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: Understanding the Question:
We need to find the shortest distance between two parallel lines in three-dimensional space given their symmetric equations.

Step 2: Key Formula or Approach:
The shortest distance $d$ between two parallel lines passing through points $A(\vec{a}_1)$ and $B(\vec{a}_2)$ and parallel to the vector $\vec{b}$ is given by: $$d = \frac{| \vec{b} \times (\vec{a}_2 - \vec{a}_1) |}{|\vec{b}|}$$

Step 3: Detailed Explanation:
From the given equations of the lines: Line 1 passes through $A(\vec{a}_1) = (2, 4, 1) \implies \vec{a}_1 = 2\hat{i} + 4\hat{j} + \hat{k}$ Line 2 passes through $B(\vec{a}_2) = (1, -2, -3) \implies \vec{a}_2 = \hat{i} - 2\hat{j} - 3\hat{k}$ Both lines are parallel to the vector direction $\vec{b} = 3\hat{i} + 5\hat{j} + 2\hat{k}$ 1. Compute the difference vector $(\vec{a}_2 - \vec{a}_1)$: $$\vec{a}_2 - \vec{a}_1 = (1-2)\hat{i} + (-2-4)\hat{j} + (-3-1)\hat{k} = -\hat{i} - 6\hat{j} - 4\hat{k}$$ 2. Compute the cross product $\vec{b} \times (\vec{a}_2 - \vec{a}_1)$: $$\vec{b} \times (\vec{a}_2 - \vec{a}_1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 2 \\ -1 & -6 & -4 \end{vmatrix}$$ $$\vec{b} \times (\vec{a}_2 - \vec{a}_1) = \hat{i}[5(-4) - 2(-6)] - \hat{j}[3(-4) - 2(-1)] + \hat{k}[3(-6) - 5(-1)]$$ $$\vec{b} \times (\vec{a}_2 - \vec{a}_1) = \hat{i}[-20 + 12] - \hat{j}[-12 + 2] + \hat{k}[-18 + 5] = -8\hat{i} + 10\hat{j} - 13\hat{k}$$ 3. Calculate the magnitude of the cross product: $$|\vec{b} \times (\vec{a}_2 - \vec{a}_1)| = \sqrt{(-8)^2 + (10)^2 + (-13)^2} = \sqrt{64 + 100 + 169} = \sqrt{333}$$ 4. Calculate the magnitude of direction vector $\vec{b}$: $$|\vec{b}| = \sqrt{3^2 + 5^2 + 2^2} = \sqrt{9 + 25 + 4} = \sqrt{38}$$ 5. Substitute these values back into the parallel lines distance formula: $$d = \frac{\sqrt{333}}{\sqrt{38}} = \sqrt{\frac{333}{38}}\text{ units}$$

Step 4: Final Answer:
The distance between the parallel lines is $\sqrt{\frac{333}{38}}$ units, which matches option (B).