<p>Equation of the line of the shortest distance between the lines \(\dfrac{x}{1} = \dfrac{y}{-1} = \dfrac{z}{1}\) and \(\dfrac{x-1}{0} = \dfrac{y+1}{-2} = \dfrac{z}{1}\) is</p>
<p>\(\dfrac{x}{1} = \dfrac{y}{1} = \dfrac{z}{-2}\)</p>
<p>\(\dfrac{x-1}{1} = \dfrac{y+1}{-1} = \dfrac{z}{-2}\)</p>
<p>\(\dfrac{x-1}{1} = \dfrac{y+1}{-1} = \dfrac{z}{1}\)</p>
<p>\(\dfrac{x}{-2} = \dfrac{y}{1} = \dfrac{z}{2}\)</p>
Step-by-Step Solution
Key Concept: The line of shortest distance between two skew lines is perpendicular to both lines. Its direction vector equals the cross product of the direction vectors of the given lines, and it passes through a point on the common perpendicular.
Step 1: Identify the given lines Line 1: $\frac{x}{1} = \frac{y}{-1} = \frac{z}{1}$ has direction vector $\vec{d_1} = (1, -1, 1)$ and passes through point $P_1 = (0, 0, 0)$ Line 2: $\frac{x-1}{0} = \frac{y+1}{-2} = \frac{z}{1}$ has direction vector $\vec{d_2} = (0, -2, 1)$ and passes through point $P_2 = (1, -1, 0)$ Step 2: Find the direction of the shortest distance line The direction vector of the shortest distance line is perpendicular to both $\vec{d_1}$ and $\vec{d_2}$: $\vec{d} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 0 & -2 & 1 \end{vmatrix}$ $= \hat{i}((-1)(1) - (1)(-2)) - \hat{j}((1)(1) - (1)(0)) + \hat{k}((1)(-2) - (-1)(0))$ $= \hat{i}(-1 + 2) - \hat{j}(1 - 0) + \hat{k}(-2 - 0)$ $= \hat{i}(1) - \hat{j}(1) + \hat{k}(-2) = (1, -1, -2)$ Step 3: Verify the direction vector Check: $\vec{d} \cdot \vec{d_1} = (1)(1) + (-1)(-1) + (-2)(1) = 1 + 1 - 2 = 0$ ✓ Check: $\vec{d} \cdot \vec{d_2} = (1)(0) + (-1)(-2) + (-2)(1) = 0 + 2 - 2 = 0$ ✓ Step 4: Find a point on the shortest distance line The shortest distance line passes through the origin $(0, 0, 0)$ which lies on Line 1. The equation of the shortest distance line is: $\frac{x}{1} = \frac{y}{-1} = \frac{z}{-2}$ This can also be written as: $\frac{x}{1} = \frac{y}{1} = \frac{z}{-2}$ (dividing by -1 in the middle term) ∴ Answer: A
Correct Answer: A