The shortest distance between the lines x - 3 \over 2 = y - 2 \over 3 = z - 1 \over - 1 and x + 3 \over 2 = y - 6 \over 1 = z - 5 \over 3 , is :

3D Geometry PYQ — JEE Main 2022 · 27 Jun · Shift 2 · Q31 with Solution

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Correct answer: A

The given lines are

$\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$ and $\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$

So their parametric forms are:

Line $L_1$ : $x=3+2\lambda,\, y=2+3\lambda,\, z=1-\lambda$ Hence a point on $L_1$ is $A(3,2,1)$ and its direction vector is $\vec d_1=(2,3,-1).$
Line $L_2$ : $x=-3+2\mu,\, y=6+\mu,\, z=5+3\mu$ Hence a point on $L_2$ is $B(-3,6,5)$ and its direction vector is $\vec d_2=(2,1,3).$

For two lines with direction vectors $\vec d_1, \vec d_2$ and points $A, B$ , the shortest distance is

$D=\frac{|(\vec{AB})\cdot (\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.$

Here, $\vec{AB}=B-A=(-3-3,\,6-2,\,5-1)=(-6,4,4).$

\begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 3 & -1\\ 2 & 1 & 3 \end{vmatrix}$$ $$=\hat i(3\cdot 3-(-1)\cdot 1)-\hat j(2\cdot 3-(-1)\cdot 2)+\hat k(2\cdot 1-3\cdot 2)$$ $$=\hat i(9+1)-\hat j(6+2)+\hat k(2-6)$$ $$=(10,-8,-4).$$ Its magnitude is $$|\vec d_1\times \vec d_2|=\sqrt{10^2+(-8)^2+(-4)^2} =\sqrt{100+64+16}= \sqrt{180}=6\sqrt5.$$ --- 4. **Compute the scalar triple product** $$\vec{AB}\cdot (\vec d_1\times \vec d_2)=(-6,4,4)\cdot(10,-8,-4)$$ $$=-60-32-16=-108.$$ So, $$|(\vec{AB})\cdot (\vec d_1\times \vec d_2)|=108.$$ --- 5. **Find the shortest distance** $$D=\frac{108}{6\sqrt5}=\frac{18}{\sqrt5}.$$ --- 6. **Match with options** $$\boxed{\frac{18}{\sqrt5}}$$ This is **Option A**. --- 7. **Comparison with stored answer** Stored correct answer: **A** Derived answer: **A** So the derived answer agrees with the stored correct answer.