore$ Point on line is $B(6,1,-8)$

Parallel vector to this line is,

$\overrightarrow q = 3\widehat i - 2\widehat j + 0\widehat k$

Now shortest distance between this two lines is,

$= \left| {{{\overrightarrow {AB} \,.\,\left( {\overrightarrow p \times \overrightarrow q } \right)} \over {\left| {\overrightarrow p \times \overrightarrow q } \right|}}} \right|$

Now, $\overrightarrow {AB} = 4 \widehat i + 2\widehat j -14\widehat k$

$$\overrightarrow{p}\times \overrightarrow{q}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& 2\\ 3& -2& 0\end{array}\right|$$

$= 4\widehat i +6\widehat j -12\widehat k$

$\therefore$ $|\vec{p} \times \vec{q}|=\sqrt{16+36+144}=\sqrt{196}=14$

and $\overrightarrow {AB} \,.\,\left( {\overrightarrow p \times \overrightarrow q } \right) = + 16 + 12 + 168 = +196$

$\therefore$ Shortest distance $= \left| {{{ 196} \over {14 }}} \right| = 14$