Question

The distance between the parallel lines $\frac{x-2}{3}=\frac{y-4}{5}=\frac{z-1}{2}$ and $\frac{x-1}{3}=\frac{y+2}{5}=\frac{z+3}{2}$ is

• A. $\frac{1}{\sqrt{38}}$ units
• B. $\sqrt{\frac{333}{38}}$ units
• C. $\sqrt{\frac{300}{37}}$ units
• D. $\sqrt{\frac{300}{35}}$ units

Hint:

Before doing any complex cross products, always notice that the denominator must be the magnitude of the direction vector $\vec{b}$. Since $|\vec{b}| = \sqrt{3^2+5^2+2^2} = \sqrt{38}$, the final answer must include a $\sqrt{38}$ term in its denominator, which immediately points to option (B).

Answer 1

The Correct Option is B

Step 1: Read off the line data.
Line 1 passes through $A(2,4,1)$ and line 2 through $B(1,-2,-3)$; both share direction $\vec{b} = (3,5,2)$, so they are parallel.
Step 2: State the distance formula.
For parallel lines, $d = \dfrac{|\vec{b}\times(\vec{a}_2 - \vec{a}_1)|}{|\vec{b}|}$.
Step 3: Find the joining vector.
$\vec{a}_2 - \vec{a}_1 = (1-2,\,-2-4,\,-3-1) = (-1,-6,-4)$.
Step 4: Cross with the direction vector.
\[ \vec{b}\times(\vec{a}_2-\vec{a}_1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 2 \\ -1 & -6 & -4 \end{vmatrix} = -8\hat{i} + 10\hat{j} - 13\hat{k} \] using $5(-4)-2(-6)=-8$, $-[3(-4)-2(-1)]=10$, $3(-6)-5(-1)=-13$.
Step 5: Take both magnitudes.
Numerator $= \sqrt{(-8)^2+10^2+(-13)^2} = \sqrt{64+100+169} = \sqrt{333}$; denominator $= \sqrt{3^2+5^2+2^2} = \sqrt{38}$.
Step 6: Form the distance.
$d = \dfrac{\sqrt{333}}{\sqrt{38}} = \sqrt{\dfrac{333}{38}}$ units, which is option 2 and agrees with the key.
\[ \boxed{\sqrt{\tfrac{333}{38}}\ \text{units}} \]