Question

Equation of the line of the shortest distance between the lines $\frac{x}{1} = \frac{y}{-1} = \frac{z}{1}$ and $\frac{x-1}{0} = \frac{y+1}{-2} = \frac{z}{1}$ is :

• A. $\frac{x}{1} = \frac{y}{-1} = \frac{z}{-2}$
• B. $\frac{x-1}{1} = \frac{y+1}{-1} = \frac{z}{-2}$
• C. $\frac{x-1}{1} = \frac{y+1}{-1} = \frac{z}{-1}$
• D. $\frac{x}{-2} = \frac{y}{1} = \frac{z}{2}$

Answer 1

The Correct Option is B

To find the equation of the line of shortest distance between the given skew lines, we follow these steps:

1. Identify the direction ratios of the given lines. For the lines $\frac{x}{1} = \frac{y}{-1} = \frac{z}{1}$ and $\frac{x-1}{0} = \frac{y+1}{-2} = \frac{z}{1}$, the direction ratios are (1, -1, 1) and (0, -2, 1) respectively.
2. The line of shortest distance between two skew lines is parallel to the cross product of the direction vectors of these lines. Calculate the cross product: \[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 0 & -2 & 1 \end{vmatrix} \]. Expanding the determinant, we get (1(-2) - (-1)(1), -(1(1) - 0(1)), 1(0) - 1(-2))\]. Simplifying, this gives the normal vector (-1, -1, -2).
3. We will find a vector connecting a point from each line. Let's take P(0, 0, 0) from the first line and Q(1, -1, 0) from the second line. The vector \(\mathbf{PQ} = (1-0, -1-0, 0-0) = (1, -1, 0)\).
4. For the shortest distance, this vector should be perpendicular to the normal vector. Check the dot product: \((1, -1, 0) \cdot (-1, -1, -2) = 1(-1) + (-1)(-1) + 0(-2) = -1 + 1 = 0\), confirming perpendicularity.
5. The line of shortest distance can be represented by a point, say \( Q(1, -1, 0) \), and the direction of \(\mathbf{a} \times \mathbf{b}\). Therefore, the equation of the line is: \(\frac{x-1}{1} = \frac{y+1}{-1} = \frac{z}{-2}\).

Thus, the equation of the line of the shortest distance between the given lines is: \(\frac{x-1}{1} = \frac{y+1}{-1} = \frac{z}{-2}\).