The lines `(x-1)/1=(y-2)/2=(z-3)/(3) and (x-1)/1=y/3 =z/4`
are

Options

A. Parallel
B. Intersecting
C. Skew
D. At right angles

Answer

To determine the nature of the lines given by the equations

1. \(\frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{3}\)
2. \(\frac{x-1}{1} = \frac{y}{3} = \frac{z}{4}\)

we will follow these steps:

### Step 1: Identify Direction Ratios
For the first line, we can express the direction ratios as follows:
- From \(\frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{3}\), the direction ratios are \( (1, 2, 3) \).

For the second line:
- From \(\frac{x-1}{1} = \frac{y}{3} = \frac{z}{4}\), the direction ratios are \( (1, 3, 4) \).

### Step 2: Represent Direction Vectors
Let:
- \(\mathbf{b_1} = \langle 1, 2, 3 \rangle\)
- \(\mathbf{b_2} = \langle 1, 3, 4 \rangle\)

### Step 3: Check for Parallelism
To check if the lines are parallel, we compute the cross product \(\mathbf{b_1} \times \mathbf{b_2}\):

\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 1 & 3 & 4 \end{vmatrix} \]

Calculating the determinant:

\[ = \mathbf{i} \begin{vmatrix} 2 & 3 \\ 3 & 4 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} \]

Calculating each of these determinants:

1. \(\begin{vmatrix} 2 & 3 \\ 3 & 4 \end{vmatrix} = (2)(4) - (3)(3) = 8 - 9 = -1\)

2. \(\begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} = (1)(4) - (3)(1) = 4 - 3 = 1\)

3. \(\begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} = (1)(3) - (2)(1) = 3 - 2 = 1\)

Thus,

\[ \mathbf{b_1} \times \mathbf{b_2} = -\mathbf{i} - \mathbf{j} + \mathbf{k} = \langle -1, -1, 1 \rangle \]

Since the cross product is non-zero, the lines are not parallel.

### Step 4: Check for Perpendicularity
Next, we check if the lines are perpendicular by calculating the dot product \(\mathbf{b_1} \cdot \mathbf{b_2}\):

\[ \mathbf{b_1} \cdot \mathbf{b_2} = (1)(1) + (2)(3) + (3)(4) = 1 + 6 + 12 = 19 \]

Since the dot product is non-zero, the lines are not perpendicular.

### Step 5: Check for Intersection
To check if the lines intersect, we set the equations equal to a parameter \(\lambda\) for the first line:

\[ x_1 = \lambda + 1, \quad y_1 = 2\lambda + 2, \quad z_1 = 3\lambda + 3 \]

For the second line, we can express it in terms of another parameter \(\mu\):

\[ x_2 = \mu + 1, \quad y_2 = 3\mu, \quad z_2 = 4\mu \]

For the lines to intersect, we need to find \(\lambda\) and \(\mu\) such that:

\[ \lambda + 1 = \mu + 1 \quad (1) \]
\[ 2\lambda + 2 = 3\mu \quad (2) \]
\[ 3\lambda + 3 = 4\mu \quad (3) \]

From equation (1), we have:

\[ \lambda = \mu \]

Substituting \(\lambda = \mu\) into equations (2) and (3):

From (2):
\[ 2\lambda + 2 = 3\lambda \implies 2 = \lambda \implies \lambda = 2 \]

From (3):
\[ 3\lambda + 3 = 4\lambda \implies 3 = \lambda \implies \lambda = 3 \]

Since we get different values for \(\lambda\), the lines do not intersect.

### Conclusion
Since the lines are neither parallel nor intersecting, they are classified as skew lines.