The equation of straight line passing through the point `(a,b,c)` and parallel to Z-axis, is:

Options

A. `(x-a)/(1) = (y-b)/(1) = (z-c)/(0)`
B. `(x-a)/(0) = (y-b)/(1) = (z-c)/(1)`
C. `(x-a)/(1) = (y-b)/(0) = (z-c)/(0)`
D. `(x-a)/(0) = (y-b)/(0) = (z-c)/(1)`

Answer

To find the equation of a straight line that passes through the point \((a, b, c)\) and is parallel to the Z-axis, we can follow these steps:

### Step 1: Understand the Direction Cosines
A line that is parallel to the Z-axis will have direction cosines that reflect this orientation. Specifically, the direction cosines for a line parallel to the Z-axis are:
- With respect to the X-axis: \(\cos(90^\circ) = 0\)
- With respect to the Y-axis: \(\cos(90^\circ) = 0\)
- With respect to the Z-axis: \(\cos(0^\circ) = 1\)

Thus, the direction cosines of the line can be represented as:
\[ l = 0, \quad m = 0, \quad n = 1 \]

### Step 2: Use the General Equation of a Line
The general equation of a line in three-dimensional space can be expressed as:
\[ \frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n} \]
where \((x_1, y_1, z_1)\) is a point on the line, and \(l, m, n\) are the direction cosines.

### Step 3: Substitute the Values
Since the line passes through the point \((a, b, c)\), we substitute \(x_1 = a\), \(y_1 = b\), and \(z_1 = c\). Also, we substitute the direction cosines:
\[ l = 0, \quad m = 0, \quad n = 1 \]
Substituting these values into the general equation gives:
\[ \frac{x - a}{0} = \frac{y - b}{0} = \frac{z - c}{1} \]

### Step 4: Simplify the Equation
The terms \(\frac{x - a}{0}\) and \(\frac{y - b}{0}\) indicate that \(x\) and \(y\) do not change, meaning:
- \(x = a\)
- \(y = b\)

The equation for \(z\) is:
\[ z - c = t \quad \text{(where \(t\) is a parameter)} \]
This indicates that \(z\) can take any value while \(x\) and \(y\) remain constant.

### Final Equation
Thus, the final equation of the line can be summarized as:
\[ x = a, \quad y = b, \quad z = c + t \]
This can be expressed in a parametric form as:
\[ \begin{cases} x = a \\ y = b \\ z = c + t \end{cases} \]